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Chapter 1. Problem Solutions 1.

If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than they are now?

【Sol】 All else being the same, including the rates of the chemical reactions that govern our brains and bodies, relativisitic phenomena would be more conspicuous if the speed of light were smaller. If we could attain the absolute speeds obtainable to us in the universe as it is, but with the speed of light being smaller, we would be able to move at speeds that would correspond to larger fractions of the speed of light, and in such instances relativistic effects would be more conspicuous.

3.

An athlete has learned enough physics to know that if he measures from the earth a time interval on a moving spacecraft, what he finds will be greater than what somebody on the spacecraft would measure. He therefore proposes to set a world record for the 100-m dash by having his time taken by an observer on a moving spacecraft. Is this a good idea?

【Sol】 Even if the judges would allow it, the observers in the moving spaceship would measure a longer time, since they would see the runners being timed by clocks that appear to run slowly compared to the ship's clocks. Actually, when the effects of length contraction are included (discussed in Section 1.4 and Appendix 1), the runner's speed may be greater than, less than, or the same as that measured by an observer on the ground.

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5.

Two observers, A on earth and B in a spacecraft whose speed is 2.00 x 108 m/s, both set their watches to the same time when the ship is abreast of the earth. (a) How much time must elapse by A's reckoning before the watches differ by 1.00 s? (b) To A, B's watch seems to run slow. To B, does A's watch seem to run fast, run slow, or keep the same time as his own watch?

【Sol】 Note that the nonrelativistic approximation is not valid, as v/c = 2/3. (a) See Example 1.1. In Equation (1.3), with t representing both the time measured by A and the time as measured in A's frame for the clock in B's frame to advance by to, we need 2   v 2   2    t − t 0 = t 1 − 1 − 2 = t 1 − 1 −   = t × 0.255 = 1.00 s   c   3     from which t = 3.93 s. (b) A moving clock always seems to run slower. In this problem, the time t is the time that observer A measures as the time that B's clock takes to record a time change of to.

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7.

How fast must a spacecraft travel relative to the earth for each day on the spacecraft to correspond to 2 d on the earth?

【Sol】 From Equation (1.3), for the time t on the earth to correspond to twice the time t0 elapsed on the ship’s clock, v2 1 3 1 − 2 = , so v = c = 2.60 × 108 m/s, 2 2 c relating three significant figures. 9.

A certain particle has a lifetime of 1.00 x10-7 s when measured at rest. How far does it go before decaying if its speed is 0.99c when it is created?

【Sol】 The lifetime of the particle is t0, and the distance the particle will travel is, from Equation (1.3), vt =

vt 0 1 − v /c 2

2

=

( 0.99)( 3.0 × 108 m/s)(1.00 × 10− 7 s) 1 − ( 0.99)

2

= 210 m

to two significant figures.

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Department of Physics

11. A galaxy in the constellation Ursa Major is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the corresponding wavelength measured by astronomers on the earth? 【Sol】 See Example 1.3; for the intermediate calculations, note that

c c νo 1 − v /c = = λo , ν νo ν 1 + v /c where the sign convention for v is that of Equation (1.8), which v positive for an approaching source and v negative for a receding source. For this problem, v 1.50 × 107 km/s =− = −0.050, c 3.0 × 108 m/s so that λ=

λ = λo

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1 − v /c 1 + 0.050 = (550 nm) = 578 nm 1 + v /c 1 − 0.050

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13. A spacecraft receding from the earth emits radio waves at a constant frequency of 109 Hz. If the receiver on earth can measure frequencies to the nearest hertz, at what spacecraft speed can the difference between the relativistic and classical Doppler effects be detected? For the classical effect, assume the earth is stationary. 【Sol】 This problem may be done in several ways, all of which need to use the fact that when the frequencies due to the classical and relativistic effects are found, those frequencies, while differing by 1 Hz, will both be sufficiently close to vo = 109 Hz so that vo could be used for an approximation to either. In Equation (1.4), we have v = 0 and V = -u, where u is the speed of the spacecraft, moving away from the earth (V < 0). In Equation (1.6), we have v = u (or v = -u in Equation (1.8)). The classical and relativistic frequencies, vc and vr respectively, are ν0 νc = , 1 + ( u /c )

νr = νo

1 − (u /c ) 1 − (u /c )2 = νo 1 + (u /c ) 1 + (u /c )

The last expression for vo, is motivated by the derivation of Equation (1.6), which essentially incorporates the classical result (counting the number of ticks), and allows expression of the ratio νc 1 = . 2 νr 1 − (u /c )

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Use of the above forms for the frequencies allows the calculation of the ratio ∆ν νc − ν r 1 − 1 − (u /c )2 1 Hz = = = 9 = 10−9 νo νo 1 + (u /c ) 10 Hz Attempts to solve this equation exactly are not likely to be met with success, and even numerical solutions would require a higher precision than is commonly available. However, recognizing that the numerator 1 − 1 − (u /c )2 is of the form that can be approximated using the methods outlined at the beginning of this chapter, we can use 1 − 1 − (u /c )2 ≈ (1/ 2)(u /c )2 . The denominator will be indistinguishable from 1 at low speed, with the result 1 u2 = 10−9 , 2 2c which is solved for

u = 2 ×10−9c = 1.34×104 m/s= 13.4km/s.

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15. If the angle between the direction of motion of a light source of frequency vo and the direction from it to an observer is 0, the frequency v the observer finds is given by 1 − v 2 /c 2 ν = νo 1 − (v /c ) cos θ where v is the relative speed of the source. Show that this formula includes Eqs. (1.5) to (1.7) as special cases. 【Sol】 The transverse Doppler effect corresponds to a direction of motion of the light source that is perpendicular to the direction from it to the observer; the angle θ = ±π/2 (or ±90o), so cos θ = 0, and ν = νo 1 − v 2 /c 2 , which is Equation (1.5). For a receding source, θ = π (or 180o), and cos θ = 1. The given expression becomes ν = νo

1 − v 2 /c 2 1 − v /c = νo , 1 + v /c 1 + v /c

which is Equation (1.8). For an approaching source, θ = 0, cos θ = 1, and the given expression becomes 1 − v 2 /c 2 1 + v /c ν = νo = νo , 1 − v /c 1 − v /c which is Equation (1.8).

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17. An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a spacecraft moving at 0.90c relative to the earth. What is his height as measured by an observer in the same spacecraft? By an observer on the earth? 【Sol】 The astronaut’s proper length (height) is 6 ft, and this is what any observer in the spacecraft will measure. From Equation (1.9), an observer on the earth would measure L = Lo 1 − v 2 /c 2 = ( 6 ft ) 1 − (0.90)2 = 2.6 ft 19. How much time does a meter stick moving at 0.100c relative to an observer take to pass the observer? The meter stick is parallel to its direction of motion. 【Sol】 The time will be the length as measured by the observer divided by the speed, or

L Lo 1 − v 2 /c 2 (1.00 m) 1 − (0.100)2 t= = = = 3.32 × 10− 8 s 8 v v (0.100)(3.0 × 10 m/s)

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21. A spacecraft antenna is at an angle of 10o relative to the axis of the spacecraft. If the spacecraft moves away from the earth at a speed of 0.70c, what is the angle of the antenna as seen from the earth? 【Sol】 If the antenna has a length L' as measured by an observer on the spacecraft (L' is not either L or LO in Equation (1.9)), the projection of the antenna onto the spacecraft will have a length L'cos(10o), and the projection onto an axis perpendicular to the spacecraft's axis will have a length L'sin(10o). To an observer on the earth, the length in the direction of the spacecraft's axis will be contracted as described by Equation (1.9), while the length perpendicular to the spacecraft's motion will appear unchanged. The angle as seen from the earth will then be

   tan(10o )  L ′ sin(10o ) arctan  = arctan  = 14o. L ′ cos(10o ) 1 − v 2 /c 2   1 − (0.70)2  The generalization of the above is that if the angle is 00 as measured by an observer on the spacecraft, an observer on the earth would measure an angle θ given by tan θ =

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tan θo 1 − v 2 /c 2

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23. A woman leaves the earth in a spacecraft that makes a round trip to the nearest star, 4 lightyears distant, at a speed of 0.9c. 【Sol】 The age difference will be the difference in the times that each measures the round trip to take, or ∆t = 2

(

)

(

)

Lo 4 yr 1 − 1 − v 2 /c 2 = 2 1 − 1 − 0.9 2 = 5 yr. v 0.9

25. All definitions are arbitrary, but some are more useful than others. What is the objection to defining linear momentum as p = mv instead of the more complicated p = γ mv? 【Sol】 It is convenient to maintain the relationship from Newtonian mechanics, in that a force on an object changes the object's momentum; symbolically, F = dp/dt should still be valid. In the absence of forces, momentum should be conserved in any inertial frame, and the conserved quantity is p = -γmv, not mv 27. Dynamite liberates about 5.4 x 106 J/kg when it explodes. What fraction of its total energy content is this? 【Sol】 For a given mass M, the ratio of the mass liberated to the mass energy is

M × (5.4 × 106 J/kg) −11 = 6 . 0 × 10 . 8 2 M × (3.0 × 10 m/s)

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29. At what speed does the kinetic energy of a particle equal its rest energy? 【Sol】 If the kinetic energy K = Eo = mc2, then E = 2mc2 and Equation (1.23) reduces to 1 =2 2 2 1 − v /c (γ = 2 in the notation of Section 1.7). Solving for v, v =

3 c = 2.60 × 108 m/s 2

31. An electron has a kinetic energy of 0.100 MeV. Find its speed according to classical and relativistic mechanics. 【Sol】 Classically, v =

2K = me

2 × 0.200 MeV × 1.60 × 10−19 J/eV 9.11 × 10− 31 kg

= 1.88 × 108 m/s.

Relativistically, solving Equation (1.23) for v as a function of K, 2

2

2

 me c 2   me c 2    1  = c 1−   = c 1−   v = c 1 −   m c 2 + K   1 + K /(m c 2 )  . E    e   e 

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With K/(mec2) = (0.100 MeV)/(0.511 MeV) = 0.100/0.511, 2

 1  v = 3.0 × 108 m/s × 1 −   = 1.64 × 108 m/s.  1 + ( 0.100) /(0.511)  The two speeds are comparable, but not the same; for larger values of the ratio of the kinetic and rest energies, larger discrepancies would be found. 33. A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c. 【Sol】 Using Equation (1.22) in Equation (1.23) and solving for v/c, v E  = 1−  o  c E 

2

With E = 21Eo, that is, E = Eo + 20Eo, 2

1  v = c 1 −   = 0.9989c .  21 

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35. How much work (in MeV) must be done to increase the speed of an electron from 1.2 x 108 m/s to 2.4 X 108 m/s? 【Sol】 The difference in energies will be, from Equation (1.23),   1 1 m ec  −  2 2 2 2  1 − v 2 /c 1 − v1 /c  2

  1 1 = ( 0.511 MeV ) −  = 0.294 MeV 2 2  1 − ( 2.4 / 3.0) 1 − (1.2 / 3.0)  37. Prove that ½γmv2, does not equal the kinetic energy of a particle moving at relativistic speeds. 【Sol】 Using the expression in Equation (1.20) for the kinetic energy, the ratio of the two quantities is 1 γmv 2 2

K

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 1 v2  γ  1 v2  1 =  =  . 2 c 2  γ − 1  2 c 2 1 − 1 − v 2 /c 2   

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39. An alternative derivation of the mass-energy formula EO = mc2, also given by Einstein, is based on the principle that the location of the center of mass (CM) of an isolated system cannot be changed by any process that occurs inside the system. Figure 1.27 shows a rigid box of length L that rests on a frictionless surface; the mass M of the box is equally divided between its two ends. A burst of electromagnetic radiation of energy Eo is emitted by one end of the box. According to classical physics, the radiation has the momentum p = Eo/c, and when it is emitted, the box recoils with the speed v ≈ E01Mc so that the total momentum of the system remains zero. After a time t ≈ L/c the radiation reaches the other end of the box and is absorbed there, which brings the box to a stop after having moved the distance S. If the CM of the box is to remain in its original place, the radiation must have transferred mass from one end to the other. Show that this amount of mass is m = EO 1c2. 【Sol】 Measured from the original center of the box, so that the original position of the center of mass is 0, the final position of the center of mass is M  L  M  L   − m  + S  −  + m  − S  = 0.  2  2   2  2  Expanding the products and canceling similar terms [(M/2)(L/2), mS], the result MS = mL is obtained. The distance 5 is the product vt, where, as shown in the problem statement, v ≈ E/Mc (approximate in the nonrelativistic limit M >> Elc2) and t ≈ L/c. Then, MS M E L E m= = = . L L Mc c c 2

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41. In its own frame of reference, a proton takes 5 min to cross the Milky Way galaxy, which is about 105 light-years in diameter. (a) What is the approximate energy of the proton in electronvolts?. (b) About how long would the proton take to cross the galaxy as measured by an observer in the galaxy's reference frame? 【Sol】 To cross the galaxy in a matter of minutes, the proton must be highly relativistic, with v ≈ c (but v < c, of course). The energy of the proton will be E = Eoγ, where EO is the proton's rest energy and γ = 1/ 1 − v 2 /c 2. However, γ, from Equation (1.9), is the same as the ratio LO/L, where L is the diameter of the galaxy in the proton's frame of reference, and for the highly-relativistic proton L ≈ ct, where t is the time in the proton's frame that it takes to cross the galaxy. Combining, Lo Lo 105 ly 9 E = Eoγ = Eo ≈ Eo ≈ (10 eV ) × (3 × 107 s/yr ) = 1019 eV L ct c ( 300 s ) 43. Find the momentum (in MeV/c) of an electron whose speed is 0.600c. 【Sol】 Taking magnitudes in Equation (1.16), p=

mev 1 − v /c

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2

2

=

( 0.511 MeV /c 2 )( 0.600c ) 1 − (0.600)

2

= 0.383 MeV /c

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45. Find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV 【Sol】 When the kinetic energy of an electron is equal to its rest energy, the total energy is twice the rest energy, and Equation (1.24) becomes 4me4c 4 = me4c 4 + p 2c 2 ,

or

p=

3(mec 2 ) /c =

3( 511 keV /c ) = 1.94 GeV /c

The result of Problem 1-29 could be used directly; γ = 2, v = ( p= m3ec, as above.

/2)c, 3 and Equation (1.17) gives

47. Find the speed and momentum (in GeV/c) of a proton whose total energy is 3.500 GeV 【Sol】 Solving Equation (1.23) for the speed v in terms of the rest energy EO and the total energy E, v = c 1 − ( Eo / E ) = c 1 − ( 0.938 /3.500) 2 = 0.963c numerically 2.888 x 108 m/s. (The result of Problem 1-32 does not give an answer accurate to three significant figures.) The value of the speed may be substituted into Equation (1.16) (or the result of Problem 1-46), or Equation (1.24) may be solved for the magnitude of the momentum,

p = (E /c )2 − (Eo /c )2 = (3.500 GeV /c )2 − (0.938 GeV/c )2 = 3.37 GeV/c

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49. A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its mass (in MeV/c2) and speed (as a fraction of c). 【Sol】 From E = mc2 + K and Equation (1.24),

(mc 2 + K )2 = m 2c 4 + p 2c 2

Expanding the binomial, cancelling the m2c4 term, and solving for m, m=

( pc )2 − K 2

=

( 335 MeV )2 − ( 62 MeV )2

= 874 MeV /c 2 .

2c 2 K 2c 2 ( 62 MeV ) The particle's speed may be found any number of ways; a very convenient result is that of Problem 1-46, giving v = c2

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p pc 335 MeV =c = c = 0.36c . E 874 MeV + 62 MeV mc 2 + K

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51. An observer detects two explosions, one that occurs near her at a certain time and another that occurs 2.00 ms later 100 km away. Another observer finds that the two explosions occur at the, same place. What time interval separates the explosions to the second observer? 【Sol】 The given observation that the two explosions occur at the same place to the second observer means that x' = 0 in Equation (1.41), and so the second observer is moving at a speed x 1.00 × 105 m 7 v= = = 5 . 00 × 10 m/s − 3 t 2.00 × 10 s with respect to the first observer. Inserting this into Equation (1.44), x2 x2 t− 2 1− 2 2 ( x /t )2 tc c t t′ = =t = t 1− 2 2 2 2 c2 1 − ( x /ct ) 1 − x /c t = ( 2.00 ms ) 1 −

(5.00 × 107 m/s)2

= 1.97 ms. (2.998 × 108 m/s) 2 (For this calculation, the approximation 1 − (x /ct)2 ≈ 1 − (x 2 / 2c 2tis2 )valid to three significant figures.) An equally valid method, and a good cheek, is to note that when the relative speed of the observers (5.00 x 107 m/s) has been determined, the time interval that the second observer measures should be that given by Equation (1.3) (but be careful of which time it t, which is to). Algebraically and numerically, the different methods give the same result.

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53. A spacecraft moving in the +x direction receives a light signal from a source in the xy plane. In the reference frame of the fixed stars, the speed of the spacecraft is v and the signal arrives at an angle θ to the axis of the spacecraft. (a) With the help of the Lorentz transformation find the angle θ ' at which the signal arrives in the reference frame of the spacecraft. (b) What would you conclude from this result about the view of the stars from a porthole on the side of the spacecraft? 【Sol】 (a) A convenient choice for the origins of both the unprimed and primed coordinate systems is the point, in both space and time, where the ship receives the signal. Then, in the unprimed frame (given here as the frame of the fixed stars, one of which may be the source), the signal was sent at a time t = -r/c, where r is the distance from the source to the place where the ship receives the signal, and the minus sign merely indicates that the signal was sent before it was received. Take the direction of the ship's motion (assumed parallel to its axis) to be the positive x-direction, so that in the frame of the fixed stars (the unprimed frame), the signal arrives at an angle 0 with respect to the positive x-direction. In the unprimed frame, x = r cos θ and y = r sin θ . From Equation (1.41),

x′ =

x − vt

1 − v /c and y’ = y = r sin θ. Then,

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2

2

=

r cos θ − ( −r /c ) 1 − v /c 2

2

=r

cosθ + (v /c ) 1 − v /c 2

2

,

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 sinθ 1 − v 2 /c 2  and θ ′ = arctan . cos θ + ( v / c )   (b) From the form of the result of part (a), it can be seen that the numerator of the term in square brackets is less than sinθ , and the denominator is greater than cosθ , and so tan θ and θ’ < θ when v ≠ 0. Looking out of a porthole, the sources, including the stars, will appear to be in the directions close to the direction of the ship’s motion than they would for a ship with v = 0. As v àc, θ’ à0, and all stars appear to be almost on the ship’s axis(farther forward in the field of view). y′ sinθ tanθ ′ = = , x ′ (cos θ + (v /c ))/ 1 − v 2 /c 2

55. A man on the moon sees two spacecraft, A and B, coming toward him from opposite directions at the respective speeds of 0.800c and 0.900c. (a) What does a man on A measure for the speed with which he is approaching the moon? For the speed with which he is approaching B? (b) What does a man on B measure for the speed with which he is approaching the moon? For the speed with which he is approaching A ? 【Sol】 (a) If the man on the moon sees A approaching with speed v = 0.800 c, then the observer on A will see the man in the moon approaching with speed v = 0.800c. The relative velocities will have opposite directions, but the relative speeds will be the same. The speed with which B is seen to approach A, to an observer in A, is then Vx′ + v 0.800 + 0.900 Vx = = c = 0.988c . 2 1 + ( 0 . 800 )( 0 . 900 ) ′ 1 + vV x /c

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(b) Similarly, the observer on B will see the man on the moon approaching with speed 0.900 c, and the apparent speed of A, to an observer on B, will be

0.900 + 0.800 c = 0.988c. 1 + (0.900)(0.800) (Note that Equation (1.49) is unchanged if Vx’ and v are interchanged.) S’(moon)

B Vx’

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v

S

A

O’

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Chapter 2 Problem Solutions 1.

If Planck's constant were smaller than it is, would quantum phenomena be more or less conspicuous than they are now?

【Sol】 Planck’s constant gives a measure of the energy at which quantum effects are observed. If Planck’s constant had a smaller value, while all other physical quantities, such as the speed of light, remained the same, quantum effects would be seen for phenomena that occur at higher frequencies or shorter wavelengths. That is, quantum phenomena would be less conspicuous than they are now.

Is it correct to say that the maximum photoelectron energy KEmax is proportional to the frequency ν of the incident light? If not, what would a correct statement of the relationship between KEmax and ν be? 【Sol】 No: the relation is given in Equation (2.8) and Equation (2.9),

3.

KEmax = hν − φ = h(ν − νo ), So that while KEmax is a linear function of the frequency ν of the incident light, KEmax is not proportional to the frequency.

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5. Find the energy of a 700-nm photon. 【Sol】 From Equation (2.11), 1.24 × 10− 6 eV ⋅ m E = = 1.77 eV. -9 700 × 10 m Or, in terms of joules, ( 6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s) E = = 2.84 × 10−19 J −9 700 × 10 m 7.

A 1.00-kW radio transmitter operates at a frequency of 880 kHz. How many photons per second does it emit? 【Sol】 The number of photons per unit time is the total energy per unit time(the power) divided by the energy per photon, or P P 1.00 × 103 J/s = = = 1.72 × 1030 photons/s . − 34 3 E hν (6.63 × 10 J ⋅ s)(880 × 10 Hz)

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Light from the sun arrives at the earth, an average of 1.5 x 1011 m away, at the rate of 1.4 x 103 W/m2 of area perpendicular to the direction of the light. Assume that sunlight is monochromatic with a frequency of 5.0 x 1014 Hz. (a) How many photons fall per second on each square meter of the earth's surface directly facing the sun? (b) What is the power output of the sun, and how many photons per second does it emit? (c) How many photons per cubic meter are there near the earth? 【Sol】 (a) The number of photons per unit time per unit are will be the energy per unit time per unit area (the power per unit area, P/A), divided by the energy per photon, or P /A 1.4 × 103 W/m 2 21 2 = = 4 . 2 × 10 photons/( s ⋅ m ). - 34 14 hν (6.63 × 10 J ⋅ s)(5.0 × 10 Hz)

9.

(b) With the reasonable assumption that the sun radiates uniformly in all directions, all points at the same distance from the sun should have the same flux of energy, even if there is no surface to absorb the energy. The total power is then, ( P / A )4πRE2 −S = (1.4 × 103 W/m 2 )4π (1.5 × 1011 m )2 = 4.0 × 1026 W, where RE-S is the mean Earth-Sun distance, commonly abbreviated as “1 AU,” for “astronomical unit.” The number of photons emitted per second is this power divided by the energy per photon, or 4.0 × 1026 J/s (6.63 × 10- 34 J ⋅ s)( 5.0 × 1014 Hz )

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= 1.2 × 1045 photons/s .

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(c) The photons are all moving at the same speed c, and in the same direction (spreading is not significant on the scale of the earth), and so the number of photons per unit time per unit area is the product of the number per unit volume and the speed. Using the result from part (a),

4.2 × 1021 photons/(s ⋅ m 2 ) = 1.4 × 1013 photons/m3. 8 3.0 × 10 m/s

11. The maximum wavelength for photoelectric emission in tungsten is 230 nm. What wavelength of light must be used in order for electrons with a maximum energy of 1.5 eV to be ejected? 【Sol】 Expressing Equation (2.9) in terms of λ = c/ν and λ0 = c/ν0, and performing the needed algebraic manipulations, hc λ   K λ= = λ0 1 + max o  (hc / λo ) + K max hc  

−1

 (1.5 eV )( 230 × 10− 9 m )  = ( 230 nm) 1 +  −6 1 . 24 × 10 eV ⋅ m  

−1

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= 180 nm.

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13. What is the maximum wavelength of light that will cause photoelectrons to be emitted from sodium? What will the maximum kinetic energy of the photoelectrons be if 200-nm light falls on a sodium surface? 【Sol】 The maximum wavelength would correspond to the least energy that would allow an electron to be emitted, so the incident energy would be equal to the work function, and hc 1.24 × 10−6 eV ⋅ m λmax = = = 539 nm φ 2.3 eV where the value of φ for sodium is taken from Table 2.1. From Equation (2.8), hc 1.24 × 10− 6 eV ⋅ m K max = hν − φ = −φ = − 2.3 eV = 3.9 eV. − 9 λ 200 × 10 m 15. 1.5 mW of 400-nm light is directed at a photoelectric cell. If 0.10 percent of the incident photons produce photoelectrons, find the current in the cell. 【Sol】 Because only 0.10% of the light creates photoelectrons, the available power is (1.0x10-3)(1.5x10-3W) = 1.5x10-6 W. the current will be the product of the number of photoelectrons per unit time and the electron charge, or P P Pλ (1.5 × 10− 6 J/s )( 400 × 10−9 m ) I =e =e =e = (1e ) = 0.48 µA E hc / λ hc 1.24 × 10- 6 eV ⋅ m

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17. A metal surface illuminated by 8.5 x 1014 Hz light emits electrons whose maximum energy is 0.52 eV The same surface illuminated by 12.0 x 1014 Hz light emits electrons whose maximum energy is 1.97 eV From these data find Planck's constant and the work function of the surface . 【Sol】 Denoting the two energies and frequencies with subscripts 1 and 2, K max,1 = h ν1 − φ , K max,2 = hν 2 − φ . Subtracting to eliminate the work function φ and dividing by ν1 - ν2, K max, 2 − K max,1 19.7 eV − 0.52 eV −15 h= = = 4 . 1 × 10 eV ⋅ s ν 2 − ν1 12.0 × 1014 Hz − 8.5 × 1014 Hz to the allowed two significant figures. Keeping an extra figure gives h = 4.14 × 10−15 eV ⋅ s = 6.64 × 10- 34 J ⋅ s The work function φ may be obtained by substituting the above result into either of the above expressions relating the frequencies and the energies, yielding φ = 3.0 eV to the same two significant figures, or the equations may be solved by rewriting them as K max,1 ν 2 = hν1ν 2 − φν 2 , K max,2 ν1 = hν 2ν1 − φν1, subtracting to eliminate the product hν1ν2 and dividing by ν1 - ν2 to obtain K max,2 ν1 − K max,1 ν 2 (19.7 eV)(8.5 × 1014 Hz) − (0.52 eV)(12.0 × 1014 Hz) φ= = = 3.0 eV ν 2 − ν1 (12.0 × 1014 Hz − 8.5 × 1014 Hz) (This last calculation, while possibly more cumbersome than direct substitution, reflects the result of solving the system of equations using a symbolic-manipulation program; using such a program for this problem is, of course, a case of "swatting a fly with a sledgehammer".)

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19. Show that it is impossible for a photon to give up all its energy and momentum to a free electron. This is the reason why the photoelectric effect can take place only when photons strike bound electrons. 【Sol】 Consider the proposed interaction in the frame of the electron initially at rest. The photon's initial momentum is po = Eo/c, and if the electron were to attain all of the photon's momentum and energy, the final momentum of the electron must be pe = po = p, the final electron kinetic energy must be KE = Eo = pc, and so the final electron energy is Ee = pc + mec2. However, for any electron we must have Ee2 = (pc)2 + (mec2)2. Equating the two expressions for Ee2

(

or

) ( ( ) 2

Ee2 = ( pc )2 + mec 2 = pc + mec 2 0 = 2( pc ) me c 2 .

)

2

(

) (

)

2

= ( pc )2 + 2( pc ) mec 2 + mec 2 ,

This is only possible if p = 0, in which case the photon had no initial momentum and no initial energy, and hence could not have existed. To see the same result without using as much algebra, the electron's final kinetic energy is p 2c 2 + me2c 4 − mec 2 ≠ pc for nonzero p. An easier alternative is to consider the interaction in the frame where the electron is at rest after absorbing the photon. In this frame, the final energy is the rest energy of the electron, mec2, but before the interaction, the electron would have been moving (to conserve momentum), and hence would have had more energy than after the interaction, and the photon would have had positive energy, so energy could not be conserved.

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21. Electrons are accelerated in television tubes through potential differences of about 10 kV. Find the highest frequency of the electromagnetic waves emitted when these electrons strike the screen of the tube. What kind of waves are these? 【Sol】 For the highest frequency, the electrons will acquire all of their kinetic energy from the accelerating voltage, and this energy will appear as the electromagnetic radiation emitted when these electrons strike the screen. The frequency of this radiation will be E eV (1e )(10 × 103 V) 18 ν = = = = 2 . 4 × 10 Hz −15 h h 4.14 × 10 eV ⋅ s which corresponds to x-rays. 23. The distance between adjacent atomic planes in calcite (CaCO3) is 0.300 nm. Find the smallest angle of Bragg scattering for 0.030-nm x-rays. 【Sol】 Solving Equation (2.13) for θ with n = 1,  λ   0.030 nm  o θ = arcsin   = arcsin   = 2.9  2d   2 × 0.300 nm 

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25. What is the frequency of an x-ray photon whose momentum is 1.1 x 10-23 kg m/s? 【Sol】 From Equation (2.15), cp ( 3.0 × 108 m/s)(1.1 × 10- 23 kg ⋅ m/s) 18 ν = = = 5 . 0 × 10 Hz h 6.63 × 10− 34 J ⋅ s 27. In See. 2.7 the x-rays scattered by a crystal were assumed to undergo no change in wavelength. Show that this assumption is reasonable by calculating the Compton wavelength of a Na atom and comparing it with the typical x-ray wavelength of 0.1 nm. 【Sol】 Following the steps that led to Equation (2.22), but with a sodium atom instead of an electron,

λC ,Na

h 6.63 × 10−34 J ⋅ s −17 = = = 5 . 8 × 10 m, cM Na (3.0 × 108 m/s)(3.82 × 10- 26 kg)

or 5.8 x 10-8 nm, which is much less than o.1 nm. (Here, the rest mass MNa =3.82 x 10-26 kg was taken from Problem 2-24.)

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29. A beam of x-rays is scattered by a target. At 45o from the beam direction the scattered x-rays have a wavelength of 2.2 pm. What is the wavelength of the x-rays in the direct beam? 【Sol】 Solving Equation (2.23) for λ, the wavelength of the x-rays in the direct beam, λ = λ ′ − λC (1 − cos φ ) = 2.2 pm − (2.426 pm)(1 − cos 45o ) = 1.5 pm to the given two significant figures. 31. An x-ray photon of initial frequency 3.0 x 1019 Hz collides with an electron and is scattered through 90o. Find its new frequency. 【Sol】 Rewriting Equation (2.23) in terms of frequencies, with λ = c/ν and λ’ = c/ν’ , and with cos 90o = 0, c c = + λC ν′ ν and solving for ν’ gives −1

−1  1 λC  1 2.43 × 10−12 m   19 ν′ =  + = + = 2 . 4 × 10 Hz   19 8 c  ν 3 . 0 × 10 Hz 3 . 0 × 10 m/s   The above method avoids the intermediate calculation of wavelengths.

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33. At what scattering angle will incident 100-keV x-rays leave a target with an energy of 90 keV? 【Sol】 Solving Equation (2.23) for cos φ,  mc 2 mc 2  λ λ′  = 1 +  511 keV − 511 keV  = 0.432 cos φ = 1 + − = 1 +  − λC λC E ′   100 keV 90 keV   E from which φ = 64o to two significant figures. 35. A photon of frequency ν is scattered by an electron initially at rest. Verify that the maximum kinetic energy of the recoil electron is KEmax = (2h2 ν2/mc2)/(1 + 2hν/mc2). 【Sol】 For the electron to have the maximum recoil energy, the scattering angle must be 1800, and Equation (2.20) becomes mc2 KEmax = 2 (hv) (hv'), where KEmax = (hv - hv') has been used. To simplify the algebra somewhat, consider λ ν ν ν , ν′ =ν = = = λ′ 1 + ( ∆λ / λ ) 1 + ( 2λC / λ ) 1 + ( 2νλC /c ) where ∆λ = 2λC for φ = 180o. With this expression, 2(h ν )(hν ′) 2(hν )2 /(mc 2 ) KE max = = . 1 + ( 2νλC /c ) mc 2 Using λC = h/(mc) (which is Equation (2.22)) gives the desired result.

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37. A photon whose energy equals the rest energy of the electron undergoes a Compton collision with an electron. If the electron moves off at an angle of 40o with the original photon direction, what is the energy of the scattered photon? 【Sol】 As presented in the text, the energy of the scattered photon is known in terms of the scattered angle, not the recoil angle of the scattering electron. Consider the expression for the recoil angle as given preceding the solution to Problem 2-25: sin φ sin φ sin φ tan θ = = = . λC  ( ∆λ / λ ) + (1 − cos φ ) ( λC /λ )(1 − cos φ ) + (1 − cos φ )  1 + (1 − cos φ ) λ   For the given problem, with E = mc2, λ = hc/E = h/(mc) = λC, so the above expression reduces to sin φ tanθ = . 2(1 − cos φ ) At this point, there are many ways to proceed; a numerical solution with θ = 40 o gives φ = 61.6 0 to three significant figures. For an analytic solution which avoids the intermediate calculation of the scattering angle φ, one method is to square both sides of the above relation and use the trigonometric identity sin2 φ = 1 - cos2 φ = (1 + cos φ)(1 – cos φ) to obtain 1 + cos φ 4 tan 2 θ = 1 − cos φ (the factor 1 - cos φ may be divided, as cos φ = 1, φ = 0, represents an undeflected photon, and hence no interaction). This may be re-expressed as

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(1 − cos φ )( 4 tan 2 θ ) = 1 + cos φ = 2 − (1 − cos φ ),

or

2 3 + 4 tan2 θ 1 − cos φ = , 2 − cos φ = . 1 + 4 tan2 θ 1 + 4 tan2 θ Then with λ’ = λ + λC(1 – cos φ) = λC(2 – cos φ), λ 1 + 4 tan 2 θ 1 + 4 tan 2 ( 40o ) ′ E =E =E = ( 511 keV) = 335 eV λ′ 3 + 4 tan 2 θ 3 + 4 tan 2 ( 40o ) An equivalent but slightly more cumbersome method is to use the trigonometric identities φ φ φ sin φ = 2 sin cos , 1 − cos φ = 2 sin 2 2 2 2 in the expression for tan θ to obtain 1 φ  1  cot , φ = 2 arctan   2 2  2 tan θ  yielding the result θ = 61.6 o more readily. tan θ =

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39. A positron collides head on with an electron and both are annihilated. Each particle had a kinetic energy of 1.00 MeV Find the wavelength of the resulting photons. 【Sol】 The energy of each photon will he the sum of one particle's rest and kinetic energies, 1.511 MeV (keeping an extra significant figure). The wavelength of each photon will be hc 1.24 × 10−6 eV ⋅ m −13 λ= = = 8 . 21 × 10 m = 0.821 pm E 1.51 × 106 eV 41. Show that, regardless of its initial energy, a photon cannot undergo Compton scattering through an angle of more than 60o and still be able to produce an electron-positron pair. (Hint: Start by expressing the Compton wavelength of the electron in terms of the maximum photon wavelength needed for pair production.) 【Sol】 Following the hint, h 2hc 2hc λC = = = , mc 2mc 2 E min where Emin = 2mc2 is the minimum photon energy needed for pair production. The scattered wavelength (a maximum) corresponding to this minimum energy is λ’max = (h/Emin ), so λC = 2λ’max . At this point, it is possible to say that for the most energetic incoming photons, λ ~ 0, and so 1 - cos φ = ½ for λ ' = λC /2, from which cos φ = ½ and φ = 60o. As an alternative, the angle at which the scattered photons will have wavelength λ’max can m be found as a function of the incoming photon energy E; solving Equation (2.23) with λ ' = λ'max)

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′ ′ λmax −λ λmax hc / E 1 mc 2 cos φ = 1 − = 1− + = + . λC λC λC 2 E This expression shows that for E >> mc2, cos φ = ½ and so φ = 60o, but it also shows that, because cos φ must always be less than 1, for pair production at any angle, E must be greater than 2mc2, which we know to be the case. 43. (a) Show that the thickness x1/2, of an absorber required to reduce the intensity of a beam of radiation by a factor of 2 is given by x1/2 = 0.693/µ. (b) Find the absorber thickness needed to produce an intensity reduction of a factor of 10. 【Sol】 (a) The most direct way to get this result is to use Equation (2.26) with Io/I = 2, so that ln 2 0.693 I = I oe − µx ⇒ x 1/2 = = . µ µ (b) Similarly, with Io/I = 10, ln10 2.30 x1/10 = = . µ µ

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45. The linear absorption coefficient for 1-MeV gamma rays in lead is 78 m-1. find the thickness of lead required to reduce by half the intensity of a beam of such gamma rays. 【Sol】 From either Equation (2.26) or Problem 2-43 above, ln 2 0.693 x 1/ 2 = = = 8.9 mm µ 78 m -1 47. The linear absorption coefficients for 2.0-MeV gamma rays are 4.9 m-1 in water and 52 in -1 in lead. What thickness of water would give the same shielding for such gamma rays as 10 mm of lead? 【Sol】 Rather than calculating the actual intensity ratios, Equation (2.26) indicates that the ratios will be the same when the distances in water and lead are related by µ H 2O x H2O = µ Pb x Pb , or x H2O = x Pb

µ Pb 52 m -1 −3 = (10 × 10 m ) = 0.106 m µ H2O 4.9 m -1

or 11 cm two significant figures.

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49. What thickness of copper is needed to reduce the intensity of the beam in Exercise 48 by half. 【Sol】 Either a direct application of Equation (2.26) or use of the result of Problem 2-43 gives ln 2 x 1/ 2 = = 1.47 × 10−5 m, 4 -1 4.7 × 10 m which is 0.015 mm to two significant figures. 51. The sun's mass is 2.0 x 1030 kg and its radius is 7.0 x 108 m. Find the approximate gravitational red shift in light of wavelength 500 nm emitted by the sun. 【Sol】 In Equation (2.29), the ratio

GM (6.67 × 10−11 N ⋅ m2 / kg)(2.0 × 1030 kg) = = 2.12 × 10− 6 2 8 2 4 -1 c R (3.0 × 10 m/s) (7.0 × 10 m ) (keeping an extra significant figure) is so small that for an “approximate” red shift, the ratio ∆λ/λ will be the same as ∆ν/ν, and ∆λ = λ

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GM c 2R

= (500 × 10− 9 m)(2.12 × 10- 6 ) = 1.06 × 10-12 m = 1.06 pm.

Department of Physics

53. As discussed in Chap. 12, certain atomic nuclei emit photons in undergoing transitions from "excited" energy states to their “ground” or normal states. These photons constitute gamma rays. When a nucleus emits a photon, it recoils in the opposite direction. (a) The 57 27 Co nucleus 57 decays by K capture to 26 Fe , which then emits a photon in losing 14.4 keV to reach its ground -26 kg. By how much is the photon energy reduced from state. The mass of a 57 26 Fe atom is 9.5 x 10 the full 14.4 keV available as a result of having to share energy and momentum with the recoiling atom? (b) In certain crystals the atoms are so tightly bound that the entire crystal recoils when a gamma-ray photon is emitted, instead of the individual atom. This phenomenon is known as the Mössbauer effect. By how much is the photon energy reduced in this situation if the ex- cited 2576Fe nucleus is part of a 1.0-g crystal? (c) The essentially recoil-free emission of gamma rays in situations like that of b means that it is possible to construct a source of virtually monoenergetic and hence monochromatic photons. Such a source was used in the experiment described in See. 2.9. What is the original frequency and the change in frequency of a 14.4-keV gamma-ray photon after it has fallen 20 m near the earth's surface? 【Sol】 (a) The most convenient way to do this problem, for computational purposes, is to realize that the nucleus will be moving nonrelativistically after the emission of the photon, and that the energy of the photon will be very close to E∞ = 14.4 keV, the energy that the photon would have if the nucleus had been infinitely massive. So, if the photon has an energy E, the recoil momentum of the nucleus 2 2 2 is E/c, and its kinetic energy is p / 2M = E /(2Mc ) , here M is the rest mass of the nucleus. Then, conservation of energy implies

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E2

+ E = E∞ . 2Mc 2 This is a quadratic in E, and solution might be attempted by standard methods, but to find the change in energy due to the finite mass of the nucleus, and recognizing that E will be very close to E∞, the above relation may be expressed as E2 E ∞2 E∞ − E = ≈ 2Mc 2 2Mc 2 (14.4 keV) 2 (1.60 × 10−16 J/keV) = = 1.9 × 106 keV = 1.9 × 103 eV. - 26 8 2 2(9.5 × 10 kg)(3.0 × 10 m/s) If the approximation E ≈ E∞, is not made, the resulting quadratic is E 2 + 2Mc 2E − 2Mc 2E ∞ = 0, which is solved for   E E = Mc 2  1 + 2 ∞2 − 1. Mc   However, the dimensionless quantity E∞/(Mc2) is so small that standard calculators are not able to determine the difference between E and E∞. The square root must be expanded, using (1 + x)1/2 ≈ 1 + (x/2) - (x2/8), and two terms must be kept to find the difference between E and E∞. This approximation gives the previous result.

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It so happens that a relativistic treatment of the recoiling nucleus gives the same numerical result, but without intermediate approximations or solution of a quadratic equation. The relativistic form expressing conservation of energy is, with pc = E and before,

E 2 + (Mc2 )2 + E = Mc 2 + E∞ , or E 2 + (Mc 2 )2 = Mc 2 + E∞ − E . Squaring both sides, canceling E2 and (Mc2)2, and then solving for E,  1 + ( E ∞ /(2Mc 2 ))  E ∞2 + 2Mc 2E ∞ . E = = E ∞  2 2  2(Mc + E ∞ )  1 + (E ∞ /(Mc ))  From this form,  E ∞2  1  E ∞ − E =  , 2 2 2 Mc 1 + E /( Mc )   ∞ giving the same result. (b) For this situation, the above result applies, but the nonrelativistic approximation is by far the easiest for calculation; E∞2 (14.4 × 103 eV)2 (1.6 × 10−19 J/eV) E∞ − E = = = 1.8 × 10− 25 eV. 2 -3 8 2 2Mc 2(1.0 × 10 kg)(3.0 × 10 m/s) E∞ 14.4 × 103 eV (c) The original frequency is ν = = = 3.48 × 1018 Hz. − 15 h 4.14 × 10 eV ⋅ s From Equation (2.28), the change in frequency is gH  (9.8 m/s 2 )( 20 m )  ∆ ν = ν ′ − ν =  2 ν = ( 3.48 × 1018 Hz) = 7.6 Hz. 8 2 c  (3.0 × 10 m/s)

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55. The gravitational potential energy U relative to infinity of a body of mass m at a distance R from the center of a body of mass M is U = -GmM/R. (a) If R is the radius of the body of mass M, find the escape speed v, of the body, which is the minimum speed needed to leave it permanently. (b) Obtain a formula for the Schwarzschild radius of the body by setting vc = c, the speed of light, and solving for R. (Of course, a relativistic calculation is correct here, but it is interesting to see what a classical calculation produces.) 【Sol】 (a) To leave the body of mass M permanently, the body of mass m must have enough kinetic energy so that there is no radius at which its energy is positive. That is, its total energy must be nonnegative. The escape velocity ve is the speed (for a given radius, and assuming M >> m) that the body of mass m would have for a total energy of zero; 1 GMm 2GM mv e2 − = 0, or v e = . 2 R R (b) Solving the above expression for R in terms of ve, R=

2GM v e2

,

and if ve = c, Equation (2.30) is obtained.

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Chapter 3. Problem Solutions 1.

A photon and a particle have the same wavelength. Can anything be said about how their linear momenta compare? About how the photon's energy compares with the particle's total energy? About how the photon’s energy compares with the particle's kinetic energy?

【Sol】 From Equation (3.1), any particle’s wavelength is determined by its momentum, and hence particles with the same wavelength have the same momenta. With a common momentum p, the photon’s energy is pc, and the particle’s energy is ( pc )2 + (mc 2 )2 , which is necessarily greater than pc for a massive particle. The particle’s kinetic energy is K = E − mc 2 =

(pc )2 + (mc 2 )

2

− mc 2

For low values of p (p (mc2)2, then pc >> mc2 and E ≈ pc. For a photon with the same energy, E = pc, so the momentum of such a particle would be nearly the same as a photon with the same energy, and so the de Broglie wavelengths would be the same.

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Chapter 3. Problem Solutions 13. An electron and a proton have the same velocity Compare the wavelengths and the phase and group velocities of their de Broglie waves. 【Sol】 For massive particles of the same speed, relativistic or nonrelativistic, the momentum will be proportional to the mass, and so the de Broglie wavelength will be inversely proportional to the mass; the electron will have the longer wavelength by a factor of (mp/me) = 1838. From Equation (3.3) the particles have the same phase velocity and from Equation (3.16) they have the same group velocity. 15. Verify the statement in the text that, if the phase velocity is the same for all wavelengths of a certain wave phenomenon (that is, there is no dispersion), the group and phase velocities are the same. 【Sol】 Suppose that the phase velocity is independent of wavelength, and hence independent of the wave number k; then, from Equation (3.3), the phase velocity vp = (ω/k) = u, a constant. It follows that because ω = uk, dω vg = = u = v p. dk

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Chapter 3. Problem Solutions 17. The phase velocity of ocean waves is gλ / 2π , where g is the acceleration of gravity. Find the group velocity of ocean waves 【Sol】 The phase velocity may be expressed in terms of the wave number k = 2π/λ as ω g vp = = , or ω = gk or ω 2 = gk. k k Finding the group velocity by differentiating ω(k) with respect to k, dω 1 1 1 g 1ω 1 vg = = g = = = v p. dk 2 k 2 k 2k 2 Using implicit differentiation in the formula for ω2(k), dω 2ω = 2ωv g = g , dk g gk ω2 ω 1 v = = = = = vp, so that g 2ω 2ωk 2ωk 2k 2 the same result. For those more comfortable with calculus, the dispersion relation may be expressed as 2 ln(ω ) = ln(k ) + ln(g ), dω dk 1ω 1 = , and v g = = v p. from which 2 ω k 2k 2

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Chapter 3. Problem Solutions 19. Find the phase and group velocities of the de Broglie waves of an electron whose kinetic energy is 500 keV. 【Sol】 1 K + mc 2 500 + 511 = = = 1.978. For a kinetic energy of 500 keV, γ = 2 2 2 511 mc 1 − v /c Solving for v, v = c 1 − (1/γ )2 = c 1 − (1/1.978)2 = 0.863c , and from Equation (3.16), vg = v = 0.863c. The phase velocity is then vp = c2 /vg = 1.16 c. 21. (a) Show that the phase velocity of the de Broglie waves of a particle of mass m and de Broglie wavelength λ is given by 2 mc λ   vp = c 1 +    h  (b) Compare the phase and group velocities of an electron whose de Broglie wavelength is exactly 1 x 10-13 m. 【Sol】 (a) Two equivalent methods will be presented here. Both will assume the validity of Equation (3.16), in that vg = v. First: Express the wavelength x in terms of vg, v g2 h h h λ= = = 1− 2. p mv g γ mv g c

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Multiplying by mvg, squaring and solving for vg2 gives −1

  mλc  2  h 2 2 vg = = c 1 +    . 2 2 2 ( λm ) + (h /c )   h   Taking the square root and using Equation (3.3), vp = c2/vg, gives the desired result. 2

Second: Consider the particle energy in terms of vp = c2 lvg;

( )2 (mc 2 )2

E 2 = ( pc )2 + mc 2

( )

γ mc 2

Dividing by

2 2

=

1−c

2

/v 2p

2

( )

2  hc  =   + mc 2 .  λ 

(mc2)2

leads to c2 1 1− 2 = , so that v p 1 + h 2 /(mc λ )2 c2 1 h 2 (mcλ )2 1 − 1 = = = , 2 2 2 2 2 2 2 vp 1 + h /(mcλ ) h (mcλ ) + 1 1 + (mcλ ) /h

which is an equivalent statement of the desired result. It should be noted that in the first method presented above could be used to find λ in terms of vp directly, and in the second method the energy could be found in terms of vg. The final result is, or course, the same.

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(b) Using the result of part (a), 2

 (9.1 × 10− 31 kg)(3.0 × 108 m/s)(1.0 × 10-13 m)   = 1.00085c , v p = c 1 +  −34  6.63 × 10 J⋅s   and vg = c2/vp = 0.99915c. For a calculational shortcut, write the result of part (a) as 2

2

 mc 2λ   (511 × 103 eV)(1.00 × 10-13 m)   = c 1+   = 1.00085c . v p = c 1 +  − 6    1.24 × 10 eV ⋅ m  hc    In both of the above answers, the statement that the de Broglie wavelength is “exactly” 10-13 m means that the answers can be given to any desired precision.

23. What effect on the scattering angle in the Davisson-Germer experiment does increasing the electron energy have? 【Sol】 Increasing the electron energy increases the electron's momentum, and hence decreases the electron's de Broglie wavelength. From Equation (2.13), a smaller de Broglie wavelength results in a smaller scattering angle.

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Chapter 3. Problem Solutions 25. In Sec. 3.5 it was mentioned that the energy of an electron entering a crystal increase, which reduces its de Broglie wavelength. Consider a beam of 54-eV electrons directed at a nickel target. The potential energy of an electron that enters the target changes by 26 eV. (a) Compare the electron speeds outside and inside the target. (b) Compare the respective de Broglie wavelengths. 【Sol】 (a) For the given energies, a nonrelativistic calculation is sufficient; 2K 2( 54 eV)(1.60 × 10-19 J/eV) v = = = 4.36 m/s − 31 m 9.1 × 10 kg outside the crystal, and (from a similar calculation, with K = 80 eV), v = 5.30 x 106 m/s inside the crystal (keeping an extra significant figure in both calculations). (b) With the speeds found in part (a), the de Brogile wavelengths are found from h h 6.63 × 10− 34 J ⋅ s λ= = = = 1.67 × 10−10 m, − 31 6 p mv ( 9.11 × 10 kg)(4.36 × 10 m/s) or 0.167 nm outside the crystal, with a similar calculation giving 0.137 nm inside the crystal.

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Chapter 3. Problem Solutions 27. Obtain an expression for the energy levels (in MeV) of a neutron confined to a one-dimensional box 1.00 x 10 -14 m wide. What is the neutron's minimum energy? (The diameter of an atomic nucleus is of this order of magnitude.) 【Sol】 From Equation (3.18), En = n

2

h2

=n

2

(6.63 × 10−34 J ⋅ s) 2

8mL 2 8(1.67 × 10− 27 kg)(1.00 × 10-14 m) 2 The minimum energy, corresponding to n = 1, is 20.5 MeV

= n 2 3.28 × 10−13 J = n 2 20.5 MeV.

29. A proton in a one-dimensional box has an energy of 400 keV in its first excited state. How wide is the box? 【Sol】 The first excited state corresponds to n = 2 in Equation (3.18). Solving for the width L, h2 ( 6.63 × 10− 34 J ⋅ s) 2 L =n =2 8mE 2 8(1.67 × 10− 27 kg)(400 × 103 eV)(1.60 × 10-19 J/eV) = 4.53 × 10−14 m = 45.3 fm.

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Chapter 3. Problem Solutions 31. The atoms in a solid possess a certain minimum zero-point energy even at 0 K, while no such restriction holds for the molecules in an ideal gas. Use the uncertainty principle to explain these statements. 【Sol】 Each atom in a solid is limited to a certain definite region of space - otherwise the assembly of atoms would not be a solid. The uncertainty in position of each atom is therefore finite, and its momentum and hence energy cannot be zero. The position of an ideal-gas molecule is not restricted, so the uncertainty in its position is effectively infinite and its momentum and hence energy can be zero. 33. The position and momentum of a 1.00-keV electron are simultaneously determined. If its position is located to within 0.100 nm, what is the percentage of uncertainty in its momentum? 【Sol】 The percentage uncertainty in the electron's momentum will be at least ∆p h h hc = = = p 4πp∆x 4π∆x 2mK 4π∆x 2(mc )2 K =

(1.24 × 10− 6 eV ⋅ m)

= 3.1 × 10− 2 = 3.1 %.

4π (1.00 × 10−10 m) 2(511 × 103 eV)(1.00 × 103 eV) Note that in the above calculation, conversion of the mass of the electron into its energy equivalent in electronvolts is purely optional; converting the kinetic energy into joules and using h = 6.626 x 10-34 J·s will of course give the same percentage uncertainty.

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Chapter 3. Problem Solutions 35. How accurately can the position of a proton with v = ∫−∞ Ψ * Ψdx = ∫−∞ Ψ* Ψdx = i i i for Ψ(x, t) normalized.

Thus

( pˆxˆ − xˆp ˆ )Ψ =

11. Obtain Schrödinger’s steady-state equation from Eq.(3.5) with the help of de Broglie’s relationship λ = h/mv by letting y = ψ and finding ∂2ψ/∂x2 . 【Sol】 Using λν = vp in Equation (3.5), and using ψ instead of y,   x   x  ψ = A cos 2π  t − = A cos 2πνt − 2π .   v p   λ    Differentiating twice with respect to x using the chain rule for partial differentiation (similar to Example 5.1), ∂ψ x  2π  2π x   = − A sin  2πνt − 2π  − A sin  2πνt − 2π , = ∂x λ  λ  λ λ   2 2 ∂ 2ψ 2π x   2π   2π  x    2π  = A cos  2πνt − 2π   − =  A cos 2πνt − 2π  = −  ψ λ λ  λ   λ  λ    λ  ∂x 2

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The kinetic energy of a nonrelativistic particle is 2

p2 h  1 KE = E − U = =  , 2m  λ  2m

1

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=

2m

(E − U ) 2 2 λ h ∂ 2ψ 1 and ψ Substituting the above expression relating ∂x 2 λ2 2 ∂ 2ψ 8π 2m 2m  2π  = − ψ = − ( E − U ) ψ = − ( E − U )ψ , which is Equation (5.32)   2 2 2 λ   ∂x h h so that

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13. One of the possible wave functions of a particle in the potential well of Fig. 5.17 is sketched there. Explain why the wavelength and amplitude of &P vary as they do.

【Sol】 The wave function must vanish at x = 0, where V →∞. As the potential energy increases with x, the particle's kinetic energy must decrease, and so the wavelength increases. The amplitude increases as the wavelength increases because a larger wavelength means a smaller momentum (indicated as well by the lower kinetic energy), and the particle is more likely to be found where the momentum has a lower magnitude. The wave function vanishes again where the potential V →∞; this condition would determine the allowed energies.

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15. An important property of the eigenfunctions of a system is that they are orthogonal to one another, which means that +∞ ψ ψ dV −∞ n m

∫

=0

n ≠m

Verify this relationship for the eigenfunctions of a particle in a one-dimensional box given by Eq. (5.46). 【Sol】 The necessary integrals are of the form +∞ 2 L nπx mπx ∫−∞ψ nψ mdx = L ∫0 sin L sin L dx for integers n, m, with n ≠ m and n ≠ -m. (A more general orthogonality relation would involve the integral of ψn*ψm, but as the eigenfunctions in this problem are real, the distinction need not be made.) To do the integrals directly, a convenient identity to use is sin α sin β = 12 [cos(α − β ) − cos(α + β )], as may be verified by expanding the cosines of the sum and difference of α and β. To show orthogonality, the stipulation n ≠ m means that α ≠ β and α ≠ -β and the integrals are of the form

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+∞

∫−∞ψ nψ mdx

=

1 L

L

∫0

(n − m )πx (n + m )πx   cos − cos  dx L L L

 L (n − m )πx L (n + m )πx  = sin − sin  = 0, ( n − m ) π L ( n + m ) π L  o where sin(n - m)π = sin(n - m)π = sin 0 = 0 has been used. 17. As shown in the text, the expectation value of a particle trapped in a box L wide is L/2, which means that its average position is the middle of the box. Find the expectation value . 【Sol】 Using Equation (5.46), the expectation value is 2 L  n πx  < x 2 >n = ∫0 x 2 sin 2  dx . L  L  See the end of this chapter for an alternate analytic technique for evaluating this integral using Leibniz’s Rule. From either a table or repeated integration by parts, the indefinite integral is 3

nπx  L   L  3 ∫ x sin L dx =  nπ  ∫ u sin udu =  n π  where the substitution u = (nπ/L)x has been made. 2

2

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3 3

 u u2 u 1 − sin 2 u − cos 2 u + sin 2 u  . 6 4 4 8  

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This form makes evaluation of the definite integral a bit simpler; when x = 0 u = 0, and when x = L u = nπ. Each of the terms in the integral vanish at u = 0, and the terms with sin 2u vanish at u = nπ, cos 2u = cos 2nπ = 1, and so the result is 3 2  L  (nπ )3 nπ  1  2 2 1 < x >n =  − .   =L  − 2 2 L  nπ   6 4  3  2n π  As a check, note that L2 2 lim < x >n = , n →∞ 3 which is the expectation value of in the classical limit, for which the probability distribution is independent of position in the box. 19. Find the probability that a particle in a box L wide can be found between x = 0 and x = L/n when it is in the nth state. 【Sol】 This is a special case of the probability that such a particle is between x1 and x2, as found in Example 5.4. With x1 = 0 and x2 = L, L

P0L

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1 2nπx  1 x = − sin = . L 0 n  L 2nπ

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21. A particle is in a cubic box with infinitely hard walls whose edges are L long (Fig. 5. 18). The wave functions of the particle are given by n x = 1, 2, 3,K n xπ x n z πy n z πz ψ = A sin sin sin n y = 1, 2, 3,K L L L n z = 1, 2, 3, K Find the value of the normalization constant A. 【Sol】 The normalization constant, assuming A to be real, is given by

∫ψ * ψ dV

= 1 = ∫ψ * ψdxdydz

n yπy  L n πx  L n πz   L = A 2  ∫0 sin 2 x dx  ∫0 sin 2 dy  ∫0 sin 2 z dz . L L L     Each integral above is equal to L/2 (from calculations identical to Equation (5.43)). The result is 2 L

3

 A   =1 2

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or

2 A=  L 

3/ 2

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23. (a) Find the possible energies of the particle in the box of Exercise 21 by substituting its wave function ψ in Schrödinger's equation and solving for E. (Hint: inside the box U = 0.) (b) Compare the ground-state energy of a particle in a one-dimensional box of length L with that of a particle in the three-dimensional box. 【Sol】 (a) For the wave function of Problem 5-21, Equation (5.33) must be used to find the energy. Before substitution into Equation (5.33), it is convenient and useful to note that for this wave function n y2π 2 ∂ 2ψ n x2 π 2 ∂ 2ψ ∂ 2ψ n z2π 2 = − 2 ψ, = − 2 ψ, = − 2 ψ. ∂x 2 L ∂y 2 L ∂z 2 L Then, substitution into Equation (5.33) gives −

π2

(n x2 + n y2 + n z2 )ψ +

2m

E ψ = 0, 2 L2 h 2 2 π h and so the energies are En x ,n y ,nz = (n x2 + n y2 + n z2 ). 2 2mL (b) The lowest energy occurs when nx = ny = nz = 1. None of the integers nx, ny, or nz can be zero, as that would mean ψ = 0 identically. The minimum energy is then E min =

3π 2h 2 2

,

2mL which is three times the ground-state energy of a particle in a one-dimensional box of length L.

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25. A beam of electrons is incident on a barrier 6.00 eV high and 0.200 nm wide. Use Eq. (5.60) to find the energy they should have if 1.00 percent of them are to get through the barrier. 【Sol】 Solving equation (5.60) for k2, 1 1 1 10 -1 k2 = ln = ln( 100 ) = 1 . 15 × 10 m 2L T 2( 0.200 × 10− 9 m ) Equation (5.86), from the appendix, may be solved for the energy E, but a more direct expression is p2 ( hk 2 )2 E = U − KE = U − =U − 2m 2m

2 ( (1.05 × 10− 34 J ⋅ s)(1.15 × 1010 m −1 )) = 6.00 eV − = 0.95 eV

2( 9.1 × 10−31 kg)(1.6 × 10−19 J/eV )

27. What bearing would you think the uncertainty principle has on the existence of the zero-point energy of a harmonic oscillator? 【Sol】 If a particle in a harmonic-oscillator potential had zero energy, the particle would have to be at rest at the position of the potential minimum. The uncertainty principle dictates that such a particle would have an infinite uncertainty in momentum, and hence an infinite uncertainty in energy. This contradiction implies that the zero-point energy of a harmonic oscillator cannot be zero.

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29. Show that for the n = 0 state of a harmonic oscillator whose classical amplitude of motion is A, y = 1 at x = A, where y is the quantity defined by Eq. (5.67). 【Sol】 When the classical amplitude of motion is A, the energy of the oscillator is 1 1 hν kA 2 = hν , so A = . 2 2 k Using this for x in Equation (5.67) gives 2πmν hν mν 2 y= = 2π = 1, h k k where Equation (5.64) has been used to relate ν, m and k. 31. Find the expectation values and for the first two states of a harmonic oscillator. 【Sol】 The expectation values will be of the forms ∞

∫−∞ xψ * ψ dx

and

∞ 2 ∫−∞ x ψ * ψ dx

It is far more convenient to use the dimensionless variable y as defined in Equation (5.67). The necessary integrals will be proportional to ∞ −y2 ye dy , −∞

∫

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∞ 2 −y 2 y e dy , −∞

∫

∞ 3 −y2 y e dy , −∞

∫

∞ 4 −y 2 y e dy , −∞

∫

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The first and third integrals are seen to be zero (see Example 5.7). The other two integrals may be found from tables, from symbolic-manipulation programs, or by any of the methods outlined at the end of this chapter or in Special Integrals for Harmonic Oscillators, preceding the solutions for Section 5.8 problems in this manual. The integrals are ∞ 1 ∞ 3 2 −y 2 4 −y2 y e dy = π , y e dy = π. ∫−∞ ∫−∞ 2 4 An immediate result is that = 0 for the first two states of any harmonic oscillator, and in fact = 0 for any state of a harmonic oscillator (if x = 0 is the minimum of potential energy). A generalization of the above to any case where the potential energy is a symmetric function of x, which gives rise to wave functions that are either symmetric or antisymmetric, leads to = 0. To find for the first two states, the necessary integrals are ∞ x 2ψ o *ψ 0dx −∞

∫

1/2

 2mν  =   h  =

∞ 2 x ψ 1* ψ1dx −∞

∫

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3/ 2

∞ 2 −y2 y e dy −∞

∫

π (1/ 2)hν E0 = = ; k 2π 3/ 2mν 2 4π 2mν 2 h

1/2

 2mν  =   h  =

 h     2πmν 

h 2π 3 /2mν

 h     2πmν 

2

3/ 2

∞ 4 −y2 2 y e dy −∞

∫

3 π ( 3/ 2)hν E1 = = . 2 k 4π 2mν 2

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In both of the above integrals, dx dx = dy = dy

h dy 2πmν

has been used, as well as Table 5.2 and Equation (5.64). 33. A pendulum with a 1.00-g bob has a massless string 250 mm long. The period of the pendulum is 1.00 s. (a) What is its zero-point energy? Would you expect the zero-point oscillations to be detectable? (b) The pendulum swings with a very small amplitude such that its bob rises a maximum of 1.00 mm above its equilibrium position. What is the corresponding quantum number? 【Sol】 (a) The zero-point energy would be 1 h 4.14 × 10−15 eV ⋅ s E 0 = hν = = = 2.07 × 10−15 eV, 2 2T 2(1.00 s) which is not detectable. (b) The total energy is E = mgH (here, H is the maximum pendulum height, given as an uppercase letter to distinguish from Planck's constant), and solving Equation (5.70) for n, E 1 mgH (1.00 × 10− 3 kg )( 9.80 m/s 2 )(1.00 s ) 1 n= − = = − = 1.48 × 1028. − 34 hν 2 h /T 2 6.63 × 10 J ⋅ s

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37. Consider a beam of particles of kinetic energy E incident on a potential step at x = 0 that is U high, where E > U (Fig. 5.19). (a) Explain why the solution De-ik’x (in the notation of appendix) has no physical meaning in this situation, so that D = 0. (b) Show that the transmission probability here is T = CC*v‘/AA*v1 = 4k12/(k1 + k’)2. (c) A 1.00-mA beam of electrons moving at 2.00x106 m/s enters a region with a sharply defined boundary in which the electron speeds are reduced to 1.00x106 m/s by a difference in potential. Find the transmitted and reflected currents. 【Sol】 (a) In the notation of the Appendix, the wave function in the two regions has the form ψ I = Aeik1x + Be −ik1x ,

ψ II = Ce ik ′x + De −ik ′x ,

where

2mE 2m ( E − U ) , k′ = . h h The terms corresponding to exp(ik1x) and exp(ik’x) represent particles traveling to the left; this is possible in region I, due to reflection at the step at x = 0, but not in region II (the reasoning is the same as that which lead to setting G = 0 in Equation (5.82)). Therefore, the exp(-ik’x) term is not physically meaningful, and D = 0. k1 =

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(b) The boundary condition at x= 0 are then

k′ ik1A − ik1B = ik ′C or A − B = C . k1  k′  Adding to eliminate B, 2 A = 1 + C , so  k1  C 2k1 CC * 4k12 = , and = . 2 ′ A k1 + k AA * (k1 + k ′) (c) The particle speeds are different in the two regions, so Equation (5.83) becomes 2 ψ II v ′ CC * k ′ 4k1k ′ 4(k1 /k ′) T = = = = . 2 2 2 AA * k (k1 + k ′ ) ((k1 /k ′) + 1) ψ I v1 1 A + B = C,

For the given situation, k1/k’ = v1/v’ = 2.00, so T = (4x2)/(2+1)2 = 8/9. The transmitted current is (T)(1.00 mA) = 0.889 mA, and the reflected current is 0.111mA. As a check on the last result, note that the ratio of the reflected current to the incident current is, in the notation of the Appendix, 2

R=

ψ I − v1 2

ψ I + v1

=

BB * AA *

Eliminating C from the equations obtained in part (b) from the continuity condition as x = 0,    (k /k ′) − 1  1 k′  k′  A 1 −  = B 1 + , so R =  1  = = 1 − T ′ k k ( k / k ) + 1  1  1  1  9

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Chapter 6 Problem Solutions 1. Why is it natural that three quantum numbers are needed to describe an atomic electron (apart from electron spin)? 【sol】 Whether in Cartesian (x, y, z) or spherical coordinates, three quantities are needed to describe the variation of the wave function throughout space. The three quantum numbers needed to describe an atomic electron correspond to the variation in the radial direction, the variation in the azimuthal direction (the variation along the circumference of the classical orbit), and the variation with the polar direction (variation along the direction from the classical axis of rotation). 3. Show that R10 (r ) =

2

−r / 2ao e 3 /2

a0 is a solution of Eq. (6.14) and that it is normalized. 【sol】 For the given function, d 2 R10 = − 5 / 2 e −r /ao , dr a0

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and

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1 d  2 dR10  2 1  r 2  −r /ao e r  = − 5 /2 2  2r − dr  ao  r 2 dr  ao r   1 2  R10 =  2 −  r a o  ao This is the solution to Equation (6.14) if l=0 ( as indicated by the index of R10), 2 2me 2 = , ao h 2 4πεo

or

ao =

4π 2εoh 2 me 2

which is the case, and 2m

1

e2 E =− = E1 8πεoao

E = − 2 , or h2 ao again as indicated by the index of R10. To show normalization, 4 ∞ 2 − 2r /ao 1 ∞ ∞ 2 2 dr = ∫0 u 2e −udu, ∫0 R10 r dr = a 3 ∫0 r e 2 o where the substitution u=2r/ao has been made. The improper definite integral in u is known to have the value 2 and so the given function is normalized.

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5. In Exercise 12 of Chap. 5 it was stated that an important property of the eigenfunctions of a system is that they are orthogonal to one another, which means that ∞ * ∫−∞ψ nψm dV = 0 n ≠ m

Verify that this is true for the azimuthal wave functions Φml of the hydrogen atom by calculating 2π * ∫ Φml Φml′ dφ for ml ≠ ml′ 0

【sol】 From Equation (6.15) the integral, apart from the normalization constants, is 2π * 2π −im φ im ′φ ∫0 Φml Φml′dφ = ∫0 e l e l dφ ,

It is possible to express the integral in terms of real and imaginary parts, but it turns out to be more convenient to do the integral directly in terms of complex exponentials: 2π

∫0

′

e −imlφe imlφdφ = =

2π i (ml′ −ml′ )φ

∫0

dφ

e

[

]

1 ′ ′ 2π e i(ml −ml )φ 0 = 0 i (ml′ − ml )

The above form for the integral is valid only for ml ≠ ml’, which is given for this case. In evaluating the integral at the limits, the fact that ei2πn = 1 for any integer n ( in this case (ml’ – ml)) has been used.

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7. Compare the angular momentum of a ground-state electron in the Bohr model of the hydrogen atom with its value in the quantum theory. 【sol】 In the Bohr model, for the ground-state orbit of an electron in a hydrogen atom, λ = h/mv = 2πr, and so L = pr = . In the quantum theory, zero-angular-momentum states (ψ spherically symmetric) are allowed, and L = 0 for a ground-state hydrogen atom. 9. Under what circumstances, if any, is Lz equal to L? 【sol】 From Equation (6.22), Lz must be an integer multiple of ; for L to be equal to Lz, the product l(1+1), from Equation(6.21), must be the square of some integer less than or equal to l. But, l 2 ≤ l (l + 1) < (l + 1)2 or any nonnegative l, with equality holding in the first relation only if l = 0. Therefore, l(l + 1) is the square of an integer only if l = 0, in which case Lz = 0 and L = Lz = 0.

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11. What are the possible values of the magnetic quantum number ml of an atomic electron whose orbital quantum number is l = 4? 【sol】 From Equation (6.22), the possible values for the magnetic quantum number ml are ml = 0, ±1, ±2, ±3, ±4, a total of nine possible values. 13. Find the percentage difference between L and the maximum value of Lz for an atomic electron in p, d, and f states. 【sol】 The fractional difference between L and the largest value of Lz, is, for a given l, L − L z ,max L

=

l (l + 1) − l l =1− . l (l + 1) l +1

For a p state, l = 1 and 1 -

1 2

= 0.29 = 29%

For a d state, l = 2 and 1 -

2 3

= 0.18 = 18%

For a f state, l = 3 and 1 -

3 4

= 0.13 = 13%

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15. In Sec. 6.7 it is stated that the most probable value of r for a 1s electron in a hydrogen atom is the Bohr radius ao. Verify this. 【sol】 Using R10(r) from Table 6.1 in Equation (6.25), 4r 2 − 2r /ao P (r ) = 3 e . ao The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero, d 4  2r 2  − 2r /ao e P (r ) = 3  2r − = 0,   dr a ao  o  r2 r = ao

and

or

r = 0, ao

for an extreme. At r = 0, P(r) = 0, and because P(r) is never negative, this must be a minimum. dp/dr → 0 as r → ∞, and this also corresponds to a minimum. The only maximum of P(r) is at r = ao, which is the radius of the first Bohr orbit.

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17. Find the most probable value of r for a 3d electron in a hydrogen atom. 【sol】 Using R20 (r) from Table 6.1 in Equation (6.25), and ignoring the leading constants (which would not affect the position of extremes), P (r ) = r 6e − 2r / 3ao The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero, d P (r ) = dr 2r 6 6r = 3ao 5

 5 2r 6  − 2r /3ao  6r − e = 0,   3 a o  and

or

r = 0, 9a o

for an extreme. At r = 0, P(r) = 0, and because P(r) is never negative, this must be a minimum. dP/dr → 0 as r → ∞, and this also corresponds to a minimum. The only maximum of P(r) is at r = 9ao, which is the radius of the third Bohr orbit.

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19. How much more likely is the electron in a ground-state hydrogen atom to be at the distance ao from the nucleus than at the distance 2ao? 【sol】 For the ground state, n = 1, the wave function is independent of angle, as seen from the functions Φ(φ) and Θ(θ) in Table 6.1, where for n = 1, l = ml, = 0 (see Problem 6-14). The ratio of the probabilities is then the ratio of the product r2 (R10 (r))2 evaluated at the different distances. Specially, − 2a /a

P (ao )dr a o2e o o e −2 4 = = = = 1.47 P (ao / 2)dr (ao / 2)2 e − 2(ao / 2) /ao (1/ 4)e −1 e Similarly, − 2a /a

P (ao )dr ao2e o o e −2 e2 = = = = 1.85 P ( 2ao )dr ( 2a o )2 e − 2( 2ao )/ao ( 4)e − 4 4

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21. The probability of finding an atomic electron whose radial wave function is R(r) outside a sphere of radius ro centered on the nucleus is ∞ ro

∫

2

R (r ) r 2dr

(a) Calculate the probability of finding a 1s electron in a hydrogen atom at a distance greater than ao from the nucleus. (b) When a 1s electron in a hydrogen atom is 2ao from the nucleus, all its energy is potential energy. According to classical physics, the electron therefore cannot ever exceed the distance 2ao from the nucleus. Find the probability r > 2ao for a 1s electron in a hydrogen atom. 【sol】 (a) Using R10(r) for the 1s radial function from Table 6.1, ∞ 4 ∞ 2 − 2r /ao 1 ∞ 2 2 dr = ∫2 u 2e −udu , ∫ao R(r ) r dr = a 3 ∫ao r e 2 o where the substitution u = 2r/a0 has been made. Using the method outlined at the end of this chapter to find the improper definite integral leads to

[

(

)]

1 ∞ 2 −u 1 u e du = − e −u u 2 + 2u + 2 ∫ 2 2 2

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∞ 2

=

[

]

1 −2 e 10 = 0.68 = 68%, 2

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(b) Repeating the above calculation with 2 a0 as the lower limit of the integral, ∞ 1 ∞ 2 −u 1 1 −4 −u 2 u e du = − e u + 2 u + 2 = e 26 = 0.24 = 24% , 4 2 ∫4 2 2

[

)]

(

[

]

23. Unsold's theorem states that for any value of the orbital quantum number l, the probability densities summed over all possible states from ml = -1 to ml = +1 yield a constant independent of angles θ or φ that is, +l

∑Θ

ml = −l

2

2

Φ = constant

This theorem means that every closed subshell atom or ion (Sec. 7.6) has a spherically symmetric distribution of electric charge. Verify Unsold's theorem for l = 0, l = 1, and l = 2 with the help of Table 6. 1. 【sol】For l = 0, only ml = 0 is allowed, Φ(φ) and Θ(θ) are both constants (from Table 6.1)), and the theorem is verified. For l = 1, the sum is 1 3 1 3 1 3 3 sin 2 θ + cos 2 θ + sin 2 θ = , 2π 4 2π 2 2π 4 4π

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In the above, Φ*Φ= 1/2π, which holds for any l and ml, has been used. Note that one term appears twice, one for ml = -1 and once for ml = 1. For l = 2, combining the identical terms for ml = ±2 and ml = ±1, and again using Φ*Φ= 1/2π, the sum is

2

1 15 1 15 1 10 sin4 θ + 2 sin2 θ cos2 θ + (3 cos2 θ − 1)2 . 2π 16 2π 4 2π 16

The above may he simplified by extracting the commons constant factors, to 5 [(3 cos2 θ − 1)2 + 12 sin2 θ cos2 θ + 3 sin4 θ ]. 16π Of the many ways of showing the term in brackets is indeed a constant, the one presented here, using a bit of hindsight, seems to be one of the more direct methods. Using the identity sin2 θ = 1 - cos2 θ to eliminate sin θ , ( 3 cos 2 θ − 1)2 + 12 sin 2 θ cos2 θ + 3 sin 4 θ = ( 9 cos4 θ − 6 cos 2 θ + 1) + 12(1 − cos 2 θ ) cos 2 θ + 3(1 − 2 cos 2 θ + cos4 θ ) = 1, and the theorem is verified.

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25. With the help of the wave functions listed in Table 6.1 verify that ∆l = ±1 for n = 2 à n = 1 transitions in the hydrogen atom. 【sol】 In the integral of Equation (6.35), the radial integral will never. vanish, and only the angular functions Φ(φ) and Θ(θ) need to be considered. The ∆l = 0 transition is seen to be forbidden, in that the product 1 ( Φ0 (φ )Θ00(θ ))∗ (Φ 0 (φ )Θ00 (θ )) = 4π is spherically symmetric, and any integral of the form of Equation (6.35) must vanish, as the argument u = x, y or z will assume positive and negative values with equal probability amplitudes. If l = 1 in the initial state, the integral in Equation (6.35) will be seen to to vanish if u is chosen appropriately. If ml = 0 initially, and u = z = r cos θ is used, the integral (apart from constants) is π 2 2 cos θ sin θ d θ = ≠0 ∫0 3 If ml = ±1 initially, and u = x = r sin θ cos φ is used, the θ -integral is of the form

π

π 2 ∫0 sin θdθ = 2 ≠ 0

and the φ -integral is of the form

2π ±i φ e cos φdφ 0

∫

=

2π 2 cos φdφ 0

∫

=π ≠0

and the transition is allowed.

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27. Verify that the n = 3 → n = 1 transition for the particle in a box of Sec. 5.8 is forbidden whereas the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. 【sol】 The relevant integrals are of the form nπx mπx L x sin sin dx. ∫0 L L The integrals may be found in a number of ways, including consulting tables or using symbolicmanipulation programs (see; for instance, the solution to Problem 5-15 for sample Maple commands that are easily adapted to this problem). One way to find a general form for the integral is to use the identity

sin α sin β = 12 [cos(α − β ) − cos(α + β )] and the indefinite integral (found from integration by parts) x sin kx 1 x sin kx cos kx ∫ x cos kxdx = k − k ∫ sin kxdx = k + k 2 to find the above definite integral as L  Lx  (n − m )πx L2 (n − m )πx sin + cos   2 2 ( n − m ) L L ( n − m ) π 1  , 2 Lx (n + m )πx L2 (n + m )πx   −  sin + cos 2 2 (n + m )π L L  (n + m ) π 0

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where n ≠ m2 is assumed. The terms involving sines vanish, with the result of L2  cos(n − m )π − 1 cos(n + m )π − 1 −  . 2π 2  (n − m )2 (n + m )2  If n and m axe both odd or both even, n + m and n - m are even, the arguments of the cosine terms in the above expression are even-integral multiples of π, and the integral vanishes. Thus, the n = 3 → n = 1 transition is forbidden, while the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. To make use of symmetry arguments, consider that L L nπx mπx L nπx mπx x − sin sin dx = x sin sin dx   ∫0  ∫0 2 L L L L for n ≠ m, because the integral of L times the product of the wave functions is zero; the wave functions were shown to be orthogonal in Chapter 5 (again, see Problem 5-15). Letting u=L/2 – x, sin

nπx nπ (( L / 2) − u )  nπ nπu  = sin = sin  −  L L 2 L  

This expression will be ± cos ( nπu/L ) for n odd and ±sin ( nπu/L ) for n even. The integrand is then an odd function of u when n and m are both even or both odd, and hence the integral is zero. If one of n or m is even and the other odd, the integrand is an even function of u and the integral is nonzero.

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29. Show that the magnetic moment of an electron in a Bohr orbit of radius rn is proportional to rn 【sol】 From Equation (6.39), the magnitude of the magnetic moment of an electron in a Bohr orbit is proportional to the magnitude of the angular momentum, and hence proportional to n. The orbital radius is proportional to n2 (See Equation (4.13) or Problem 4-28), and so the magnetic moment is proportional to rn . 31. Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400-nm wavelength when a spectrometer whose resolution is 0.010 nm is used. 【sol】 See Example 6.4; solving for B, ∆ λ 4πmc 0.010 × 10−9 m 4π ( 9.1 × 10−31 kg )( 3.0 × 108 m/s) B = 2 = = 1.34 T 9 2 − 19 e λ (400 × 10 m) (1.6 × 10 C)

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Chapter 7

Problem Solutions

1.

A beam of electrons enters a uniform 1.20-T magnetic field. (a) Find the energy difference between electrons whose spins are parallel and antiparallel to the field. (b) Find the wavelength of the radiation that can cause the electrons whose spins are parallel to the field to flip so that their spins are antiparallel. 【sol】 (a) Using Equations (7.4) and (6.41), the energy difference is,

∆E = 2µsz B = 2µB B = 2(5.79 × 10−5 eV/T)(1.20 T) = 1.39 × 10−4 eV (b) The wavelength of the radiation that corresponds to this energy is hc 1.24 × 10− 6 eV ⋅ m λ= = = 8.93 mm ∆E 1.39 × 10− 4 eV Note that a more precise value of AB was needed in the intermediate calculation to avoid roundoff error. 3. Find the possible angles between the z axis and the direction of the spin angular-momentum vector S. 【sol】 For an electron, s = ( 3 /2)h, sz = ±(1/ 2)h, and so the possible angles axe given by  ± (1/ 2)h   1  o o arccos   = arccos   = 54.7 , 125.3  3  ( 3 / 2)h 

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5. Protons and neutrons, like electrons, are spin- ½ particles. The nuclei of ordinary helium atoms, 42 He , contain two protons and two neutrons each; the nuclei of another type of helium atom, 23 He, contain two protons and one neutron each. The properties of liquid 42 He and liquid 3 2 He are different because one type of helium atom obeys the exclusion principle but the other does not. Which is which, and why? 【sol】 4 2 He atoms contain even numbers of spin-½ particles, which pair off to give zero or integral spins for the atoms. Such atoms do not obey the exclusion principle. 23 He atoms contain odd numbers of spin- ½ particles, and so have net spins of 12 , 32 or 52 , and they obey the exclusion principle. 7. In what way does the electron structure of an alkali metal atom differ from that of a halogen atom? From that of an inert gas atom? 【sol】 An alkali metal atom has one electron outside closed inner shells: A halogen atom lacks one electron of having a closed outer shell: An inert gas atom has a closed outer shell.

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9. How many electrons can occupy an f subshell? 【sol】 For f subshell, with l = 3, the possible values of ml are ±3, ±2, ±1 or 0, for a total of 2l +1=7 values of ml. Each state can have two electrons of opposite spins, for a total of 14 electrons. 11. If atoms could contain electrons with principal quantum numbers up to and including n = 6, how many elements would there be? 【sol】 The number of elements would be the total number of electrons in all of the shells. Repeated use of Equation (7.14) gives 2n2 + 2 (n - 1)2 +... + 2 (1)2 = 2 (36 + 25 + 16 + 9 + 4 + 1) = 182. In general, using the expression for the sum of the squares of the first n integers, the number of elements would be

2(16 n(2n + 1)(n + 1)) = 13 [n( 2n + 1)(n + 1)],

which gives a total of 182 elements when n = 6.

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13. The ionization energies of Li, Na, K, Rb, and Cs are, respectively, 5.4, 5.1, 4.3, 4.2, and 3.9 eV. All are in group 1 of the periodic table. Account for the decrease in ionization energy with increasing atomic number. 【sol】 All of the atoms are hydrogenlike, in that there is a completely filled subshell that screens the nuclear charge and causes the atom to "appear" to be a single charge. The outermost electron in each of these atoms is further from the nucleus for higher atomic number, and hence has a successively lower binding energy. 15. (a) Make a rough estimate of the effective nuclear charge that acts on each electron in the outer shell of the calcium (Z = 20) atom. Would you think that such an electron is relatively easy or relatively hard to detach from the atom? (b) Do the same for the sulfur (Z = 16) atom. 【sol】 (a) See Table 7.4. The 3d subshell is empty, and so the effective nuclear charge is roughly +2e, and the outer electron is relatively easy to detach. (b) Again, see Table 7.4. The completely filled K and L shells shield +10e of the nuclear charge of = 16e; the filled 3s2 subshell will partially shield the nuclear charge, but not to the same extent as the filled shells, so +6e is a rough estimate for the effective nuclear charge. This outer electron is then relatively hard to d etach.

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17. Why are Cl atoms more chemically active than Cl- ions? Why are Na atoms more chemically active than Na+ ions? 【sol】 Cl- ions have closed shells, whereas a Cl atom is one electron short of having a closed shell and the relatively poorly shielded nuclear charge tends to attract an electron from another atom to fill the shell. Na+ ions have closed shells, whereas an Na atom has a single outer electron that can be detached relatively easily in a chemical reaction with another atom. 19. In each of the following pairs of atoms, which would you expect to be larger in size? Why? Li and F; Li and Na; F and Cl; Na and Si. 【sol】 The Li atom (Z = 3) is larger because the effective nuclear charge acting on its outer electron is less than that acting on the outer electrons of the F atom (Z = 9). The Na atom (Z = 11) is larger because it has an additional electron shell (see Table 7.4). The Cl atom (Z = 17) atom is larger because has an additional electron shell. The Na atom is larger than the Si atom (Z = 14) for the same reason as given for the Li atom.

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21. Why is the normal Zeeman effect observed only in atoms with an even number of electrons? 【sol】 The only way to produce a normal Zeeman effect is to have no net electron spin; because the electron spin is ± ½, the total number of electrons must be even. If the total number of electrons were odd, the net spin would be nonzero, and the anomalous Zeeman effect would be observable. 23. The spin-orbit effect splits the 3P → 3S transition in sodium (which gives rise to the yellow light of sodium-vapor highway lamps) into two lines, 589.0 nm corresponding to 3P3/2 → 3S1/2 and 589.6 nm corresponding to 3P1/2 → 3S1/2 . Use these wavelengths to calculate the effective magnetic field experienced by the outer electron in the sodium atom as a result of its orbital motion. 【sol】 See Example 7.6. Expressing the difference in energy levels as 1 1  ∆ E = 2µ B B = hc  − ; solving for B ,  λ1 λ2  hc  1 1 B =  −  2 µB  λ1 λ2  1.24 × 10−6 eV ⋅ m  1 1  = −   = 18.5 T 2 × 5.79 × 10- 5 eV/T  589.0 × 10−9 m 589.6 × 10−9 m 

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25. If

j=

5 2

, what values of l are possible?

【sol】 The possible values of l are j +

1 2

= 3 and j −

1 2

= 2.

27. What must be true of the subshells of an atom which has a 1S0 ground state? 【sol】 For the ground state to be a singlet state with no net angular momentum, all of the subshells must be filled. 29. The lithium atom has one 2s electron outside a filled inner shell. Its ground state is 2S1/2 . (a) What are the term symbols of the other allowed states, if any? (b) Why would you think the 2S 1/2 state is the ground state? 【sol】 For this doublet state, L = 0, S = J = ½. There axe no other allowed states. This state has the lowest possible values of L and J, and is the only possible ground state.

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31. The aluminum atom has two 3s electrons and one 3p electron outside filled inner shells. Find the term symbol of its ground state. 【sol】 The two 3s electrons have no orbital angular momentum, and their spins are aligned oppositely to give no net angular momentum. The 3p electron has l = 1, so L = 1, and in the ground state J = ½ . The term symbol is 2P1/2 . 33. Why is it impossible for a 2 2D3/2 state to exist? 【sol】 A D state has L = 2; for a 2 2D3/2 state, n = 2 but L must always be strictly less than n, and so this state cannot exist. 35. Answer the questions of Exercise 34 for an f electron in an atom whose total angular momentum is provided by this electron. 【sol】 (a) From Equation (7.17),

j =l ±

1 2

= 52 , 72 .

(b) Also from Equation (7.17), the corresponding angular momenta are

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35 2

h and

63 2

h

Department of Physics

(c) The values of L and S are 12h and 23 h. The law of cosines is J 2 − L2 − S 2 cos θ = , 2LS where θ is the angle between L and S; then the angles are,  ( 35 / 4) − 12 − ( 3/ 4)   2 arccos   = arccos  −  = 132o 2 12( 3 / 2)  3   and  (63/ 4) − 12 − ( 3/ 4)  1 arccos   = arccos   = 60.0o 2 12( 3 / 2)  2   (d) The multiplicity is 2(1/2) + 1 = 2, the state is an f state because the total angular momentum is provided by the f electron, and so the terms symbols are 2F5/2 and 2F7/2 . 37. The magnetic moment µ J of an atom in which LS coupling holds has the magnitude µJ = J( J + 1)g J µ B where µ B = eħ/2m is the Bohr magneton and gJ = 1 +

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J(J + 1) − L(L + 1) + S(S + 1) 2J(J + 1)

Department of Physics

is the Landé g factor. (a) Derive this result with the help of the law of cosines starting from the fact that averaged over time, only the components of µ L and µ S parallel to J contribute to µ L . (b) Consider an atom that obeys LS coupling that is in a weak magnetic field B in which the coupling is preserved. How many substates are there for a given value of J? What is the energy difference between different substates? 【sol】 (a) In Figure 7.15, let the angle between J and S be α and the angle between J and L be β. Then, the product µJ has magnitude   S 2 µB S cos α + µB L cos β = µB J + µB S cos α = µB J  1 + cos α  J   In the above, the factor of 2 in 2µB relating the electron spin magnetic moment to the Bohr magneton is from Equation (7.3). The middle term is obtained by using |S| cos α + |S| cos β = |J|. The above expression is equal to the product µJ because in this form, the magnitudes of the angular momenta include factors of h. From the law of cosines, 2

2

L − J −S cos α = − 2J S

2

and so 2

2

2

S L −J −S J(J + 1) − L(L + 1) + S(S + 1) cos α = = 2 J 2J(J + 1) 2J

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and the expression for µJ in terms of the quantum numbers is µJ h = J gJ µ B , or

µJ = J(J + 1)g J µB ,

whe re

J(J + 1) − L (L + 1) + S(S + 1) 2J(J + 1) (b) There will be one substate for each value of MJ, where MJ = -J ... J, for a total of 2J + 1 substates. The difference in energy between the substates is ∆ E = g J µB M J B gJ = 1 +

39. Explain why the x-ray spectra of elements of nearby atomic numbers are qualitatively very similar, although the optical spectra of these elements may differ considerably. 【sol】 The transitions that give rise to x-ray spectra are the same in all elements since the transitions involve only inner, closed-shell electrons. Optical spectra, however, depend upon the possible states of the outermost electrons, which, together with the transitions permitted for them, are different for atoms of different atomic number.

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41. Find the energy and the wavelength of the Kα x-rays of aluminum. 【sol】 From either of Equations (7.21) or (7.22), E = (10.2 eV) (Z - 1)2 = (10.2 eV) (144) = 1.47 keV. The wavelength is hc 1.24 × 10−6 eV ⋅ m λ= = = 8.44 × 10−10 m = 0.844 nm 3 E 14.7 × 10 eV 43. Distinguish between singlet and triplet states in atoms with two outer electrons. 【sol】 In a singlet state, the spins of the outer electrons are antiparrallel. In a triplet state, they are parallel

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Chapter 8 Problem Solutions 1. The energy needed to detach the electron from a hydrogen atom is 13.6 eV, but the energy needed to detach an electron from a hydrogen molecule is 15.7 eV. Why do you think the latter energy is greater? 【sol】 The nuclear charge of +2e is concentrated at the nucleus, while the electron charges' densities are spread out in (presumably) the 1s subshell. This means that the additional attractive force of the two protons exceeds the mutual repulsion of the electrons to increase the binding energy. 3. At what temperature would the average kinetic energy of the molecules in a hydrogen sample be equal to their binding energy? 【sol】 Using 4.5 eV for the binding energy of hydrogen, 3 2 4.5 eV kT = 4.5 eV or T = = 3.5 × 104 K 5 2 3 8.62 × 10 eV/K

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5. When a molecule rotates, inertia causes its bonds to stretch. (This is why the earth bulges at the equator.) What effects does this stretching have on the rotational spectrum of the molecule? 【sol】 The increase in bond lengths in the molecule increases its moment of inertia and accordingly decreases the frequencies in its rotational spectrum (see Equation (8.9)). In addition, the higher the quantum number J (and hence the greater the angular momentum), the faster the rotation and the greater the distortion, so the spectral lines are no longer evenly spaced. Quantitatively, the parameter I (the moment of inertia of the molecule) is a function of J, with I larger for higher J. Thus, all of the levels as given by Equation (8.11) are different, so that the spectral lines are not evenly spaced. (It should be noted that if I depends on J, the algebraic steps that lead to Equation (8.11) will not be valid.) 7. The J=0àJ=1 rotational absorption line occurs at 1.153x1011 Hz in 12C16O and at 1.102x10 11 Hz in ?C16O. Find the mass number of the unknown carbon isotope. 【sol】 From Equation (8.11), the ratios of the frequencies will be the ratio of the moments of inertia. For the different isotopes, the atomic separation, which depends on the charges of the atoms, will be essentially the same. The ratio of the moments of inertia will then be the ratio of the reduced masses. Denoting the unknown mass number by x and the ratio of the frequencies as r, r in terms of x is

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x ⋅ 16 r = x + 16 12 ⋅ 16 12 + 16 Solving for x in terms of r,

48r 7 − 3r Using r = (1.153)/(1.102) in the above expression gives x = 13.007, or the integer 13 to three significant figures. x=

9. The rotational spectrum of HCI contains the following wavelengths: 12.03 x 10-5 m, 9.60 x 10-5 m, 8.04 x 10-5 m, 6.89 x 10-5 m, 6.04 x 10-5 m If the isotopes involved are 1H and 35Cl, find the distance between the hydrogen and chlorine nuclei in an HCl molecule.

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【sol】 The corresponding frequencies are, from ν = c/λ , and keeping an extra significant figure, in multiplies of 1012 Hz: 2.484, 3.113, 4.337, 4.947 The average spacing of these frequencies is ∆v = 0.616 x 1012 Hz. (A least-squares fit from a spreadsheet program gives 0.6151 if c = 2.998 x 108 m/s is used.) From Equation (8.11), the spacing of the frequencies should be ∆v = /2πI ; Solving for I and using ∆v as found above, h 1.055 × 10−34 J ⋅ s − 47 2 I = = = 2 . 73 × 10 kg ⋅ m 2π∆ν 2π ( 0.6151 × 1012 Hz ) The reduced mass of the HCI molecule is (35/36)rnH, and so the distance between the nuclei is R=

I = µ

36 × ( 2.73 × 10−47 kg ⋅ m 2 ) 35 × (1.67 × 10

− 27

kg)

= 0.129 nm

(keeping extra significant figures in the intermediate calculation gives a result that is rounded to 0.130 nm to three significant figures).

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11. A 200Hg35Cl Molecule emits a 4.4-cm photon when it undergoes a rotational transition from j = 1 to j = 0. Find the interatomic distance in this molecule. 【sol】 Using ν1→0 = c/λ and I = m’ R2 in Equation (8.11) and solving for R, hλ R2 = 2πm ′c For this atom, m’ = mH(200x35)/(200 + 35), and

(1.055 × 10− 34 J ⋅ s)(4.4 × 10− 2 m) R= = 0.223 nm −27 8 2π (1.67 × 10 kg)(3.0 × 10 m/s) or 0.22 nm to two significant figures.

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13. In Sec. 4.6 it was shown that, for large quantum numbers, the frequency of the radiation from a hydrogen atom that drops from an initial state of quantum number n to a final state of quantum number n - 1 is equal to the classical frequency of revolution of an electron in the n-th Bohr orbit. This is an example of Bohr's correspondence principle. Show that a similar correspondence holds for a diatomic molecule rotating about its center of mass. 【sol】 Equation (8.11) may be re-expressed in terms of the frequency of the emitted photon when the molecule drops from the J rotational level to the J - 1 rotational level, hJ ν J →J −1 = . 2πI For large J, the angular momentum of the molecule in its initial state is L = h J ( J + 1) = hJ 1 + 1/ J ≈ hJ Thus, for large J, L ν ≈ , or L = ωI , 2πI the classical expression.

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15. The hydrogen isotope deuterium has an atomic mass approximately twice that of ordinary hydrogen. Does H2 or HD have the greater zero-point energy? How does this affect the binding energies of the two molecules? 【sol】 The shape of the curve in Figure 8.18 will be the same for either isotope; that is, the value of k in Equation (8.14) will be the same. HD has the greater reduced mass, and hence the smaller frequency of vibration vo and the smaller zero- point energy. HD is the more tightly bound, and has the greater binding energy since its zero-point energy contributes less energy to the splitting of the molecule. 17. The force constant of the 1H19F molecule is approximately 966 N/m. (a) Find the frequency of vibration of the molecule. (b) The bond length in 1H19F is approximately 0.92 nm. Plot the potential energy of this molecule versus internuclear distance in the vicinity of 0.92 nm and show the vibrational energy levels as in Fig. 8.20. 【sol】 (a) Using m'= (19/20)mH in Equation (8.15), νo =

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1 2π

966 N/m

20 = 1.24 × 1014 Hz - 27 1.67 × 10 kg 19

Department of Physics

k = 4.11 X 10-20 J. The levels are shown below, where the vertical m′ scale is in units of 10-20 J and the horizontal scale is in units of 10-11 m. (b) Eo = 12 h

19. The lowest vibrational states of the 23Na35Cl molecule are 0.063 eV apart. Find the approximate force constant of this molecule. 【sol】 From Equation (8.16), the lower energy levels are separated by ∆E = hvo, and vo = ∆E /h. Solving Equation (8.15) for k,  ∆E  k = m ′( 2πνo )2 = m ′   h 

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Using m’ = mH (23·35)/(23 + 35),

 (0.063 eV)(1.60 × 10-19 J/eV)  23 ⋅ 35 − 27  = 213 N/m k= (1.67 × 10 kg) 15  58 4.14 × 10 eV ⋅ s   or 2.1 x102 N/m to the given two significant figures. 21. The bond between the hydrogen and chlorine atoms in a 1H35Cl molecule has a force constant of 516 N/m. Is it likely that an HCl molecule will be vibrating in its first excited vibrational state at room temperature? Atomic masses are given in the Appendix. 【sol】 Using

∆E = hνo = h

k m′

and

m ′ = mH

35 , 36

36 − 20 = 5 . 94 × 10 J = 0.371 eV − 27 35 1.67 × 10 kg At room temperature of about 300 K, ∆ E = (1.055 × 10−34 J ⋅ s)

516 N/m

k T = (8.617 x 10-5 eV/K) (300 K) = 0.026 eV. An individual atom is not likely to he vibrating in its first ex cited level, but in a large collection of atoms, it is likely that some of these atoms will be in the first excited state. It's important to note that in the above calculations, the symbol "k" has been used for both a spring constant and Boltzmann's constant, quantities that are not interchangeable.

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Chapter 9 Problem Solutions 1. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in the n=2 energy level? 【sol】 g(ε 2 ) = 8, g (ε1 ) = 2 Then,

n(ε 2 ) 1 = = 4 e −(ε2 −ε1)/kT = 4 e 3ε1 /kT n(ε1 ) 1000 (3/ 4)(13. 6 eV )  1  (3/ 4)(−ε 1 ) T =  = = 1.43 × 10 4 K − 5  k  ln 4000 (8.62 × 10 eV/K )(ln 4000 )

where 3.

ε 2 = ε1 / 4, and ε1 = − 13.6 eV

The 3 2Pl/2 first excited sate in sodium is 2.093 eV above the 3 2S1/2 ground state. Find the ratio between the numbers of atoms in each state in sodium vapor at l200 K. (see Example 7.6.) 【sol】 multiplicity of P-level : 2L+1=3, multiplicity of S-level : 1 The ratio of the numbers of atoms in the states is then,   2.09 eV  3   = 4.86 × 10 − 9 exp −    (8.62 × 10 − 5 eV/K )(1200 K )   1  

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5. The moment of inertia of the H2 molecule is 4.64×10-48 kg·m2. (a) Find the relative populations of the J=0,1,2,3, and 4 rotational states at 300 K. (b) can the populations of the J=2 and J=3 states ever be equal? If so, at what temperature does this occur? 【sol】 g (J ) = 2J + 1, (a)

εJ

J (J + 1)h 2 = 2I

ε J =0 = 0

2   J (J + 1)h 2   N (J )  = ( 2J + 1) exp − h = (2J + 1) exp −  2IkT N (J = 0 ) 2IkT    

    

J ( J +1)

   (1.06 × 10 −34 J ⋅ s)2  = (2J + 1) exp − − 48 2 − 23  2 ( 4 . 64 × 10 kg ⋅ m )( 1 . 38 × 10 J/K )( 300 K )    

J ( J +1)

= (2J + 1)[0. 749]J ( J +1)

Applying this expression to J=0, 1, 2, 3, and 4 gives, respectively, 1 exactly, 1.68, 0.880, 0.217, and 0.0275. (b) Introduce the dimensionless parameter . Then, for the populations of the J=2 and J=3 states to be equal, 5x 6 = 7x 12 ,

x6 =

5 7

and

6 ln x = ln

Using , ln x = − h 2 / 2IkT and ln ( 5/7 ) = - ln ( 7/5)

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5 7

and solving for T,

Department of Physics

6h2 T = 2Ik ln(7 / 5) 6(1.05 × 10 − 34 J ⋅ s)2 = = 1.55 × 10 3 K − 48 2 − 23 2( 4.64 × 10 kg ⋅ m )(1.38 × 10 J/K) ln(1.4) 7. Find v and vrms for an assembly of two molecules, one with a speed of 1.00 m/s and the other with a speed of 3.00 m/s. 【sol】

v = 12 (1.00 + 3.00) = 2.00 (m/s) vrms =

1 [1.00 2 2

+ 3.00 2 ] = 2 .24 (m/s)

9. At what temperature will the average molecular kinetic energy in gaseous hydrogen equal the binding energy of a hydrogen atom? 【sol】 For a monatomic hydrogen, the kinetic energy is all translational and KE = 32 kT solving for T with KE = − E1

T =

2  E1  (2 / 3)(13.6 eV) 5 = 1 . 05 × 10 K −  = 3  k  (8.62 × 10− 5 eV/K)

Inha University

Department of Physics

11. Find the width due to the Doppler effect of the 656.3-nm spectral line emitted by a gas of atomic hydrogen at 500 K. 【sol】 For nonrelativistic atoms, the shift in wavelength will be between +λ(v/c) and -λ(v/c) and the width of the Doppler-broadened line will be 2λ(v/c). Using the rms speed from KE=(3/2)kT = (1/2)mv2, v = 3kT /m , and ∆λ = 2λ

3kT /m c

3(1.38 × 10−23 J/K)(500 K)/(1.67 × 10−27 kg) = 2(656.3 × 10 m) 3.0 × 108 m/s −9

= 1.54 × 10−11 m = 15.4 pm

13. Verify that the average value of 1/v for an ideal-gas molecule is 2m /πkT . 2 ∞ [Note : ∫0 ve −av dv = 1/( 2a )] 【sol】 1 1 ∞1 = n (v )dv The average value of 1/v is v N ∫0 v 1  m  = 4πN   N  2πkT  m  = 4π    2πkT 

Inha University

3/ 2

3/ 2

∞ − mv 2 / 2kT ve dv 0

∫

 kT  =   m 

2m 1 =2 πkT

Department of Physics

17. How many independent standing waves with wavelengths between 95 and 10.5 mm can occur in a cubical cavity 1 m on a side? How many with wavelengths between 99.5 and 100.5 mm? (Hint: First show that g(λ)dλ = 8πL3 dλ/λ4.) 【sol】 The number of standing waves in the cavity is 8πL3ν 2 g(ν )dν = dν c3 2 g ( λ)dλ = g (ν)dν =

8πL3  c  c 8πL3 d λ = dλ   c 3  λ  λ2 λ4

Therefore the number of standing waves between 9.5mm and 10.5mm is 8π (1 m)3 g( λ )dλ = (1.0 mm ) = 2.5 × 10 6 4 (10 mm ) Similarly, the number of waves between99.5mm and 100.5mm is 2.5x102, lower by a factor of 104. l9. A thermograph measures the rate at which each small portion of a persons skin emits infrared radiation. To verify that a small difference in skin temperature means a significant difference in radiation rate, find the percentage difference between the total radiation from skin at 34o and at 35oC.

Inha University

Department of Physics

【sol】 By the Stefan-Boltzmann law, the total energy density is proportional to the fourth power of the absolute temperature of the cavity walls, as R = σT 4 The percentage difference is σT14 − σT 24 σT14

=

T14 − T 24 T14

4

T   307 K  = 1 −  2  = 1 −   = 0.013 = 1. 3% T 308 K    1

For temperature variations this small, the fractional variation may be approximated by ∆ R ∆(T 4 ) 3T 3∆T ∆T 1K = = = 3 = 3 = 0.013 R T 308 K T4 T4 21. At what rate would solar energy arrive at the earth if the solar surface had a temperature 10 percent lower than it is? 【sol】 Lowering the Kelvin temperature by a given fraction will lower the radiation by a factor equal to the fourth power of the ratio of the temperatures. Using 1.4 kW/m2 as the rate at which the sun’s energy arrives at the surface of the earth

(1.4 kW/m 2 )(0.90)4 = 0.92 kW/m 2 (= 66%)

Inha University

Department of Physics

23. An object is at a temperature of 400oC. At what temperature would it radiate energy twice as fast? 【sol】 To radiate at twice the radiate, the fourth power of the Kelvin temperature would need to double. Thus, 2[( 400 + 273) K]4 = T 4 T = 673 × 21/ 4 K = 800 K(527 o C) 25. At what rate does radiation escape from a hole l0 cm2 in area in the wall of a furnace whose interior is at 700oC? 【sol】 The power radiated per unit area with unit emissivity in the wall is P=σT4. Then the power radiated for the hole in the wall is P ' = σT 4 A = (5.67 × 10−8 W/(m 2 ⋅ K 4 ))( 973 K )4 (10 × 10− 4 m 2 ) = 51 W 27. Find the surface area of a blackbody that radiates 100 kW when its temperature is 500oC. If the blackbody is a sphere, what is its radius? 【sol】 The radiated power of the blackbody (assuming unit emissivity) is

P = Ae σT

4

P 100 × 103 W A= = eσT 4 (1)(5.67 × 10−8 W/(m 2 ⋅ K 4 ))((500 + 273)K )4 = 4. 94 × 10− 2 m 2 = 494 cm2

Inha University

Department of Physics

The radius of a sphere with this surface area is, then, A = 4πr 2

r =

A / 4π = 6.27 cm

31. The brightest part of the spectrum of the star Sirius is located at a wavelength of about 290 nm. What is the surface temperature of Sirius? 【sol】 From the Wien’s displacement law, the surface temperature of Sirius is 2.898 × 10− 3 m ⋅ K 2.898 × 10− 3 m ⋅ K T = = = 1.0 × 104 K − 9 λmax 290 × 10 m

33. A gas cloud in our galaxy emits radiation at a rate of 1.0x1027 W. The radiation has its maximum intensity at a wavelength of 10 µm. If the cloud is spherical and radiates like a blackbody, find its surface temperature and its diameter. 【sol】 From the Wien’s displacement law, the surface temperature of cloud is 2.898 × 10 − 3 m ⋅ K 2 o T = = 2 . 9 × 10 K = 290 K = 17 C -6 10 × 10 m P P Assuming unit emissivity, the radiation rate is R = σT 4 = = A πD 2 where D is the cloud’s diameter. Solving for D, D=

Inha University

1 /2

P πσT

4

  1.0 × 10 27 W  =  -8 2 4 4 π (5.67 × 10 W/m ⋅ K )( 290 K )  

= 8.9 × 1011 m

Department of Physics

35. Find the specific heat at constant volume of 1.00 cm3 of radiation in thermal equilibrium at 1000 K. 【sol】 The total energy(U) is related to the energy density by U=Vu, where V is the volume. In terms of temperature,

U = Vu = VaT 4 = V

4σ T 4 c

The specific heat at constant volume is then ∂U 16σ 3 cV = = T V ∂T c 16( 5.67 × 10 − 8 W/m 2 ⋅ K 4 ) 3 −6 3 = ( 1000 K ) ( 1 . 0 × 10 m ) 2 .998 × 108 m/s = 3 .03 × 10 −12 J/K

37. Show that the median energy in a free-electron gas at T=0 is equal to ε F/22/3 =0.630εF. 【sol】 At T=0, all states with energy less than the Fermi energy εF are occupied, and all states with energy above the Fermi energy are empty. For 0≤ε≤εF, the electron energy distribution is proportional to ε . The median energy is that energy for which there are many occupied states below the median as there are above. The median energy εM is then the energy such that εM

∫0

Inha University

ε dε =

1 εF 2 0

∫

ε dε

Department of Physics

Evaluating the integrals, 2 (ε )3/ 2 3 M

= 13 (ε F )3/ 2 ,

or ε M = ( 12 )3/ 2 ε F = 0.63 ε F

39. The Fermi energy in silver is 5.51 eV. (a)What is the average energy of the free electrons in silver at O K? (b)What temperature is necessary for the average molecular energy in an ideal gas to have this value? (c)What is the speed of an electron with this energy? 【sol】 (a) The average energy at T=0 K is ε 0 = 35 ε F = 3. 31 eV (b) Setting (3/2)kT=(3/5)εF and solving for T, 2 εF 2 5.51 eV T = = = 2.56 × 104 K 5 5 k 5 8.62 × 10 eV/K (c) The speed in terms of the kinetic energy is v =

2KE = m

6ε F = 5m

6(5.51 eV )(1.602 × 10−19 J/eV ) 5(9.11 × 10

− 31

kg)

= 1.08 × 10 6 m/s

43. Show that, if the average occupancy of a state of energy εF+∆ε is fl at any temperature, then the average occupancy of a state of energy εF-∆ε is f2=1-f1. (This is the reason for the symmetry of the curves in Fig.9.10 about εF.)

Inha University

Department of Physics

【sol】 Using the Fermi-Dirac distribution function

f 1 = f FD (ε F + ∆ε ) =

1

f 2 = f FD (ε F − ∆ε ) =

1

e ∆ε /kT + 1 e − ∆ε /kT + 1 1 1 1 e ∆ε /kT f1 + f 2 = ∆ε /kT + − ∆ε /kT = ∆ε /kT + ∆ε /kT =1 e +1 e +1 e +1 e +1

45. The density of zinc is 7.l3 g/cm3 and its atomic mass is 65.4 u. The electronic structure of zinc is given in Table 7.4, and the effective mass of an electron in zinc is 0.85 me. Calculate the Fermi energy in zinc. 【sol】 Zinc in its ground state has two electrons in 4 s subshell and completely filled K, L, and M shells. Thus, there are two free electrons per atom. The number of atoms per unit volume is the ratio of the mass density ρZn to the mass per atom mZn . Then,

εF

h2 = 2m *

 3( 2) ρ Zn   8π mZn

  

2 /3

 (6.626 × 10 −34 J ⋅ s)2  3(2)(7.13 × 103 kg/m 3 )    =  - 31 - 27     2(0.85)(9. 11 × 10 kg)  8π (65.4u)(1.66 × 10 kg/u ) 

2 /3

= 1.78 × 10−18 J = 11 eV

Inha University

Department of Physics

47. Find the number of electron states per electronvolt at ε=εF/2 in a 1.00-g sample of copper at O K. Are we justified in considering the electron energy distribution as continuous in a metal? 【sol】 3N (ε F )− 3 /2 ε At T=0, the electron distribution n(ε) is n (ε ) = 2 3 N At ε=εF/2, n ε2F = 8 εF

( )

The number of atoms is the mass divided by the mass per atom, (1.00 × 10− 3 kg) 21 N = = 9 . 48 × 10 ( 63.55 u)(1.66 × 10− 27 kg/u ) with the atomic mass of copper from the front endpapers and εF =7.04 eV. The number of states per electronvolt is 3 9.48 × 1021  εF  n  = = 1.43 × 1021 states/eV 8 7.04 eV  2  and the distribution may certainly be considered to be continuous.

Inha University

Department of Physics

49. The Bose-Einstein and Fermi-Dirac distribution functions both reduce to the MaxwellBoltzmann function when eαeε/kT>>1. For energies in the neighborhood of kT, this approximation holds if eα>>1. Helium atoms have spin 0 and so obey Bose-Einstein statistics verify that f(ε)=1/eαeε/kT≈Ae-ε/kT is valid for He at STP (20oC and atmospheric pressure, when the volume of 1 kmol of any gas is=22.4 m3) by showing that of A1 in copper if f(ε)≈Aexp(ε/kT). As calculated in Sec. 9.9 N/V=8.48x1028 electrons/m3 for copper. Note that Eq.(9.55) must be used unchanged here. 【sol】 Here, the original factor of 8 must be retained, with the result that 1N 3 h ( 2πm ekT ) − 3 / 2 2V = 12 (8.48 × 1026 m −3 )( 6.63 ×10 −34 J ⋅ s ) 3 × [ 2π(9.11 ×10 −31 )(1.38 ×10 −23 J/K )( 293 K)]− 3 / 2

A=

= 3.50 ×103 ,

Which is much greater than one, and so the Fermi-Dirac distribution cannot be approximated by a Maxwell-Boltzmann distribution.

Inha University

Department of Physics

Physics 107

Problem 1.1

If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than they are now? All of the phenomena of special relativity depend upon the factor G(v):

G( v ) ≡ 1 −

v c

2

The square box at the end of this equation indicates it is just text; i.e., not "live."

2

If c were smaller, then G(v) would differ from unity at much lower velocities v thus making relativistic effects more conspicuous than they are now. Let's plot G(v) as a function of v/c to see the effect. This column is for c=3x108.

This column is for c=3x107.

Set the range of v:

Set the range of v:

8

7

c := 3⋅ 10

cc := 3⋅ 10

v := 0⋅ c , 0.01⋅ c .. c G( v ) := 1 −

v c

2

GG( v ) := 1 −

2

v

2

cc

Plot:

2

Plot: 1

1

G( v) 0.5

GG( v) 0.5

0 0

1 .10

8

2 .10 v

8

3 .10

8

Notice how G(v) falls rather slowly.

0 0

1 .10

8

2 .10 v

8

3 .10

8

Notice the rapid dropoff in GG(v)

When G(v) is "near" 1, relativistic effects are not obvious. When G(v) deviates significantly from 1, relativistic effects are more obvious. ========================================================================================

Physics 107

Problem 1.2

It is possible for the electron beam in a television picture tube to move across the screen at a speed faster than the speed of light. Why does this not contradict special relativity? The image of a moving object created by the successive impacts of the electrons on the screen is not an object. The information carried by the electrons moves from the electron gun to the screen at a speed below that of light. Adjacent dots on the screen may light up as if the dot had moved from one position to the next at a speed greater than the speed of light, but in fact, no dot moved. Each lit dot is produced by a different electron. ======================================================================================== 1

Physics 107

Problem 1.3

Is it a good idea for an athlete trying to set a world record 100-m dash tie to have his time taken by an observer on a moving spacecraft? If the observer in the spacecraft times the run by watching a clock on earth, nothing is gained because the clock and athlete are in the same reference frame (the athlete's speed is so small compared to c that we can ignore his motion relative to the clock). If the observer in the spacecraft times the run by watching a clock in the spacecraft, it appears as if the clock on the earth ran slow, so that, in fact, MORE time elapsed during the run. The spacecraft observer would actually measure a longer time. This answer ignores the complication of length contraction, which we will get into later in this chapter. Attempting to account for length contraction would introduce the problem of simultaneity (see the pole-barn paradox in the class notes). The time measured by the spacecraft observer would not be useable by an earth observer. ========================================================================================

Physics 107

Problem 1.4

An observer on a spacecraft moving at 0.7c relative to the earth finds that a car takes 40 minutes to make a trip. How long does the trip take to the driver of the car? Solution: we apply the time dilation equation. An observer on the spacecraft measures a dilated time t = 40 min (why is it the dilated time?). We need to calculate the proper time, as measured by the driver of the car. t0

t :=

1 − 

v

2

 c

t0 := t⋅ 1 − 

v

2

 c

Assign values and solve: c := 1

v := 0.7⋅ c

t0 := t⋅ 1 − 

v

t := 40

2

 c

t0 = 28.566 Something to think about: what time would an observer who remained stationary on earth measure? ========================================================================================

2

Physics 107

Problem 1.5

Two observes, A on earth and B in a spacecraft whose speed is 2x108 m/s, both set their watches to the same time when the ship is abreast of the earth. (a) How much time must elapse by A's reckoning before the watches differ by 1 s? (b) To A, B's watch seems to run slow. To bB, does A's watch seem to run fast, slow, or keep the same time as his own watch? Solution (a): by A's reckoning, B's watch runs slow. Suppose A has two identical watches. Watch 1 is used as the timer and watch 2 is used to provide a time interval. If both watches are in A's frame of reference, watch 1 will measure the proper time t 0 for the time interval. If A gives watch 2 to B, and B moves relative to A, A can use watch 1 to measure the time t it takes for watch 2 to tick off the time interval. Since the moving watch 2 ticks more slowly (according to A, who is doing the measuring), A's watch 1 must record a longer time for watch 2 to indicate the same time interval that it indicated when it was in A's reference frame.Thus, when the watches differ by 1 s, A's watch 1, which is being used to measure the time t, has ticked 1 more second. Thus, t=t 0+1; t is always greater than t 0. The two equations we need to solve are the time dilation equation and the relationship t=t 0+1. t0

t :=

1 − 

v

2

 c

and

t := t0 + 1

Solve the above two equations for t (Let Γ be the square root): t :=

t−1

t⋅ Γ := t − 1

Γ

Result:

t :=

t⋅ Γ − t := −1

1 1−Γ

Assign values to parameters: c := 300000000

Or I could say c=1 and v=2c/3.

v := 200000000 v Γ := 1 −   c Calculated t:

t :=

2

1 1−Γ

t = 3.927 seconds Solution (b): According to B, A is in motion relative to B. Moving watches (relative to the observer) always run slow. Therefore, B claims A's watch runs slow. ========================================================================================

3

Physics 107

Problem 1.6

An airplane is flying at 300 m/s (672 mi/h). How much time must elapse before a clock in the airplane and one on the ground differ by 1 s? Solution: this is just problem 1.5 with an airplane replacing a rocket ship. Just plug in the new numbers. As in problem 1.5, t :=

1

1−Γ Assign values to parameters: c := 300000000 v := 300 Γ := 1 − 

v

2

 c

Calculate t:

t :=

1 1−Γ 12

t = 1.9998 × 10

seconds

You can use c=2.998x108 if you want and get a slightly different answer. ========================================================================================

Physics 107

Problem 1.7

How fast must a spacecraft travel relative to the earth for each day on the spacecraft to correspond to 2 d on the earth? Solution: an observer on the spacecraft measures a proper time t 0=1 d, and a dilated time t=2 d for the same event as it takes place on the "moving" earth. We solve the time dilation equation for v. t :=

t0 1 − 

v

2

 c

2 v  t0  1 −   :=   c  t 

2

 t0    t

2

v := c⋅ 1 − 

Define the values and plug them in: c := 1

t0 := 1

t := 2

 t0  v := c⋅ 1 −    t

2

v = 0.866

Or v=0.866 c, because I let c=1.

========================================================================================

4

Physics 107

Problem 1.8

The Apollo 11 spacecraft that landed on the moon in 1969 traveled there at a speed relative to the earth of 1.08x104 m/s. To an observer on the earth, how much longer than his own day was a day on the spacecraft? This is another time dilation problem. We are given a relative velocity. We are given a time interval t 0 of one day on the spacecraft moving relative to an observer. We want to find the dilated time t measured by the observer. Let's define our variables first, so I don't have to make the time dilation equation into text and then later on re-enter it as an equation. We don't need to worry about units 8 4 c := 3⋅ 10 v := 1.08⋅ 10 here because velocities have same units. t0 := 1 The time here is in days, so my answer t will be in days.

t :=

t0 1−

v c

2

This is the equation for t. Below I will type "t=" to see the answer.

2

There are not enough digits to show any effect. Below I will type "t=" again and then type "f" to allow me to show more digits.

t=1

t = 1.000000000648

Click on the number and select Math, Numerical Format to see that I picked a precision of 12 to display t.

The problem asks how much longer t is than t, so I'd better calculate t-t 0. − 10

t − t0 = 6.48 × 10

days −5

or ( t − t0) ⋅ 24⋅ 3600 = 5.599 × 10

seconds.

========================================================================================

Physics 107

Problem 1.9

A certain particle has a lifetime of 1x10-7 s when measured at rest. How far does it go before decaying if its speed is 0.99c when it is created? Solution: our first job is to figure out that the problem is really asking us to calculate how far a human observer would observe this particle to travel. The observer sees the particle moving at a speed of 0.99c, and sees the particles "clock" dilated according to equation 1.3, where t 0 is the time the observer sees the particle in motion. We need to calculate d=vt, where v=0.99c and t is given by eq. 1.3. I'm going to include units in this solution. You could append the file "units.mcd" to for use with this problem. Instead, I will do the units here for you to see. Define units:

m ≡ 1L

Define parameters:

kg ≡ 1M −7

8 m

c := 3⋅ 10 ⋅

Pertinent equations: t :=

s ≡ 1T

t0 := 10

s

⋅s

v := 0.99⋅ c

t0 v 1 −   c

2

d := v ⋅ t

d = 210.5 m

========================================================================================

5

Physics 107

Problem 1.17

An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a spacecraft moving at 0.9c relative to the earth. What is his height as measured by an observer in the same spacecraft? By an observer on earth? An observer in the same spacecraft, at rest relative to the astronaut, measures the proper length, 6 ft, of the astronaut. If the astronaut were lying perpendicular to the velocity vector of the spacecraft, the observer on earth would also measure his proper length. But with the astronaut parallel to the direction of relative motion, the observer on earth measures a contracted length. Define parameters, then do calculation. L0 := 6

c := 1

v L := L0⋅ 1 −   c

v := 0.9⋅ c

2

L = 2.615 feet

========================================================================================

Physics 107

Problem 1.18

An astronaut is standing in a spacecraft parallel to its direction of motion. An observer on the earth finds that the spacecraft speed is 0.6c and the astronaut is 1.3 m tall. What is the astronaut's height as measured in the spacecraft? Solution: this is just problem 1.17 but solving for L0 instead of L. Define parameters: L := 1.3

c := 1

v := 0.6⋅ c L

L0 :=

v

2

L0 = 1.625

meters

1−

 c

========================================================================================

Physics 107

Problem 1.19

How much time does a meter stick moving at 0.1c relative to an observer take to pass the observer? The meter stick is parallel to its direction of motion. The relative speed is v:

8

c := 3⋅ 10 m/s

v := 0.1⋅ c

L0 := 1

We need to solve for the contracted length L, and then determine how long it takes to travel this length at a speed of 0.1*c. L := L0⋅ 1 − 

v

 c

2

L = 0.995

The equation d=vt still works in relativity, so we can solve it for t: t :=

L v

−8

t = 3.317 × 10

seconds

======================================================================================== 6

Physics 107

Problem 1.20

A meter stick moving with respect to an observer appears only 500 mm long to her. What is its relative speed? How long does it take to pass her? The meter stick is parallel to its direction of motion. L0 := 1

8

L := 0.5 L := L0⋅ 1 − 

2

 L  := 1 −  v  c L     0 8

v

2

m/s

2

 c

First we need to solve

v = 2.6 × 10

c := 3⋅ 10

for relative speed v.

 v  := 1 −  L  c L     0

2

v := c⋅ 1 − 

  L 0   L

2

m/s

Time to pass observer: (I call it t 0 because the event is timed in the observer's reference frame.

t0 :=

L v

−9

t0 = 1.92 × 10

s

========================================================================================

Physics 107

Problem 1.21

A spacecraft antenna is at an angle of 10 degrees relative to the axis of the spacecraft. If the spacecraft moves away from the earth at a speed of 0.7c, what is the angle as seen from the earth? This problem is not assigned or "testable" and the solution is included only for "interest." This problem is tricky because when the spacecraft moves, the projection of the antenna along the spacecraft contracts, which means that the apparent length of the antenna also contracts. You have to be careful to take both contractions into account. It also helps to have a drawing. Let L0 be the length of the antenna when the spacecraft is at rest, and x0 be the projection of the antenna parallel to the direction of spacecraft travel when the spacecraft is at rest. Then when the spacecraft is at rest, the angle of the antenna is found from

( )

x0 := L0⋅ cos θ 0

When the spacecraft is in motion, both L0 and x0 appear contracted to an observer on earth, and the angle can be found from x := L⋅ cos ( θ ) except that you can't apply the length contraction formula to L0 in order to calculate L, because the antenna has a component of length perpendicular to the direction of motion which has not been contracted. Let's let y be the projection of the antenna length perpendicular to the motion. When the spacecraft is in motion, the antenna has a length L, a projection x=L cos(θ) along the direction of motion, and a projection y=L sin(θ) perpendicular to the direction of motion. The angle θ is given by tan( θ ) :=

y

:=

L⋅ sin ( θ )

L⋅ cos ( θ ) That last equation looks "circular" (we already know tan=sin/cos) except that we can use our original angle to calculate the numerator, and we can apply the length contraction equation to the denominator, because it represents the component along the direction of relative motion. x

7

( )

L0⋅ sin θ 0

tan( θ ) :=

(L0⋅ cos(θ 0))⋅

v

1−

c tan( θ ) :=

2

v c

(L0⋅ cos( 10) )⋅

2

1−

v c

θ := atan

2

2

  2  1− v   2  c  

tan( 10) 1−

L0⋅ sin ( 10)

tan( θ ) :=

2

2

tan( 10)

To solve, plug in values, and remember to convert angles to radians when you are using Mathcad. c := 1

v := 0.7⋅ c

 tan( θ 0)   2  v   1− 2  c  

θ := atan

θ := θ ⋅

360 2⋅ π

θ 0 := 10⋅

2⋅ π 360

θ = 0.242 radians

θ = 13.869 degrees

========================================================================================

Physics 107

Problem 1.27

Dynamite liberates about 5.4x106 J/kg when it explodes. What fraction of its total energy content is this? The total energy content of this kg of dynamite is mc 2. The fraction is simply 6

f :=

5.4⋅ 10 c

8

We need to use c := 3⋅ 10 (SI units).

2

Here's how the units work out: if you choose the mass to be 1 kg, the units on top are joules, and the m=1kg times c2 also gives joules in the denominator, so the result is a pure number, or fraction. 6

Then

f :=

5.4⋅ 10 c

2

− 11

f = 6 × 10

========================================================================================

Physics 107

Problem 1.28

A certain quantity of ice at 0 degrees C melts into water at 0 degrees C and in so doing gains 1 kg of mass. What was its initial mass? It takes 80 calories to melt a gram of ice, and a calorie is equivalent to 4.19 joules, so in SI units, it takes L=335.2 joules to melt a gram of ice, or 3.35x105 joules to melt a kilogram of ice. Let M be the initial mass of the ice, and let L be the latent heat of fusion, as above. The energy added to melt the ice is ML. The mass equivalent of this energy is given in the statement of the problem as m=1 kg. 8

2

5

m⋅ c := M ⋅ L

L := 3.352⋅ 10

M ( m) :=

(m⋅c2)

11

M ( 1) = 2.685 × 10

L

kg

Using the density of ice, you can calculate that this mass would require a block of ice 16.5 km long, 16.5 km wide, and 1 km high. (Or about 10 miles by 10 miles by 0.6 miles high, if I remember my conversion factors correctly). Yes, ice really does gain mass when it melts. Or rather, the energy that "goes into" the ice is manifested as mass. Similarly, water would "lose mass" when it freezes. No, that's not the main reason why ice floats on water. ========================================================================================

Physics 107

Problem 1.29

At what speed does the kinetic energy of a particle equal its rest energy? rest_energy := m⋅ c

2

2

KE := γ ⋅ m⋅ c − m⋅ c

2

When KE=rest_energy, 2

2

γ ⋅ m⋅ c − m⋅ c := m⋅ c m 1−

2

2

γ ⋅ m⋅ c := 2⋅ m⋅ c

2

:= 2⋅ m v c

2

Cancel the m's and re-arrange:

γ ⋅ m := 2⋅ m

I cancelled out the c2's.

v

1 := 2⋅ 1 −

c

2

0.25 := 1 −

divide by 2 and square both sides

v c

2

2

2

v := 0.75⋅ c

2

2

2

v := 0.866⋅ c

========================================================================================

Physics 107

Problem 1.30

How many joules of energy per kilogram of rest mass are needed to bring a spacecraft from rest to a speed of 0.9c? Again, we use

2

2

γ ⋅ m⋅ c := m⋅ c + K

At a speed of 0.9c,

m :=

m 1−

where v/c=0.9. v c

Solving for K gives

K :=

m⋅ c 1−

8

c := 3⋅ 10

m := 1

v := 0.9⋅ c

2

2 2

− m⋅ c v c

2

17

K = 1.165 × 10

joules per kilogram.

2

2

======================================================================================== 9

========================================================================================

Physics 107

Problem 1.31

An electron has a kinetic energy of 0.1 MeV. Find its speed according to classical and relativistic mechanics. Let's begin by converting MeV to mks units. − 19

eV := 1.602⋅ 10

6

MeV := eV⋅ 10

This converts eV to Joules

This converts eV to MeV

Classical calculation. 2

Kcl := m⋅

We use our familiar equations of classical mechanics. − 31

m := 9.109⋅ 10 v cl := 2⋅

v cl 2

Kcl := 0.1⋅ MeV

kg

Kcl

8

v cl = 1.875 × 10

m

m/s 8

Krel := 0.1⋅ MeV

Relativistic calculation.

c := 3⋅ 10

If you start with γmc2=mc2+K and solve for v/c, you get an "unofficial" but extremely useful equation that is sure easier to use than solving "by hand" every time. Make sure this equation is on your 3x5 card.

1

v rel := c⋅ 1 −

8

Krel   1 + 2 m⋅ c  

v rel = 1.644 × 10

2

Let's put the two answers side-by-side for comparison:

m/s

8

8

v cl = 1.875 × 10

v rel = 1.644 × 10

Classically-calculated speeds are always too large, although the error is not significant for low speeds and low energies. ========================================================================================

Physics 107

Problem 1.33

A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c. Be lazy and use our "unofficial" equation (it's unofficial because it is derived, not fundamental).

(

1

v ( K , m) := c⋅ 1 −

1 +   2 m⋅ c   K

Expressed "in terms of c:"

(

2

) = 0.999

v 20⋅ m⋅ c , m c

2

)

8

v 20⋅ m⋅ c , m = 2.997 × 10

2

or v=0.999c

======================================================================================== 10

Physics 107

Problem 1.34

The speed of a proton is increased from 0.2c to 0.4c. (a) By what factor does its kinetic energy increase? (b) The speed of the proton is again doubled, this time to 0.8c. By what factor does its kinetic energy increase now? Here's another handy "unofficial" variant of our KE equation: m⋅ c

2

1−

v

K( m, v ) :=

− m⋅ c

c

K( m, v ) := m⋅ c ⋅ 

2

1

2

2

2  1− v  2 c 

2

− 1 

  

We could do this algebraically, without using the proton mass, but here it is for those who like numbers: − 27

mproton := 1.67⋅ 10 K( mproton , 0.4⋅ c) K( mproton , 0.2⋅ c)

K( mproton , 0.8⋅ c)

= 4.417

K( mproton , 0.4⋅ c)

= 7.319

Doubling the speed does not increase KE by the classically-expected factor of 4, and the discrepancy is larger as the speeds get greater. Mathcad made this solution very easy. If you're doing this "by hand," it will take several lines of algebra and computations. ========================================================================================

Physics 107

Problem 1.35

How much work (in MeV) must be done to increase the speed of an electron from 1.2x108 m/s to 2.4x108 m/s? Solution: a body in motion has a total energy E := γ ⋅ m⋅ c

2

E :=

m⋅ c

2

1−

v c

2

2

The work done on the electron is just the difference in the energies E at the two different velocities: W := E2 − E1

Remember Ef-Ei=[W other]i-->f from Phys. 23?

To get the work in units of MeV, I could calculate the numbers out and convert joules to eV to MeV. Or I could be clever and use the electron mass in energy units.

MeV := 1 me :=

0.511⋅ MeV c

W := me⋅ c ⋅ 

1

2

(

)2

 8  1 − 2.4⋅ 10 2  ( c) 

−

 2 8  1.2⋅ 10  1− 2  ( c) 

2

1

(

)

W := 0.294⋅ MeV

Setting MeV=1 above let me stick in my expression for me so that it looked like a unit. ======================================================================================== 11

========================================================================================

Physics 107

Problem 1.41

In its own frame of reference, a proton takes 5 min to cross the Milky Way galaxy, which is about 105 light-years across. (a) What is the approximate energy of the proton in eV? (b) About how long would the proton take to cross the galaxy as measured by an observer in the galaxy's frame of reference? Part (a). The only way for the proton to "think" it crosses the galaxy in 5 minutes is for the proton to see the galaxy's length contracted: L := L0⋅ 1 − 

v

c

2

 

This equation lets us solve for v, and we could plug v into E=γmc2.

Or we could be clever and note that L0 L

(

E

:= γ

mc − 27

E 1.67⋅ 10

2

so that

:= γ

E m⋅ c

)

5

, 5⋅ 60 , 10 ⋅ 365⋅ 24⋅ 3600 = 1.58

2

:=

L0 L

 L0⋅ m⋅ c2    E( m, L, L0) := L

E=1.58 joules

To get the energy in eV, use the conversion factor 1 eV=1.6x10-19 joules:

1.58 − 19

18

= 9.875 × 10

1.6⋅ 10

You could be still more clever and express the proton mass in energy units (938 MeV/c2) and save multiplying by c. Part (b). The proton's energy is about 1019 eV, or 1013 MeV. That's about 1010 times its rest energy of roughly 1000 MeV. This is one fast proton. We might as well use v proton=c. I'm guessing our answer might be off in the 10th decimal place or so. Double-check by calculating the proton's speed, if you don't believe me. If the proton is moving with the speed of light, and the galaxy is 105 light years across, the observer will say it takes 105 years (not 5 minutes) for the proton to cross the galaxy. The observer say "proton, your time was way too short." The proton will say "observer, your distance across the galaxy was way too long. ========================================================================================

Physics 107

Problem 1.42

What is the energy of a photon whose momentum is the same as that of a proton whose kinetic energy is 10 MeV? 2

2 2

( 2)2

2

E := p ⋅ c + m⋅ c

Eproton := p ⋅ c +  mproton⋅ c 2 2

2

p ⋅ c := Eproton −  mproton⋅ c

2

2

2 2

2

2



2

both proton and photon have same momentum p

2



Ephoton := Eproton −  mproton⋅ c 2

2 2

Ephoton := p ⋅ c

this is p 2c2 for photon, which is equal to p 2c2 for photon

2



2

The E in here is total energy; you are given proton KE. 12

Kproton := Eproton − mproton⋅ c

2

Eproton := Kproton + mproton⋅ c

Ephoton :=  Kproton + mproton⋅ c 2

2

2



−  mproton⋅ c

2

2

2



here are the specific values for this problem: mproton := 938⋅ 

MeV   2   c 

Kproton := 10⋅ MeV

2

Ephoton :=

 Kproton + mproton⋅ c2  −  mproton⋅ c2     

2

Ephoton := 137⋅ MeV ========================================================================================

Physics 107

Problem 1.43

Find the momentum (in MeV/c) of an electron whose speed is 0.6c. p := γ ⋅ m⋅ v If I use mass in "energy units" of MeV/c2 and v in terms of c, the answer will come out in units of MeV/c, because γ is unitless. Or you can work the problem in SI units and convert to "energy units." melectron := 0.511⋅

MeV c

p electron :=

2

v electron := 0.6⋅ c

1 2   v electron    1− c       −9

p electron = 1.278 × 10

⋅ melectron⋅ v electron

I have c defined above numerically as 3x108. When I combine symbolic and numerical work, like I do here, Mathcad actually divides by a c2 to calculate the mass of the electron. When I multiply by v=0.6c, Mathcad "puts back" one of those "two" c's I divided out. To get the answer in "energy units" I have to put the other factor of c back in "by hand." You won't encounter this confusion on an exam.

This answer has the factor of 1/c numerically embedded in the result. To see the answer expressed in units of MeV/c, I have to multiply through by c. If you do this using pen and paper (see the student solutions manual, available on reserve at the library), you will get the textbook answer "automatically." p electron := 0.383⋅

MeV

See lecture notes lect03.ppt for better "pencil-and-paper" version.

c

========================================================================================

13

Physics 107

Problem 1.44

Find the total energy and kinetic energy (in GeV) and the momentum (in GeV/c) of a proton whose speed is 0.9c. The mass of the proton is 0.938 GeV/c2. Let's do the momentum first, because it is "like" the previous calculation. GeV := 1

This definition just lets me include "GeV" in a problem as if I were writing the units by hand.

mproton := 0.938⋅

GeV c

v proton := 0.9⋅ c

2

1

p proton :=

2   v proton    1− c      

⋅ mproton⋅ v proton

Eproton := p proton ⋅ c +  mproton⋅ c 2

2 2

Eproton = 2.152

GeV

2

p proton := 1.937⋅

2

GeV c

Eproton := p proton ⋅ c +  mproton⋅ c 2 2



2

2



I could have also used E=γmc2, but I already had the Eproton equation available for cut and paste from problem 42. Kproton := Eproton − mproton⋅ c

2

Kproton := 1.214⋅ GeV

========================================================================================

Physics 107

Problem 1.45

Find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV. First I'll do the algebra. 2

( 2)2

2 2

2 2

E := p ⋅ c + m⋅ c

E := K + m⋅ c

2

2

( 2)2

p ⋅ c := E − m⋅ c

2 2

(

p ⋅ c := K + m⋅ c

) − (m⋅c )

2

2

2

K := E − m⋅ c

2

p :=

2

(K + m⋅c2)2 − (m⋅c2)2 c

Now I'll plug in the numbers. keV := 1

melectron := 511⋅

keV c

Kelectron := 511⋅ keV

2 2

 Kelectron + melectron⋅ c2  −  melectron⋅ c2      p electron :=

2

c

p electron := 885⋅

keV c

======================================================================================== 14

========================================================================================

Physics 107

Problem 1.47

Find the speed and momentum (in GeV/c) of a proton whose total energy is 3.5 GeV. First think: this proton's total energy is about 4 times its rest energy of 0.938 GeV, so the speed had better be close to c. Let's do the algebra first, working across the line and then down to save paper. E := γ ⋅ m⋅ c

2

E := γ ⋅ E0

γ :=

E

1

E0 1−

:= v c

1−

v c

2

2

E0

:=

1−

E

v c

 E0  v := c⋅ 1 −    E

2

2

2

 E0    E

2

v

:= 

c

v := c⋅ 1 − 

0.938⋅ GeV 

2

2

2

E E0

2

 E0    E

2



2  E0     E 

v := c ⋅  1 −  2

:= 1 − 

2



2

  3.5⋅ GeV 

v := 0.963⋅ c

To find the momentum, we can use this equation from the previous problem (no sense re-deriving here):

2 2

( 2)2

2

2 2

p ⋅ c := E − m⋅ c

2

2

p :=

E c

p :=

2

( 3.5⋅ GeV) c

2

−

( 0.938⋅ GeV) c

2

p :=

2

(

2

2

2

2

p ⋅ c := E − E0

)

2

c

2 GeV

3.5 − 0.938 ⋅

p :=

2

E c

p := 3.37⋅

c

2

2

E0

−

2

−

E0 c

2

GeV c

========================================================================================

Physics 107

Problem 1.48

Find the total energy of a neutron (m=0.940 GeV/c2) whose momentum is 1.2 GeV/c.

2

2 2

( )

E := p ⋅ c + m⋅ c

2

2

E :=  1.2⋅

2

2



2

E := ( 1.2⋅ Gev) + ( 0.940⋅ GeV)

2

2

E := ( 1.2) + ( 0.940) ⋅ GeV

2

GeV 

2

2 ⋅c +  c 

 0.940⋅ GeV⋅ c2    2 c  

E :=  ( 1.2) + ( 0.940) 2

2

2

2

2

 ⋅ GeV

E := 1.52⋅ GeV

========================================================================================

15

Physics 107

Problem 1.49

A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its mass (in MeV/c2) and speed (as a fraction of c). You have two knowns and two unknowns. None of our equations contain one of the unknowns expressed only in terms of the knowns. Looks like we are going to have to solve a system of equations. Here are two equations that look like candidates to me: E := K + m⋅ c

2

2

2 2

(K + m⋅c2)2 := p2⋅c2 + (m⋅c2)2 2

2

( 2)2

E := p ⋅ c + m⋅ c

Put E from the first equation into the second equation.

Makes me a bit nervous; any time you square an expression, you run the risk of introducing an extraneous root. Let's see where this takes us anyway.

( 2)2 := p2⋅c2 + (m⋅c2)2

2

K + 2⋅ K⋅ m⋅ c + m⋅ c

2

2 2

2 2 2 ( p ⋅c − K ) m :=

2

2⋅ K⋅ m⋅ c := p ⋅ c − K

2⋅ K⋅ c

m :=

(3352 − 622)⋅MeV2 2⋅ 62⋅ MeV⋅ c

m :=

2

2

2 2

K + 2⋅ K⋅ m⋅ c := p ⋅ c

2

 2 MeV2 2 2 2  335 ⋅ ⋅ c − 62 ⋅ MeV   2  c   m := 2⋅ 62⋅ MeV⋅ c

(3352 − 622)⋅MeV 2⋅ 62⋅ c

m := 874⋅

2

2

MeV c

2

Now that I have the mass (a bit less than that of a proton or neutron) I can use our handy unofficial equation introduced in problem 31. There are other ways to get v--see the student solution manual.

v := c⋅ 1 −

1

1 + K   2 m⋅ c  

v := c⋅ 1 −

Those of you who observant will notice I quit making my equations "live." This eliminates the glitch mentioned in problem 43. I'm just using Mathcad as a symbolic text processor now.

2

1

 1 + 62⋅ MeV  MeV 2   874⋅ ⋅c 2   c

2

v := c⋅ 1 −

1

 1 + 62    874  

2

v := 0.358⋅ c

See the student solution manual for a really handy algebra trick that eliminates the potential extraneous root problem. ======================================================================================== 16

Physics 107

Problem 1.50

(a) Find the mass (in GeV/c2) of a particle whose total energy is 4 GeV and whose momentum is 1.45 GeV/c. (b) Find the total energy of this particle in a reference frame in which its momentum is 2 GeV/c. Part (a): 2

( )

2 2

E := p ⋅ c + m⋅ c

2

(m⋅c )

2

2

2

2

2 2

2

:= E − p ⋅ c

2

m :=

E c

4

−

p c

2

2

E

m :=

c

2

4

−

p c

2

2

2 2 GeV

2

m :=

2

4 ⋅ GeV 4

c m := 3.73⋅

1.45 ⋅

c

− c

2

2

2

m :=

2

2

4 ⋅ GeV c

−

4

2

1.45 ⋅ GeV c

2

2 GeV

m := 4 − 1.45 ⋅

4

c

2

GeV c

2

Part (b): mass is relativistically invariant, so we can use the mass from part (a) along with the new momentum.

2

2 2

( 2)2

( 2)2

2 2

E := p ⋅ c + m⋅ c

2 2 GeV

E := p ⋅ c + m⋅ c

E := 2 ⋅

c

2

2

2

2

E := 2 ⋅ GeV + 3.73 ⋅ GeV

E :=

(

2

)

2

2 + 3.73 ⋅ GeV

2

⋅ c +  3.73⋅ 2

 

GeV 2  ⋅c  2 c 

2

E := 4.23⋅ GeV

17

Physics 107 3 .10 c

c f

8

λ E = 2.841 .10 1.6 .10

e

Problem 2.5

19

19

h

6.63 .10

E

h .f

Joules

O. A. Pringle

34

λ

700 .10

9

Note I had to set the zero tolerance here.

eV joules conversion factor

E

E eV

e

E eV = 1.776 eV ========================================================================== Physics 107 h .f

E f

Problem 2.6 6.63 .10

h

34

O. A. Pringle c

3 .10

8

E h 100 .10 eV 6

E eV e

1.6 .10

E

E eV .e

f

19

E h

f = 2.413 .10

Hz

f .λ

λ

22

c

so

c f

c

λ

f

λ = 1.243 .10

14

meters

in nanometers, λ nm

λ .10

9

λ nm = 1.243 .10

5

========================================================================== Physics 107 h

6.63 .10

Problem 2.7

34

photon energy

f E

O. A. Pringle

880 .10

3

h .f

Power is energy per time. Photons per second is energy per time divided by energy per photon. P n

1 .10

3

P E

n = 1.714 .10

30

photons per second

Note that as long as I keep everything in mks units, the units automatically work out OK. ==========================================================================

1

Physics 107 h

6.63 .10

λ

600 .10

E photon N

E

Problem 2.8

34 9

h .c

O. A. Pringle 18

E

10

c

3 .10

8

E photon = 3.315 .10

λ

19

E photon N = 3.017 The eye can detect 3 of these photons. Note again that units work out OK if we stick with the mks system. ========================================================================== Physics 107 Problem 2.9 O. A. Pringle (a) How many photons fall per second on each square meter of the earth's surface directly facing the sun? 34 14 h 6.63 .10 f photon 5 .10 3 on each square meter: P 1.4 .10 as in problem 2.5, number per second is power divided by energy n

n = 4.22 .10

21

photons/s

P h .f photon

(b) What is the power output of the sun, and how many photons per second does it emit? The power per area at a radius of 1.5*10^11 m is given as 1.4x10^3 W/m^2. To answer the first part, just multiply the power per area by the area of a sphere of the given radius. r

1.5 .10

11

P sun

A

4 .π .r

2

P .A

P sun = 3.96 .10 watts P sun N h .f photon 26

N = 1.19 .10

45

photons per second emitted by the sun

(c) How many photons per cubic meter are there near the earth? This is mainly a unit conversion problem. Look at the units. number volume number volume

=

number .

s

s

volume

=

number . s . 1 s m m2

density = ρ

n.

c

3 .10 will need this in a minute 8

number . 1 s

. 1 speed area

1 .1 c 12

ρ = 1.41 .10

13

photons per cubic meter

==========================================================================

2

Physics 107 t

20 .10

h

6.63 .10

3

P

Problem 2.10 λ

0.5

34

632 .10

9

3 .10

8

c

O. A. Pringle

The energy of a pulse is the power times the time. P .t

E pulse

The number of photons in the pulse is the energy of the pulse divided by the energy of a photon. E pulse N h .c λ N = 3.18 .10

16

========================================================================== Physics 107 Problem 2.11 The equation to use is K max h .f h .f 0 15 energies will come out in eV if I use h h 4.14 .10 in these units 8 c 3 .10 λ0

230 .10 λ0

K max f

9

c

f0

Solve for f

O. A. Pringle

1.5 h .f 0

K max

h 15 f = 1.667 .10 here's the frequency; not necessary, just wanted to look at it c λ f 7 λ = 1.8 .10 This is a wavelength of 180 nm. ========================================================================== Physics 107

Problem 2.12

h 4.14 .10 We are given the following:

15

f0

O. A. Pringle

will give K in eV

1.1 .10

15

f

1.5 .10

15

Simply plug these into equation 2.1 K max

h .f

h .f 0

K max = 1.656 eV ==========================================================================

3

Physics 107

Problem 2.13

O. A. Pringle

The equation we will use is h .f

K max

h .f 0

4.14 .10

where h

15

will give us energies in eV

Because E=hf and f=c/λ, longer wavelengths of light have lower energies. This problem is equivalent to asking "what is the minimum frequency of light that will cause photoelectrons to be emitted from sodium." That minimum frequency is just the threshold frequency for sodium, which can be found from the work function. φ 2.3 eV, from table 2.1 This is just the minimun energy needed to produce a photoelectron, so

c

3 .10

8

h .f 0

φ or

φ

f0

f 0 = 5.556 .10

14

Hz h The maximum wavelength is just, using c λ0 f0 λ 0 = 5.4 .10

7

meters, or 540 nm (to get Beiser's answer, use c=2.998*10^8 and h=4.136*10^-15)

What will the maximum kinetic energy of the photoelectrons be if 200-nm light falls on a sodium surface? This is just like problem 2.10, except we are given wavelengths instead of frequencies. λ 0 = 5.4 .10 200 .10

λ

K max

7

calculated above

9

h .c

h .c

λ

λ0

K max = 3.91eV

========================================================================== Physics 107 Problem 2.14 O. A. Pringle This sounds tricky but really isn't. Light incident on the ball will cause photoelectrons to be emitted. The ball will acquire a positive charge and therefore an electrical potential as the electrons are emitted. When work function plus the potential of the ball equals the energy of the incident light, no more electrons will be emitted. For silver

φ

4.7

Planck's constant

eV h

4.14 .10

15

eV*s

frequency of incident light: c f

3 .10

8

λ

200 .10

9

c

15 f = 1.5 .10 Hz λ In words: incident energy=V+φ

V

h .f

φ

V = 1.51 Volts ==========================================================================

4

Physics 107 Problem 2.15 O. A. Pringle The 1.5 mW (milliwatts) gives the energy per unit time in the incident light beam. 3 P 1.5 .10 joules/s 400-nm tells us the photon energy.

h

6.63 .10

c

3 .10

λ

400 .10

34

I'll work in mks units here.

8

E photon

9

h .c

λ E photon = 4.972 .10

19

joules per photon

Dividing the power (energy per time) by the energy of a photon (energy per photon) gives us the number of photons per second incident on the cell. P N E photon 15 N = 3.017 .10 photons per second

Only 0.1 percent of these photons produce photoelectrons, so the number n producing photoelectrons is N n 1000 12 n = 3.017 .10 electrons produced per second This n is actually the current, but we should express it in the more familiar units of coulombs 1.60 .10 n .e

e I

I = 4.827 .10

19

coulombs per electron

7

coulombs per second, or amps

========================================================================== Physics 107 Problem 2.16 O. A. Pringle a) Find the extinction voltage, that is, the retarding voltage at which the photoelectron current disappears. The extinction voltage occurs when the retarding voltage plus the work function equal the photoelectron energy. φ

V ext

h

4.14 .10

φ

2.50

λ

400 .10

V ext

h .f

h .f 15

9

f φ

3 .10

c

8

c λ

V ext = 0.605electron volts b) Find the speed of the fastest photoelectrons. The most energetic electrons will appear when V.ext=0, i.e., K max h .f φ K max = 0.605 electron volts; yes, the same number as in part a Experience should tell us that 0.6 eV is nonrelativistic; we will do a nonrelativistic calculation, and if the speed is too great, go back and do a relativistic calculation. 1. 2 19 K max m electron .v max e 1.6 .10 2

5

K max .e convert to Joules!

K max

9.11 .10

m electron

K max

2.

v max

31

m electron small enough to be nonrelativistic

v max = 4.61 .10

5

========================================================================== Physics 107 Problem 2.17 O. A. Pringle We are given light of frequencies f.1 and f.2, and the maximum kinetic energies K.1 and K.2 which they produce. We want to find the experimental value for h and φ. h .f 1

K1

φ

h .f 2

K2

φ

Remember, the little square means these are symbolic equations only.

It's very easy to solve these simultaneously for h. h .f 1

φ

K1

h .f 2

φ

K2

-------------------h. f 1 h

f2

K1

K2

f1

f2

Now plug in numbers

h

K1

K2 In a couple of lines, I'm going to copy this symbolic equation (F2), paste it below (F4), and turn it "on" with "eq".

K1

1.97

f1

12 .10

K1

K2

f1

f2

K2 14

f2

0.52 8.5 .10

h = 4.143 .10

14

15

This value is in units of eV*s. You can easily convert it to J*s.

Now use the value for h to solve either equation at the start for φ. Use Cut and Paste (F2 and F4) again. h .f 1

K1

φ

Now solve for φ. h .f 1

φ

K1

φ = 3.001 Since h is in eV*s, units of φ are eV ========================================================================== Physics 107 Problem 2.18 To solve this, simply take the equation K max

h .f

O. A. Pringle

φ

and solve it for h. K max φ

1.7

5.4

6

λ

175 .10

c

3 .10

f

h

9

8

c λ K max

φ

f h = 4.142 .10

15

eV*s

The actual value of h in these units is 4.136x10-15. ========================================================================== Physics 107 δf

G .M

f

2 c .R

In mks units,

λ

500 .10

9

Problem 2.51

G

6.67 .10

M

2 .10

R

7 .10

c

3 .10

O. A. Pringle

11

I'm going to work this problem the straightforward brute strength way. Nothing tricky, but it involves conversions back and forth between f and λ. You might save some time by doing the algebra first on paper.

30

8

8

the wavelength of the light f

c

gives the frequency of the light λ G .M .f 9 δf δf = 1.27 .10 2. c R The new frequency is less than the original frequency, so f prime

f

δf

Now calculate the new wavelength: c λ prime f prime

λ prime = 5.0000106 .10

7

The red shift is the amount by which the wavelength changed: δλ

λ prime

δλ = 1.059 .10

λ 12

meters

Or δλ=0.00106 nm. ==========================================================================

7

Physics 107

Problem 2.52

O. A. Pringle

This is just problem 2-51 with different numbers. I will use my 2-51 solution directly here. In mks units, 11 G 6.67 .10

λ

500 .10

9

M

2 .10

R

6.4 .10

c

3 .10

30 6

8

the wavelength of the light c f gives the frequency of the light λ G .M .f

δf

δf = 1.39 .10

11

2.

c R The new frequency is less than the original frequency, so f prime

δf

f

Now calculate the new wavelength: c λ prime f prime

λ prime = 5.0011583 .10

7

The red shift is the amount by which the wavelength changed: δλ

λ prime

δλ = 1.158 .10

λ 10

meters

Or δλ=0.116 nm. ========================================================================== Physics 107

Problem 2.54

O. A. Pringle

Find the Schwarzschild radius of the sun. G RS

6.67 .10

11

M

2 .10

30

c

3 .10

8

2 .G .M c

2

R S = 2.964 .10

3

meters

If the sun had less than this radius, it would be a black hole. ==========================================================================

8

Physics 107

Problem 2.55

O. A. Pringle

The gravitational potential energy U relative to infinity of a body of mass m at a distance R from the center of a body of mass M is U=-GmM/R. (a) If R is the radius of the body of mass M, find the escape speed v.e of the body (presumably Beiser means the body of mass m), which is the minimum speed needed to leave it permanently. To escape, the "body of mass m" must have enough kinetic energy to overcome the gravitational attraction of the body of mass M. Let me call m the small body and M the large body. The small body starts at the surface of the large body, where the gravitational potential energy is U

( G . m .M )

R and m starts with a kinetic energy of mv^2/2, so our initial total energy is ( G .m .M ) R

1. . 2 mv 2

After m has just escaped from M, m has used up all of its kinetic energy in escaping, so its kinetic energy is zero. If m had exactly enough energy to just escape, it won't have escaped until it reached a distance of r=∞. At r=∞ the gravitational potential energy is zero (1/r=0 there), so the total energy is zero. Therefore ( G .m .M ) R

1. . 2 mv =0 2

"Divide both sides by m, and solve for v to get v

2 .G .M R

(b) Obtain a formula for the Schwarzschild radius of the body by setting v.e=c, the speed of light, and solving for r. c

2 .G .M R

Square both sides, solve for R to get

R

2 .G . M c

2

9

Useful constants, etc.: − 34

h := 6.63⋅ 10

melectron := 9.11⋅ 10

− 31

8

c := 3 ⋅ 10

====================================================================================== Physics 107

Problem 3.1

Compare the linear momenta, total energy, and kinetic energy of a photon and a particle that have the same wavelength. Because

h

λ :=

regardless of whether the object is a photon

p

or a particle, if the wavelengths are the same, then the momenta p are the same. Problem 3.11 shows that the particle's total energy is nearly the same as the energy of a photon of the same wavelength PROVIDED THE PARTICLE'S TOTAL ENERGY GREATLY EXCEEDS ITS REST ENERGY; i.e. when the particle's kinetic energy is very large; it is moving very fast. 2 4

From chapter 1, E :=

2 2

m0 ⋅ c + p ⋅ c

In this problem, both the photon and the particle have the same momemtum p, so p2c2 is the same for both. Because some of the particle's total energy is tied up in its rest energy m02c4, and none of the photon's total energy is tied up in rest energy, the particle must have less kinetic energy than the photon has total energy. ====================================================================================== Physics 107

Problem 3.2

Find the de Broglie wavelength of (a) an electron whose speed is 1x10 8 m/s and (b) an electron whose speed is 2x108 m/s. γ ( v) :=

1

8

vpart1 := 1.0⋅ 10

2

1−

v c

8

vpart2 := 2.0⋅ 10

2

Part a. Do we need to worry about relativity? Let's do this non-relativistically and relativistically, and see if there is a significant difference Nonrelativistic: λ non( m , v) :=

Relativistic h

λ rel( m , v) :=

m⋅ v − 12

λ non( melectron , vpart1) = 7.278 × 10

h γ ( v) ⋅ m⋅ v − 12

λ rel( melectron , vpart1 ) = 6.861 × 10

wavelength units are meters Percent error P: P = 6.066

P :=

λ non( melectron , vpart1) − λ rel( melectron , vpart1) λ rel( melectron , vpart1 )

⋅ 100

Because a 6% error is introduced by the nonrelativistic calculation, we need to use the relativistic result.

1

Part b. Because the velocity is greater than in part a, we clearly need a relativistic calculation. − 12

λ rel( melectron , vpart2 ) = 2.712 × 10

====================================================================================== Physics 107

Problem 3.3

Find the de Broglie wavelength of a 1 mg grain of sand blown by the wind at 20 m/s. This is an extremely lightweight piece of matter. Maybe its wavelength will be observable?? λ sand( m , v) :=

(

−6

λ sand 1 ⋅ 10

h m⋅ v

)

− 29

, 20 = 3.315 × 10

(using the fact that one milligram is 10-3 grams, or 10-6 kg)

This wavelength is in meters. This is many orders of magnitude smaller than a typical size of a nucleus. We'll never "see" that a grain of sand has wave properties. ======================================================================================

Physics 107

Problem 3.4

Find the de Broglie wavelength of the 40 keV electrons used in a certain electron microscope. First let's see if a relativistic calculation is necessary. Begin by converting the 40 keV electron energy to joules: 3

KE := 40⋅ 10 ⋅ 1.60⋅ 10 KE := vcl :=

1 2

− 19

2

⋅ m⋅ v

2 ⋅ KE

if this gives a v near c, then we need to go back and calculate a relativistic velocity

melectron 8

vcl = 1.185 × 10

we need to use relativistic corrections for this magnitude of speed

The electron's total energy is its 40-keV kinetic energy plus its 0.511 MeV rest energy. In units of eV, 3

6

E := 40⋅ 10 + 0.511⋅ 10 − 19

E := E⋅ 1.60⋅ 10

5

E = 5.51 × 10 eV

convert to joules

E = 8.816 × 10

− 14

joules

We can use equation 1.22 to calculate the velocity E :=

melectron ⋅ c

2

2

1−

v c

2

2

Solving (pen-and-paper) gives this equation:

 melectron2⋅ c4    2 E  

v := c⋅ 1 −

8

v = 1.103 × 10

h

λ :=

1 2

λ = 6.139 × 10

⋅ melectron ⋅ v

− 12

meters

v

1−

c

2

====================================================================================== Physics 107

Problem 3.5

By what percentage will a nonrelativistic calculation of the de Broglie wavelength of a 100 keV electron be in error. − 19

3

KE := 100⋅ 10 ⋅ 1.60⋅ 10 Nonrelativistic: p :=

Relativistic:

2 ⋅ melectron⋅ KE

prel :=

− 22

p = 1.707 × 10 λ non :=

c

2

⋅

 KE + melectron⋅ c2 −  melectron⋅ c2    

prel = 1.789 × 10

h

λ rel :=

p − 12

Percent :=

2

− 22

h p rel − 12

λ non = 3.883 × 10 Percent error:

1

λ rel = 3.706 × 10 λ non − λ rel

⋅ 100

λ rel

Percent = 4.8

Because a 4.8% error is introduced by the nonrelativistic calculation, we probably need to use the relativistic result.

====================================================================================== Physics 107

Problem 3.6

Find the de Broglie wavelength of a 1 MeV proton. Is a relativistic calculation needed? Because the proton rest mass of 938 MeV/c 2 is much greater (by a factor of nearly 1000) than the energy of 1 MeV, the calculation may be done nonrelativistically. − 34

h := 6.63⋅ 10

− 27

mproton := 1.672⋅ 10

3

We need the proton's velocity. The 1-MeV refers to the proton's kinetic energy. There are several approaches. We could solve 1 2 KE := ⋅ mproton⋅ v 2 for v. Or we could use KE :=

p

2

2 ⋅ mproton

λ :=

and

h p

λ :=

h 2 ⋅ mproton⋅ KE

I will use the latter equation. I need to keep my units straight; otherwise, the problem is simple. − 19

6

KE := 1⋅ 10 ⋅ 1.60⋅ 10

This converts the kinetic energy to joules

h

λ :=

2 ⋅ mproton⋅ KE

λ = 2.866 × 10

− 14

nanometers

h

v :=

Note that

−5

2.863⋅ 10

meters, or

λ ⋅ mproton

7 v = 1.383 × 10 so this is a fairly fast-moving, but clearly nonrelativistic, proton

======================================================================================

Physics 107

Problem 3.7

The atomic spacing in NaCl is 0.282 nm. Find the kinetic energy (in eV) of a neutron with a de Broglie wavelength of 0.282 nm. Is a relativistic calculation needed? − 19

eV := 1.6⋅ 10 λ :=

mneutron := 1.675⋅ 10

− 27

h

(mneutron⋅ v)

v( λ ) :=

h

(λ ⋅ mneutron)

(

−9

v 0.282⋅ 10 KE( λ ) :=

(

conversion factor between eV and joules

1 2

) = 1.404 × 103

⋅ mneutron⋅ v( λ ) ⋅

KE 0.282⋅ 10

This is in units of m/s, so a relativistic calculation is not needed.

2 1

−9

) = 0.01031

The factor 1/eV converts joules to eV.

eV The answer is in eV; we would probably write it as 10.3 meV.

======================================================================================

4

Physics 107

Problem 3.8

Find the kinetic energy of an electron whose de Broglie wavelength is the same as that of a 100 keV x-ray. There's a long, tedious way to do this problem, and a quick and easy way. I'll do the long way first. First calculate the de Broglie wavelength of the x-ray (and therefore the electron): E := 100⋅ 10 E := λ :=

3

this is the 100 keV x-ray energy in eV units

h⋅ c λ h⋅ c

The eV converts energy to SI units.

E⋅ eV

λ = 1.243 × 10

− 11

So far, we have found the wavelength of the gamma ray, which we are told is the electron's wavelength. Now we use this wavelength to calculate the electron's momentum or velocity, and then energy. λ :=

v :=

h

so

melectron ⋅ v h

KE :=

2

7

 v2   = 0.038  c2  

2

⋅ melectron ⋅ v

− 15

KE = 1.561 × 10 KE

KEev :=

h melectron ⋅ λ

v = 5.854 × 10

melectron ⋅ λ 1

v :=

joules

KE :=

1 2

2

⋅ melectron ⋅ v

Is relativity important at this speed?

This could be considered relativistic; we will consider the need for a relativistic calculation below.

This is the nonrelativistic KE.

KEev = 9.757 × 10

eV

and

3

electron volts

The electron's rest mass is 0.511 MeV. If our calculated kinetic energy is much less than the rest mass, our neglect of relativity was justified. KEev 6

= 0.019

about 1.9%

.511⋅ 10

2

Also,

1−

v c

2

= 0.981

This would be 1 if relativity were completely negligible.

In other words, neglecting relativity has introduced an error of just under 2%. Whether or not relativity can be neglected depends on the precision of the measurement. It happens that for this problem, the relativistic calculation is easy if you do it "smart." The "smart" solution. E = mc2 + KE

KE = E - mc2 = (E2)1/2 - mc2 = ( p2c2 + (mc2)2 )1/2 - mc2.

The electron and the x-ray photon have the same wavelength, and therefore the same momentum p. Thus (pc)photon = (pc)electron. But the photon is "pure energy" so Ephoton = (pc)photon = 100 keV = (pc)electron.

5

KE :=

( p ⋅ c) +  melectron ⋅ c 2

2

2

2  − melectron⋅ c

If you express the electron mass in "keV energy units," you get

KE :=

2

2

( 100) + ( 511) − 511

KE = 9.693

keV

This correct relativistic KE is a bit smaller than the classical value of 9.76 keV obtained above.

================================================================================= Physics 107

Problem 3.9

O. A. Pringle

This may sound complicated, but it is just asking you to calculate the kinetic energy in eV of an electron having a de Broglie wavelength of 550 nm. The standard approach is to neglect relativity, calculate the energy, and see if it looks large enough to worry about relativity. − 34

−9

h := 6.63⋅ 10

− 31

− 19

λ := 550⋅ 10 e := 1.60⋅ 10 The "e" is the eV to joules conversion factor; this factor actually converts either way, you just need to be careful whether it goes in the numerator or the denominator.

me := 9.11⋅ 10

I'll do this slightly more "elegantly" and write KE as a function of v, so that I can calculate v later. This is a more adaptable approach than, say, in solutions to preceding problems. KE( v) := v :=

1 2

2

⋅ me⋅ v

Mathcad is happy because it knows later on, if I want to use this, I'll give it some values for v.

h me⋅ λ

KEeV( v) :=

KE( v) e

I am taking more steps than really necessary, but this will define another KE function, with units of eV.

− This 6 is a really puny energy, about 5 micro-eV. No KEeV( v) = 4.985 × 10 sense bothering with relativity here.

=================================================================================

Physics 107 Electron

O. A. Pringle Proton

velocity=v λ electron :

Problem 3.13

velocity=v h

λ proton :

h

6

λ electron :=

λ proton :=

melectron ⋅ v

mproton⋅ v

λ electron > λ proton because melectron < mproton electron: vgroup =

group velocities are same

electron: vp :=

c

proton: vgroup =

proton:

2

phase velocities are same

v

vp :=

c

2

v

=================================================================================

Physics 107

Problem 3.14

O. A. Pringle

Skip this problem. I will not give test or quiz problems on it. ================================================================================= Physics 107 This problem is trivial

Problem 3.18

O. A. Pringle

v := 0.900⋅ c The group velocity is just the electron speed: vg := 0.900⋅ c The phase velocity is calculated from vp := vp :=

c

2

v c

2

0.900⋅ c

vp := 1.111⋅ c =================================================================================

Physics 107 Problem 3.19 O. A. Pringle To get the group velocity, we need to calculate the electron particle velocity. Because 500 keV is large compared with the electron's rest energy (actually, about the same), we need to use relativity. Then from the electron's velocity, we will easily get the phase velocity.

7

From chapter 1, we know that 1

v := c⋅ 1 −

1 + K   2 m0⋅ c   Plug in:

2

6

8

c := 3 ⋅ 10

0.511⋅ 10

m0 :=

c

3

K := 500⋅ 10

if I express rest mass in keV/c^2, i can plug in K in kev

2

1

v := c⋅ 1 −

1 + K   2 m0⋅ c   v

8 v = 2.589 × 10 or

2

= 0.863

c

Phase velocity: vp :=

c

2

vp = 3.477 × 10

v

8

vp c

= 1.159

=================================================================================

Physics 107

Problem 3.27

Set a range on n:

n := 1 .. 5

Energy equation:

E :=

O. A. Pringle

2 2

n

n ⋅h

8 ⋅ m⋅ L

this energy is in Joules 2

2 2

E := n

18

n ⋅ h ⋅ 6.242⋅ 10 ⋅ 10 8 ⋅ m⋅ L

−6

this energy is in eV

2

Note the difference between the subscript n which is just a name (type in E.n to get this) and the subscript n which takes on a range of values (type E[n to get it).

h ≡ 6.626⋅ 10

key, globally defines these parameters, so I can

− 27 enter them AFTER the definition of En and get

m ≡ 1.675⋅ 10 L ≡ 10 A :=

rid of the "undefined" error message.

− 14

2

A numerical answer:

− 34 The triple equals sign, obtained by typing the ~

18

−6

h ⋅ 6.242⋅ 10 ⋅ 10 8⋅ m⋅ L

2

A = 2.045

En := 2.045⋅ n The neutron's minimum energy is

2

in MeV

Mev ≡ 1

E = 2.048 MeV 1

The minimum energy is E = 2.048 Mev That's a lot of energy; for example, it's a significant fraction 1 of the energy a neutron gets in a fusion reaction. 8

1

of the energy a neutron gets in a fusion reaction.

Just for kicks, make a plot:

Type E[n@n, then choose a graph type of "sl" to get this graph.

En

n

=================================================================================

Physics 107

Problem 3.28

O. A. Pringle

2 2

(a) A particle in a box satisfies

E( n ) :=

n ⋅h

8 ⋅ m⋅ L

2

We are given that E=1 eV for n=1. We don't know m or L, but we don't need to know them, because we do know that

2

In part b, we will plug in := 1 eV the electron mass and 2 8 ⋅ m⋅ L solve this for L. h

The energy in eV is then just 2

E( n ) := n Now get the answers: 19 E( 2 ) = 2.5 × 10 eV 19

E( 3 ) = 5.625 × 10eVeV ≡1 (b) Solve

h

2

8 ⋅ m⋅ L

2

L( m) :=

− 34

:= 1 ⋅ e for L (note I am converting eV-->J

(

h

− 19

e := 1.60⋅ 10

) = 6.14 × 10− 10

− 31

L 9.11⋅ 10

8 ⋅ m⋅ e

h := 6.63⋅ 10

=================================================================================

Physics 107

Problem 3.29 − 34

h := 6.63⋅ 10

n := 2

− 19

e := 1.6⋅ 10

O. A. Pringle − 27

mp := 1.673⋅ 10

3

E2 := 400⋅ 10 ⋅ e

2 2

Solve

En :=

n ⋅h

8 ⋅ mp⋅ L

for L: 2

2 2

L :=

n ⋅h

8 ⋅ mp⋅ E2

L = 4.531 × 10

− 14

=================================================================================

Physics 107 Problem 3.30 O. A. Pringle Particle wavelength and momentum are related by p=h/λ. The LARGEST possible wavelength for a particle trapped in a box corresponds to exactly one half wavelength of the particle wave spanning the length of the box; i.e., λ=2L. This corresponds to the SMALLEST possible momentum; i.e., p(minimum)=h/2L.

9

p(minimum)=h/2L. We can calculate the minimum energy allowed by the uncertainty principle principle by taking E(minimum)=p(minimum)^2/2m. m := 1 L := 1 h := 1 p0 := E0 :=

h 2⋅ L p0

2

2⋅ m h

E0 :=

2

8 ⋅ m⋅ L

2

2 2

E0 :=

n ⋅h

with n=1. 2

8 ⋅ m⋅ L Heisenberg's uncertainty principle requires that this be the minimum energy. This is exactly in agreement with our wave function analysis of a particle in a box. =================================================================================

Physics 107

Problem 3.31

O. A. Pringle

You can read the solution in the back of Beiser. Beiser's answer makes the underlying assumption that the ideal gas is not confined to any region in space. As soon as you put an ideal gas molecule into some kind of container, like we often do in elementary thermodynamics, then the uncertainty principle is applicable. I suppose that implies that such a gas isn't really "ideal" after all, although in the typical Physics 23 type problem the container is big enough that we never know that the uncertainty principle is there. In other words, if the container is "big enough" then δx is essentially infinite, so that δp is essentially zero, and there is no "measurable" zero point energy. =================================================================================

Physics 107 Problem 3.32 O. A. Pringle Use δxδp>=h/4π. Let's just calculate the minimum δv, so use the equality sign. h := 6.626⋅ 10

− 34

− 31

me := 9.11⋅ 10

− 27

mp := 1.672⋅ 10

−9

δx := 1 ⋅ 10

10

h

δp e :=

h

δp p :=

4 ⋅ π ⋅ δx − 26

4⋅ π ⋅ δx − 26

δp e = 5.273 × 10

δp p = 5.273 × 10

Both δp's are the same, as they ought to be. I had to fiddle with the zero tolerance above. δpe

δve :=

δp p

δvp :=

me

mp

δvp = 31.536

4

δve = 5.788 × 10

The lighter, faster-moving electron has a much greater uncertainty in its velocity. =================================================================================

Physics 107 Problem 3.33 O. A. Pringle Assume that both measurements are made with the greatest possible accuracy; i.e., δxδp is minimum. Then we use the equality sign in equation 3.21. h := 6.626⋅ 10

− 34

− 31

me := 9.11⋅ 10

δx := 0.1⋅ 10

−9

h

δp :=

− 25 This is momentum uncertainty in mks units. I had to δp = 5.273 × 10 4 ⋅ π ⋅ δx fiddle with zero tolerance to see anything. The electron's momentum can be calculated from its energy. 3

K := 1⋅ 10 ⋅ 1.602⋅ 10 p :=

− 19

2 ⋅ me⋅ K − 23

p = 1.708 × 10

The percentage uncertainty P is P=100*δp/p. P := 100⋅

δp

P = 3.086 About 3 percent.

p

Questionto think about: do we have to worry about relativity here? ve :=

p

7

ve = 1.875 × 10

me

8

c := 3 ⋅ 10

2

correction :=

1−

ve c

correction = 0.998

2

The correction factor when we include relativity is an 0.2 percent effect. Since it only makes sense to talk about uncertainties and fractional uncertainties to 1 decimal place, the relativistic correction is not important. =================================================================================

Physics 107

Problem 3.34

O. A. Pringle

(a) Use δtδE=h/4π. h := 6.626⋅ 10 m v

2

− 34

− 31

me := 9.11⋅ 10

ve := 10

11

2

Ee :=

me⋅ ve

This equation for δE specifies the 0.1 δE := Ee⋅ 0.001 percent uncertainty.

2

−3

h

δt :=

δt = 1.15759 seconds × 10

4 ⋅ π ⋅ δE During this time, the electron travels d := ve⋅ δt d = 0.012 meters (b) Do the same for the insect. vi := 10 Ei :=

mi := 1⋅ 10

2

mi⋅ vi

−3

δE := Ei⋅ 0.001

2 δt :=

h 4 ⋅ π ⋅ δE

δt = 1.05456 × 10

− 30

seconds

d := vi⋅ δt − 29 d = 1.055 × 10 meters We can safely neglect the uncertainty principle when we are talking about macroscopic objects like insects.

=================================================================================

Physics 107

Problem 3.35

O. A. Pringle

Use δxδp>=h/4π. Use the equality sign to find the maximum accuracy. h := 6.626⋅ 10

− 34

− 27

mp := 1.672⋅ 10

− 19

3

δE := 1 ⋅ 10 ⋅ 1.602⋅ 10 δp := δx :=

2 ⋅ mp⋅ δE h 4 ⋅ π ⋅ δp

δx = 7.204 × 10

− 14

m 9 δx := δx⋅ 10 convert to nm −5

δx = 7.204 × 10 nm This is comparable to the size of an atomic nucleus. The answer in the book comes from the fact that a position determination would be reported as p+/-δp, so the position would be determined to an accuracy of twice δp. ================================================================================= Physics 107 Part a solution: λ n :=

2⋅ L

Problem 3.36

equation 3.17

O. A. Pringle

12

λn :

equation 3.17 n

λ n :=

h pn

Solve for p.n: 2⋅ L n

h

:=

pn :=

pn

n⋅ h 2⋅ L

Part b solution: We are told that the minimum change in momentum that a measurement can cause corresponds in magnitude to δp minimum = δp minimum = δp minimum = Letting δx=L gives δp minimum = δpδxminimum =

which is also greater than hbar/2

================================================================================= Physics 107

Problem 3.38

O. A. Pringle

An unstable elementary particle has a rest mass of 549 MeV/c^2 and a mean lifetime of 7.00*10^-19 s. What is the uncertainty in its rest mass? Use the Uncertainty Principle to calculate the uncertainty in the particle's energy. − 19

δt := 7.00⋅ 10 δE :=

h

( 4⋅ π ⋅ δt)

The uncertainty in the particle's lifetime means there is an uncertainty in the particle's rest mass and rest energy. Use E=mc^2 to find the uncertainty in the mass. δm :=

δE c

2 − 34

δm = 8.37 × 10 This is mass uncertainty in kg. This answer is OK, or you can do it in energy units.

To compare the mass uncertainty to the rest mass, convert to units ov MeV/c^2.

 c2 

δm := δm⋅ 

e 13

δm = 470.786 This is mass uncertainty in units of eV/c^2. This answer is equivalent to the above one. How large is this uncertainty? 6 The mass is m := 549⋅ 10 δm −7 A very small fractional uncertainty. = 8.575 × 10 m

Note that the particle's mass was a "red herring." You didn't need to use it to solve the problem.

14

Useful constants, etc.: − 34

h := 6.63⋅ 10

melectron := 9.11⋅ 10

− 19

− 31

− 19

e := 1.6⋅ 10

eV := 1.6⋅ 10

8

c := 3 ⋅ 10 ε 0 := 8.85⋅ 10

− 12

====================================================================================== Physics 107

Problem 4.3

Determine the distance of closest approach of 1 MeV protons incident on gold nuclei. We can use equation 4.2, modified to account for the +1 charge on the proton. r0 :=

Z⋅ e

2

4 ⋅ π ⋅ ε 0⋅ K

Define the constants and variables: Z := 79 6

K := 1⋅ 10 ⋅ eV r0 :=

Z⋅ e

the eV converts to SI units

2

4 ⋅ π ⋅ ε 0⋅ K − 13

r0 = 1.137 × 10

meters

======================================================================================

Physics 107

Problem 4.4

Find the frequency of revolution of the electron in the classical model of the hydrogen atom. In what region of the spectrum are electromagnetic waves of this frequency? I will remove this problem from the list of assigned problems. You will not be tested on it. ======================================================================================

Physics 107

Problem 4.5

What is the shortest wavelength present in the Brackett series of spectral lines?

Coming soon. The answer is 1.46 µm. ======================================================================================

1

Physics 107

Problem 4.6

What is the shortest wavelength present in the Paschen series of spectral lines?

Coming soon. The answer is 0.820 µm.

====================================================================================== Physics 107

Problem 4.7

In the Bohr model, the electron is in constant motion. How can such an electron have a negative amount of energy? The answer to this question is best found in section 4.2. Beiser gives the kinetic and potential energies of the electron in page 1245 The kinetic energy, mv2/2, is always positive, and the electron is, indeed, in motion. The potential energy, however, results from the Coulomb attraction of the proton and electron. This negative energy is larger that the kinetic energy. The total energy E=K+V is therefore negative, indicating the electron is bound, but the kinetic energy is positive, so the electron can be in motion.

====================================================================================== Physics 107

Problem 4.8

Lacking de Broglie's hypothesis to guide his thinking, Bohr arrived at his model by postulating that the angular momentum of an orbital electron must be an integral multiple of hbar. Show that this postulate leads to equation 4.13. This problem is not "assigned" for FS 2003. I left the solution here for your interest. You will not be tested on this problem. 2 2

We want to begin with L := n ⋅ hbar

and arrive at rn :=

n ⋅h ⋅ε 0 π ⋅ m⋅ e

Begin: L := n ⋅ hbar := m⋅ v⋅ rn

rn :=

2

n⋅ h 2 ⋅ π ⋅ m⋅ v

Plug in v from eq. 4.4 2 2

n ⋅h

2

rn := 2

2

4⋅ π ⋅ m ⋅

rn

2

rn

e

2

4⋅ π ⋅ ε 0⋅ m⋅ rn

2 2

:=

n ⋅h ⋅ε 0 π ⋅ m⋅ e

2

2 2

rn :=

n ⋅h ⋅ε 0 π ⋅ m⋅ e

The desired result.

2

2

====================================================================================== Physics 107

Problem 4.11

Find the quantum number that characterizes the earth's orbit around the sun. The earth's mass, orbital radius, and orbital speed are given in the text. First, I will find the earth's wavelength. Then I will use the Bohr model equation relating wavelength and orbital radius to find the earth's quantum number. − 34

24

h := 6.63⋅ 10 λ earth :=

4

mearth := 6 ⋅ 10

vearth := 3 ⋅ 10

11

rorbit := 1.5⋅ 10

h

(mearth⋅ vearth)

λ earth = 3.683 × 10

− 63

I changed the zero tolerance to 100 to see something besides zero here.

The earth can "fit into" an orbit only if nλ=2πrn. nearth :=

2⋅ π ⋅ rorbit λ earth 74

nearth = 2.559 × 10

====================================================================================== Physics 107

Problem 4.12

I will remove this problem from the list of assigned problems. You will not be tested on it.

====================================================================================== Physics 107

Problem 4.13

Compare the uncertainty in the momentum of an electron confined to a region of linear dimension a 0 with the momentum in a ground-state Bohr orbit. − 11

a0 := 5.29⋅ 10

According to the uncertainty principle, the uncertainty in the momentum of an electron in a ground-state Bohr atom (taking a0 as an estimate of the size of the region to which the electron is confined) is at least δp :=

h

(4⋅ π ⋅ a0) − 25

δp = 9.974 × 10

In MKS units, a small number.

We can calculate the momentum of this electron in several ways. We know its speed, so we can get its momentum from p=mv.

3

6

velectron := 2.2⋅ 10

pelectron := melectron ⋅ velectron − 24

pelectron = 2.004 × 10 p electron δp

= 2.01

The actual momentum is a factor of 2 bigger than the uncertainty.

====================================================================================== Physics 107

Problem 4.14

I will remove this problem from the list of assigned problems. You will not be tested on it.

====================================================================================== Physics 107

Problem 4.15

What effect would you expect the rapid motion of the atoms of an excited gas to have on the spectral lines they produce? Coming soon. Not for quiz 4, but eligible for exam 2.

====================================================================================== Physics 107

Problem 4.16

I will remove this problem from the list of assigned problems. You will not be tested on it.

====================================================================================== Physics 107

Problem 4.17

A proton and an electron, both at rest initially, combine to form a hydrogen atom in its ground state. A single photon is emitted in this process. What is its wavelength? The equation giving the wavelength of a photon emitted by an electronic transition in hydrogen is

 E1  1 1   ⋅ −  c ⋅ h 2 2   n i    nf

λ := −

−1

(equation 4.18, page 134)

4

In this problem, the electron is initially unbound; i.e. ni := ∞ 1

or

:= 0

ni

The electron goes from the unbound state to the ground (n=1) state, so

 E1  1  λ := − ⋅  c⋅ h  n f 2    

−1

I like to manipulate Mathcad equations first to get the equation I will finally use. I like to do this before I set up the rest of the problem, so I can see what I need to know.

Now define the parameters and plug them in. − 34

8

h := 6.63⋅ 10

c := 3 ⋅ 10

nf := 1

− 19

E1 := −13.6⋅ 1.60⋅ 10

 E1  1  λ := − ⋅  c⋅ h  n f 2    

−1

λ = 9.141 × 10

−8

or about 91.4 nm

======================================================================================

Physics 107

Problem 4.18

I will remove this problem from the list of assigned problems. You will not be tested on it.

======================================================================================

Physics 107

Problem 4.19

Find the wavelength of the spectral line that corresponds to a transition in hydrogen from the n=10 state to the ground state. In what part of the spectrum is this? I "borrowed" the equation from the solution to problem 4.20, which I did before 4.19. ni := 10

We are given: λ := R⋅ 

1

  2   nf

−

 2  n i  1

nf := 1

7

R := 1.097⋅ 10

−1

−8

λ = 9.208 × 10 or 92.1 nm; this is a short wavelength, so it corresponds to high energy, or ultraviolet

======================================================================================

5

Physics 107

Problem 4.20

Find the wavelength of the spectral line that corresponds to a transition in hydrogen from the n=6 state to the n=3 state. In what part of the spectrum is this? Equation 4-18 gives λ:

 −E1  1 1   λ :=  ⋅ −   c⋅ h n f 2 n i2    

−E1

1 1  := ⋅ −  c⋅ h 2 2 λ ni   nf 1

We are given:

ni := 6 λ := R⋅ 

−1

  2   nf

−

 2  n i  1

−1

where R := 1.097⋅ 107m-1

nf := 3 1

1  1 λ := R⋅    2− 2  n i    nf

−1

−6 λ = 1.094 × 10 or 1094 nm

According to figure 2.2 on page 51, this wavelength would correspond to relatively high-energy infrared, a little below the visible (which starts at about 400 nm).

======================================================================================

Physics 107

Problem 4.21

A beam of electrons bombards a sample of hydrogen. Through what potential difference must the electrons have been accelerated if the first line of the Balmer series is to be emitted? As usual, the first step is to figure out what the problem is really asking. It's really just asking you to calculate the energy of the photon corresponding the the "first" line of the Balmer series. What does Beiser mean by "first line?" If you look at fig. 4-16, you will see that the series limit is approached as n-->∞. Thus, there is no "last line" of the Balmer series, and the "first line" must correspond to the n=3 to n=2 transition. There are several ways to solve this problem. You could use eq. 4.18 to get λ of the photon, and use E=hc/λ to get the photon (and therefore electron) energy. Or you could use eq. 4.17 to get the frequency and use E=hf. Or you could use the un-numbered equation between equations 4.16 and 4.17. That's what I will do. That's handy because I don't have to go through any energy conversions. E1 := −13.6 ∆E( n i , n f ) := −E1⋅ 

1  1  2− 2 ni   nf

∆E( 3 , 2 ) = 1.889 volts

6

======================================================================================

Physics 107

Problem 4.22

How much energy is required to remove an electron in the n=2 state from a hydrogen atom?

As usual, the first step is to figure out what the problem is really asking. The energy of an electron in hydrogen is given by E( n ) :=

E1 2

where E1 ≡ −13.6 eV

n An electron in the n=2 state has E( 2 ) = −3.4 To remove it from the hydrogen atom requires 3.4 eV: ∆E := E( ∞ ) − E( 2 )

I did this on purpose. Mathcad thinks 10307 is infinity. I tried typing in control-z (the way to get "infinity". Mathcad says "overflow" because it can't handle 1/(infinity)2. To get around this, just realize E(∞)=0.

∆E := 0 − E( 2 ) ∆E = 3.4 eV

======================================================================================

Physics 107

Problem 4.23

I will remove this problem from the list of assigned problems. You will not be tested on it.

======================================================================================

Physics 107

Problem 4.24

The longest wavelength in the Lyman series is 121.5 nm. Use this wavelength together with the values of c and h to find the ionization energy of hydrogen. The pertinent equation is 1 λ

:=

 −E1   1 1  ⋅ −    h⋅ c  n 2 n 2 i   f

7

Because E1 is the reduction in energy of the electron upon binding, the ionization energy is actually the negative of E1: 1 λ

:=

 Eionization   1 1  ⋅ −    h⋅ c  n 2 n 2 i   f

The Lyman series corresponds to transitions from n of 2 or greater down to n=1. The longest wavelength would correspond to a transition from n=2 to n=1, the smallest-energy transition in the series. Solving for Eionization gives

(

h⋅ c

)

Eionization λ , n f , n i :=

(

−9

Eionization 121.5⋅ 10 Ein_eV :=

λ

1

⋅

 1   2  nf 

−

 1   2   n i 

)

− 18

, 1 , 2 = 2.183 × 10

(

−9

Eionization 121.5⋅ 10

,1,2

Does this answer make any sense?

)

e

Ein_eV = 13.6

eV

yes, the answer does make sense

======================================================================================

Physics 107

Problem 4.25

An excited hydrogen atom emits a photon of wavelength λ in returning to the ground state. (a) Derive a formula that gives the quantum number of the initial excited state in terms of λ and R. (b) Use this formula to find ni for a 102.55 nm photon. Part (a) 1 λ

:= R ⋅ 

1  1  2− 2 ni   nf

7

R := 1.097⋅ 10

I suppose to be consistent with the other constants, I could round this to 1.10x107.

The final state is the ground state, nf=1. We assume R and λ are known, and solve for ni. 1 R⋅ λ

:=

1 − 1  1 2 ni   8

1 2 ni

:= 1 −

ni :=  1 − 2



ni( λ ) :=

1 R⋅ λ

 R⋅ λ  1

−1

1 − 1   R⋅ λ  

−1

Notice this is a "live" equation, ni as a function of λ.

(

ni 102.55⋅ 10

Part (b)

−9

)=3

====================================================================================== Physics 107

Problem 4.36

For laser action to occur, the medium used must have at least three energy levels. What must be the nature of each of these levels? Why is three the minimum number? Coming soon. Not for quiz 4, but eligible for exam 2.

======================================================================================

Physics 107

Problem 4.37

A certain ruby laser emits 1 joule pulses of light whose wavelength is 694 nm. What is the minimum number of Cr3+ ions in the ruby? − 34

h := 6.63⋅ 10

8

c := 3 ⋅ 10

λ := 694⋅ 10

−9

The energy of a 694 nm photon is E :=

h⋅ c

E = 2.866 × 10 λ In each pulse of light, there are n :=

1

− 19

Joules

photons, because 1 joule per pulse divided by joules per photon gives photons per pulse.

E 18

n = 3.489 × 10

photons

There must be at least that many chromium 3 + ions in the crystal.

9

======================================================================================

Physics 107

Problem 4.38

Steam at 100 C can be thought of as an excited state of water at 100 C. Suppose that a laser could be built based upon the transition from steam to water, with the energy lost per molecule of steam appearing as a photon. What would the frequency of such a photon be? The heat of vaporization of water is 2260 kJ/kg and its molar mass is 18.02 kg/kmol. This sounds tricky, but it is just asking: what is the frequency of a photon whose energy is equal to the energy given up by when a molecule of steam turns to water? This is really just a unit conversion problem. M := 18.02 grams per mole 23 N := 6.02⋅ 10 molecules per mole

Let m be the number of kilograms per molecule. m :=

M ⋅ 10

−3

N − 26

m = 2.993 × 10

Is this reasonable? H2O ought to be about 18 times the mass of hydrogen: − 27

18⋅ 1.67⋅ 10

L is the heat of vaporization:

− 26

= 3.006 × 10

so our m is reasonable.

3 L := 2260⋅ 10 joules/kilogram

E := m⋅ L E = 6.765 × 10 f :=

− 20

E

joules

− 34

h := 6.63⋅ 10

I really only need to define this once per document, don't I.

h

f = 1.02 × 10

14

According to figure 2.2 on page 47, this would be high-frequency infrared light, almost visible.

10

Physics 107

Problem 5.1

Which of the wave functions in Fig. 5.15 cannot have physical significance in the interval shown? Why not? Recall that

⌠  ⌡

∞

(

Φ

) 2 dV

−∞

must be finite, and that Φ must be single-valued and continuous with finite, single-valued, and continuous first derivatives. The function in figure 5.15(a) appears to satisfy all of the conditions over the interval shown, so it can be a wave function. If you want to be picky, its zero slope everywhere means the momentum is zero everywhere, which is not allowed by the uncertainty principle. I will not be that tricky on the exam. The function in figure 5.15(b) is not single-valued, so it cannot be a wave function. The function in figure 5.15(c) does not satisfy the condition for a continuous first derivative, so it cannot be a wave function. The function in figure 5.15(d) does not satisfy the condition for a continuous first derivative, so it cannot be a wave function. In addition, the function is not continuous. Beiser in his answer implies the function goes to infinity, which is not obvious from the figure as it is drawn. The function in figure 5.15(e) appears to satisfy all of the conditions, so it can be a wave function. The function in figure 5.15(f) is not continuous, so it cannot be a wave function. This is tricky, because it is possible that the derivative could be continuous and finite.

Physics 107

Problem 5.2

Which of the wave functions in Fig. 5.16 cannot have physical significance in the interval shown? Why not? "YES" means allowed, "NO" means not allowed. (a) YES. (b) YES if you assume the "interval shown" is only where the function is defined. NO if you assume the "interval shown" means the entire shown part of the positive x-axis. (c) NO. The first derivative is discontinuous in the middle. Beiser also probably intends to imply that the function goes to infinity at the middle. (d) YES. (e) NO. The function is not single valued. (f) YES. The function in figure 5.16(f) appears to satisfy all of the conditions, so it can be a wave function, although it is not clear how the amplitude can decrease with increasing x without the wavelength changing.

1

Physics 107

Problem 5.3

Which of the following wave functions cannot be solutions of Schrodinger's equation for all values of x? Why not? I will plot the functions so we can better see what is going on. A := 1 (a) The secant is 1/cosine. Let's plot the secant function for small values of x. Φ ( x) :=

1 cos( x)

 

x := −2 ⋅ π ,  −2⋅ π +

2⋅ π  100 

.. 2⋅ π

First plot Φ(x), letting Mathcad choose the plot scale.

There are singularities here, which seem to scale differently. Because of our choice of values of x, the function does not always "hit" the singularity at exactly the same position.

Φ ( x)

x

We can see the function better if we choose the plot scale ourselves.

There are singularities wherever cos(x)=0, so this is not an allowed wave function if its range includes an x-value where there is a singularity.

Φ ( x)

x

(b) Φ ( x) := A⋅ tan( x)

 

x := −2 ⋅ π ,  −2⋅ π +

2⋅ π  100 

.. 2⋅ π

This function has singularities at π/2, 3π/2, etc..

Φ ( x)

x

2

( x)

2

(d) Φ ( x) := A⋅ e

x := 0 , .05 .. 5

This function cannot be normalized because of the divergence as x goes to ∞

Φ ( x)

x

2

(e) Φ ( x) := A⋅ e

−x

x := −4 , −3.9 .. 4

This function is allowed, because the function and all of its derivatives are continuous and integrable

Φ ( x)

x

Physics 107

Problem 5.4

Find the value of the normalization constant A for the wave function 2

−

Φ ( x) := A⋅ x⋅ e

x

2

The DOS version of Mathcad is a real pain for symbolic manipulation. Let's try it anyway. To normalize, find A such that... ⌠    1 :=  ⌡

∞

2

 − x2  2 2 2 A ⋅x ⋅e  dx

−∞

⌠   2 1 := A ⋅  ⌡

∞

−∞

Yes, I originally composed these solutions using Mathcad 2.5 for DOS. Also, much of this is tutorial in nature because I used to collect student work done in Mathcad.

2

 − x2  2 2 x ⋅e  dx 3

⌠

2

∞

2

2 −x

1 := A ⋅  ⌡

x ⋅e

dx

−∞

⌠  1 := A ⋅ 2 ⋅  ⌡

∞

2

2 −x

2

x ⋅e

dx

0

 π  4 

2

1 := A ⋅ 2 ⋅ 

Solve for A: 2

2

A :=

π

1

:=

4

or

π

A :=

4  π

4

Everything from here on down is optional, for info only. Let's double-check and see if it works. ⌠     ⌡

∞

2

 − x2  2 2 2 A ⋅x ⋅e  dx = 1

−∞

Notice if you make this equation "live" Mathcad says "overflow" because it tries to calculate the integrand as x becomes infinite, and the x2 part blows up. Of course the exponential part goes to zero faster than x2 blows up.

To get around the overflow, we can try reducing the range of integrating from ∞ to something very large. ⌠     ⌡

1 2

 − x2  2 2 2 A ⋅x ⋅e  dx = 0.428

−1

Obviously -1 to 1 is not "large enough"; we have missed lots of the area under the integral. ⌠     ⌡

10 2

 − x2  2 2 2 A ⋅x ⋅e  dx = 1

− 10

It looks like -10 to 10 is nearly "infinity".

4

Physics 107

Problem 5.5

The wave function of a certain particle is φ=A cos2(x) for -π/2> exp( −k2⋅ L) = 0.438 They differ by a factor of 5.2, which is again marginal. If we use the "better" version of the transmission probability, we get T :=

   4 + 

 ⋅ exp( −2⋅ k ⋅ L) 2   k2   k   1 

16

2

T = 0.293 And because the barrier is marginally high and wide, we expect this result to be somewhat off from the full, correct expression.

Physics 107

Problem 5.25

A beam of electrons is incident on a barrier 6.00 eV high and 0.200 nm wide. Use Eq. 5.60 to find the energy they should have if 1.00 percent of them are to get through the barrier. − 31

melectron := 9.11⋅ 10 − 34

h := 6.63⋅ 10

− 19

e := 1.60⋅ 10

hbar :=

h T := 1.00⋅ 10

2⋅ π

−9

L := 0.200⋅ 10

U := 6.00⋅ e

−2

Equation 5.60 T := exp( −2 ⋅ k2⋅ L) Solve Eq. 5.60 for E −2⋅ k2⋅ L := ln( T) k2 := −

ln( T) 2⋅ L

2⋅ melectron ⋅ ( U − E) hbar

:= −

2 ⋅ melectron⋅ ( U − E) :=

U − E :=

1 2 ⋅ melectron

⋅ 



ln( T) 2⋅ L

 ln( T) ⋅ hbar    2⋅ L 

ln( T) ⋅ hbar  2⋅ L

2

2

 14

E := U −

1 2 ⋅ melectron



E = 0 joules E e

ln( T) ⋅ hbar 

⋅ 

2



2⋅ L

(The value is smaller than the zero tolerance, so it looks like zero.)

= 0.937 eV

Beiser's solution manual gives an answer of 0.95 eV, which I get if I use e=1.602x10-19 and h=6.626x10-34. On the exam, use the values of the constants which I give on the first page.

Physics 107

Problem 5.29--Not Assigned FS 2003

Show that for the n=0 state of a harmonic oscillator whose classical amplitude of motion is A, y=1 at x=A, where y is the quantity defined by Eq. 5.67. A classical amplitude of motion of A means that A is the largest value that x takes on. We can calculate the total energy from this largest value: E :=

1 2

2

⋅ k⋅ A

In the ground state E :=

1 2

⋅ h⋅ f

Equate the two energies: 1 2

2

⋅ k⋅ A :=

A :=

1 2

⋅ h⋅ f

h⋅ f k

This value of A is the largest value that x takes on (classically). The value of y corresponding to this x is y :=

y :=

2 ⋅ π ⋅ m⋅ f hbar

2 ⋅ π ⋅ m⋅ f hbar 2

y :=

⋅x

⋅

4 ⋅ π ⋅ m⋅ h⋅ f

h⋅ f k

2

h⋅ k

15

m⋅ h ⋅ f

y := 2 ⋅ π ⋅

So y := 2 ⋅ π ⋅

h⋅ k 1

f :=

But, from Eq. 5.73,

2

2⋅ π

k

⋅

m

m⋅ h ⋅ k ( h ⋅ k) ⋅ ( 2⋅ π ) ⋅ m 2

y := 1

Physics 107

Problem 5.30

Find the probability density φ.0^2 dx at x=0 and at x=+-A of a harmonic oscillator in its n=0 state. In the ground state, 1 4 −

φ 0( x) :=

 2 ⋅ m⋅ f  ⋅ e   hbar 

y

2

2

1

(

φ 0( x)

)

2

⋅ dx :=

2

 2⋅ m⋅ f  ⋅ e− y ⋅ dx   hbar  2

Eq. 5.76 relates x and y. At x=0, y=0 and 1

(

φ 0( 0 )

⋅ m⋅ f  )2⋅ dx :=  2hbar



2

⋅e



−0

2

⋅ dx

1

(

φ 0( 0 )

⋅ m⋅ f  )2⋅ dx :=  2hbar





2

⋅ dx

At x=+-A, as we showed in problem 5.29, y=1, so that 1

(

φ 0( A)

⋅ m⋅ f  ) 2⋅ dx :=  2hbar



2

2

⋅e



−1

⋅ dx

1

(

φ 0( A)

⋅ m⋅ f  ) 2⋅ dx :=  2hbar





2

1 ⋅ ⋅ dx e

16

Physics 107

Problem 5.31

Find the expectation values and for the first two states of a harmonic oscillator. This is a nice physics problem. Lots of fun math. Probably not a good physics 107 problem. I will not test or quiz you on this problem unless (1) I post the solution and (2) I announce that fact well in advance of the test or quiz. For the exam, I would expect you to be able to show that =0 for the ground state of a harmonic oscillator, and to compare that result with what you would expect from classical physics.

Physics 107

Problem 5.32

I won't test or quiz you on this problem without first posting a solution and then announcing that fact.

Physics 107

Problem 5.33

A pendulum with a 1.00 g bob has a massless string 250 mm long. The period of the pendulum is 1.00 s. − 34

h := 6.63⋅ 10

hbar :=

− 19

h

e := 1.6⋅ 10

2⋅ π

(a) What is its zero-point energy? Would you expect the zero-point oscillations to be detectable? Just as a quick double-check, let's see if the period Beiser gives agrees with the classical pendulum period .25

T := 2⋅ π ⋅

T = 1.004 seconds

9.8

close enough

The frequency is 1/T T := 1 f :=

1 T

The zero-point energy is E0 :=  0 +



E0 = 0

1 2

⋅ h⋅ f

joules

Even expressed in eV, the energy is very small: E0 e

− 15

= 2.072 × 10

eV

With such a small energy, zero-point oscillations will be undetectable.

17

With such a small energy, zero-point oscillations will be undetectable. (b) The pendulum swings with a very small amplitude such that its bob rises a maximum of 1.00 mm above its equilibrium position. What is the corresponding quantum number? The height of the rise gives us the maximum potential energy of the pendulum: we assume the pendulum is on the earth and calculate the corresponding gravitational potential energy. Then we can set the energy equal to the maximum potential energy and calculate the quantum number. −3

−3

m := 1 ⋅ 10

xmax := 1 ⋅ 10

Umax := m⋅ 9.8⋅ xmax −6

A considerably larger number than the zero-point energy.

Umax = 9.8 × 10 n :=

Umax h⋅ f

−

1 2 A variation on this problem would be a mass on a spring; to

28 calculate the quantum number given x.max, you would first n = 1.478 × 10

calculate k and then use U=kx.max^2/2.

Physics 107

Problem 5.37--Not assigned for FS 2003

Consider a beam of particles of kinetic energy E incident on a potential step at x = 0 that is U high, where E > U (Fig. 5.19). (a) Explain why the solution D*exp(-jk.2x) (in the notation of Sec. 5.8) has no physical meaning in this situation, so that D = 0. This term represents a wave travelling to the left (-x direction). There is no boundary from which to reflect such a wave beyond x = 0. Therefore D = 0. (b) Show that the transmission probability here is T = CC*/AA* and is equal to 4k.1^2/(k.1+k')^2. Set the wave function amplitudes equal at x = 0: A⋅ e

j ⋅ k1⋅ x

which gives...

+ B⋅ e

(

− j ⋅ k1⋅ x

)

:= C⋅ e

j ⋅ k2⋅ x

A + B := C

(1)

Set the wave function first derivatives equal at x = 0: j ⋅ k1⋅ A⋅ e

j ⋅ k1⋅ x

− j ⋅ k1⋅ B⋅ e

which gives...

(

− j ⋅ k1⋅ x

)

:= j ⋅ k2⋅ e

(2) ( A − B) :=

j ⋅ k2⋅ x

 k2   k ⋅C  1

Adding (2) to (1) and arranging terms gives: (3)

C A

:= 2 ⋅

k1 k1 + k2

18

The transmission probability is found by taking the ratio of the probability CURRENTS. Probability CURRENTS are found by multiplying probability densities by their respective particle velocities. Since the velocity is proportional to k (v = hbar*k/m), ratios of velocities are simply ratios of k.

k1 :=

m

2⋅

2

⋅ E

k2 :=

m

2⋅

hbar

2

⋅ E−V

hbar

Transmission probability:  C⋅ C  k2  T := ⋅  A⋅ A  k1 

T := 4⋅

k1

2

(k1 + k2)2

⋅

k2 k1

For part (c), I am going to re-write the transmission equation in terms of the particle velocities... Simplifying and keeping only the square roots: T := 4⋅

T( E , V) := 4 ⋅

k1⋅ k2

Notice that T = 1 when k.1 = k.2, as when V = 0. This equation is also valid when substituting v.1 and v.2 for k.1 and k.2.

(k1 + k2)2 E⋅ E − V

(

E+

E − V)

T( v1 , v2) := 4⋅

2

Likewise, T = 1 when V = 0, or when the energy is well above the barrier, i.e., when E >> V.

v1⋅ v2

(v1 + v2)2

c) A 1.00 ma beam of electrons moving at 2*10^6 m/s enters a region with a sharply defined boundary in which the electron speeds are reduced to 1*10^6 m/s by a difference in potential. Find the transmitted and reflected currents. Note that the powers of ten cancel in this equation. Relative currents are all that are required. T( 2 , 1 ) = 0.889

88.9% are transmitted. current =0.889 ma

R( v1 , v2) := 1 − T( v1 , v2) R( 2 , 1) = 0.111

11.1% are reflected. current =0.111 ma

19

Problem 6.3 This is not a difficult problem but it is a real pain to transfer it from paper into Mathcad. I won't give it to you on the quiz, but know how to do it for the exam. ___________________________________________________________________________ Problem 6.10 See Figure 6.4. The magnitude of L is L ( l ) .( l and the z-component is Lz

1 ) .hbar

m l .hbar

The angle between L and the z-axis is found from Lz cos ( θ ) L Let's do this first for l=1. Notice there is an hbar in both numerator and denominator of the equation for cos(θ), so the hbar's cancel. For convenience, I will set hbar=1. hbar

1

l ( l ) .( l

L( l ) θ m l, l

1

1 ) .hbar

Lz ml

ml 1 , 0 .. 1 . m l hbar

Lz ml 360 . acos L( l ) 2 .π

θ m l, l = 135

These are the angles in degrees measured clockwise from the positive z-axis to the vector L.

90 45

Now let's do it for l=2 l

2

ml

2 , 1 .. 2

θ m l, l = 144.736 114.095 90 65.905 35.264

___________________________________________________________________________

1

Problem 6.11 The magnetic quantum number, ml, is limited to the values 0, ñ 1, ñ 2, .... ñ l. There are a total of 2l + 1 values. For an orbital quantum number of l = 4, there are nine possible values for m.l: ml 4, 3, 2, 1,0,1,2,3,4 ___________________________________________________________________________ Problem 6.12 The orbital quantum number l may take the values l = 0, 1, 2, ......n-1, where n is the principal quantum number. Each orbital quantum number has an associated set of magnetic quantum numbers, m.l. The magnetic quantum number m.l is limited to the values 0, ñ 1, ñ 2, .... ñ l. There are a total of 2l + 1 values for each value of the orbital quantum number l. For a principal quantum number n = 4, l takes the values l = 0, 1, 2, 3. l=0 m.l = 0 l=1

m.l = -1, 0-, 1

l=2

m.l = -2, -1, 0, 1, 2

l=3

m.l = -3, -2, -1, 0, 1, 2, 3

The total number of orbitals with n = 4 is therefore N = 1 + 3 + 5 + 7 = 16. ___________________________________________________________________________ Problem 6.13 spdf l= 0 1 2 3 L

l .( l

1 ) .hbar

Lz

m l .hbar

because ml ranges from -l to l, the maximum value of ml is l and the maximum value of Lz is l*hbar l .hbar

L zmax

hbar cancels out in the equation for percentage difference, so for convenience, I will just set it equal to 1 hbar

1

percent ( l )

( l ) .( l

1 ) .hbar

( l ) .( l

l .hbar .

1 ) .hbar

100 2

percent ( 1 ) = 29.289 percent ( 2 ) = 18.35 percent ( 3 ) = 13.397 Beiser's definition of percentage difference is not necessarily what I would call percent difference. Beiser really asked "what percent of L is the maximum value of L.z." Here is the real definition of percentage difference between two quantities A and B: However, on an exam or quiz, do it like A B . P 100 Beiser did it. 1. (A B) 2 ___________________________________________________________________________ Problem 6.14 A spherically symmetric probability-density distribution corresponds to zero orbital angular momentum. This occurs only for l=0; i.e., for s wavefunctions. In table 6.1, when l=0, then m.l=0, and the wave functions φ and θ are independent of angle, which is why the probability-density distribution is independent of angle. ___________________________________________________________________________ Problem 6.15 The radial wave function for the 1s electron in hydrogen is r

R 1s ( r )

2 3

a0 From equation 6.24, P ( r ) .dr

.e

a0

2

2 2 r .( R )

To find the most probable r, we maximize P(r); i.e. take its derivative and set it equal to zero. d 0 P ( r) dr

3

2 .r

2

d . r dr

0

2 3

a0

.e

2 .r

a0

0

2

2 .r

4

0

3

2 .r

2 . a0 e a0

. r2 .

a0

a .d r2 .e 0 3 a 0 dr

4

2 .r .e

2 .r

a0

0

8 3

a0

. e

a0

. r

2

r a0

The only way for the right hand side above to equal zero is if 2

0

r

r a0

r a0

r. 1

The above equation has two roots: r=0 and r=a0. The answer here is r=a0, which corresponds to a maximum in P. The root at r=0 correspondes to a minimum in P. You can verify this by checking the values of the second derivative at r=0 and r=a0. With Mathcad, it is easier to just plot the function: a0

1

0 , .05 .. 5

r

In this problem, I scale a0 to be 1. Then r=2 means, for example, that r is equal to 2a0.

r

2

R 1s ( r )

3

a0

2

a0 P 1s ( r )

.e

2 2 r . R 1s ( r ) 2

P 1s ( r ) R 1s ( r )

1

0 0

1

2

3 r

4

5

6

In this plot I show both the 1s radial wave function, R(r), and the corresponding probability density, P(r). Notice that P(r) is maximum at r=1, and minimum at r=0. Remember, r=1 really means r=a0.

Mathcad also has a built-in root function and a symbolic solver which should be able to find the root for you.. Not much help on the exam, though, so be sure you can do it like I did above. ___________________________________________________________________________ 4

Problem 6.16 The 2p radial wave function corresponds to n=2, l=1, ml=0 or ml=+-1 (three choices), but all three choices for ml give the same R. From table 6.1,

1

R 2p

3

2 . 6 .a 0

2

. r .e a0

r . 2 a0

To minimize/maximize, take derivative and set it equal to zero. 0

d 2. 2 r R dr r

0

1 d 2. r 3 dr 24 .a 0

2

. r .e 2 a0

a0

r

0

a .d r4 .e 0 5 24 .a 0 d r

1

2 .r

0

1

. r4 .

24 .a 0

5

1 . a0 e a0

2 .r

e

a0

.4 .r3

2 .r

0

e

a0

.

24 .a 0

5

4

r a0

4 .r

3

The first term on the right, with the exponential in the numerator, is never zero, so 0

3 r . 4

r a0

which gives 4 roots; r=0 (three times) and r=4a0. As in problem 6.15, you can verify using calculus that the maximum is at r=4a0, or you can check it with a graph. 5

a0

1

0 , .05 .. 15

r

In this problem, I scale a0 to be 1. Then r=2 means, for example, that r is equal to 2a0. r

. r .e a0

1

R 2p ( r )

3

2 . 6 .a 0

2 .a 0

2

P 2p ( r )

2 2 r . R 2p ( r )

0.2 P 2p( r ) R 2p( r ) 0.1

0 0

4

8 r

12

16

___________________________________________________________________________

Problem 6.17 The 3d wavefunction is 2

. r .e 2 a0

4

R 3d

3

81 . 30 .a 0

2

r . 3 a0

r

0

2 C .r .e

3 .a 0

Where C is a constant.

To find the most probable r, take the derivative of P(r)=r 2R2 and set it equal to zero.

P ( r) 0

0

2 2 r .R

2 2 r .C . r .e

r 3 .a

2 0

d P ( r) dr d 2. . 2. r C r e dr

r 3 .a

2 0

6

5 C . 6 .r .e

0

C .e

0

C .e

0

2 .r 3 .a 0

2 .r 3 .a 0

2 .r 3 .a 0

2 . 6. r e 3 .a 0

. 6 .r5

2 .6 r 3 .a 0

.r5 . 6

2 . r 3 .a 0

2 .r 3 .a 0

The above equation has 6 roots; r=0 (5 times) and 6

2 . r . 3a0

The roots at r=0 all give minima, so the maximum is at the solution to the above equation, which is r 9 .a 0 Again, let's plot it. I like to SEE what is going on. 2

4

R 3d ( r )

3

81 . 30 .a 0 r

2

. r .e 2 a0

r 3 .a

0

P 3d ( r )

2 2 r .R 3d ( r )

Notice how for higher-n wave functions, I have to go further and further 0 , .05 .. 35 out in r to see most of the wave function. That's because higher-n wave functions are more extended in space.

P 3d( r ) R 3d( r )

0.1

0.05

0 0

3

6

9

12 15 18 21 24 27 r

___________________________________________________________________________ 7

Problem 6.18 Find the two values of r at which P(r) for a 2s electron has a maximum. r

1

R 2s

2 . 2 .a 0

r . e a0

. 2

3

2 .a 0

2

1

Let A

3

2 . 2 .a 0

2

Then we maximize the probability by solving this equation: d 2. 2 r R dr

0

r a0

d 2 2 A .r . 2 dr

0

2

.e

2 .r 2 .a 0

Why did I take the r 2 inside the (...)2? Answer: to simplify the application of the chain fule for the derivative. In this form, I have two terms to which I apply the chain rule. In the previous equation, I have three terms to which to apply the chain rule. The algebra for the three-term chain rule derivative is too much for an ordinary human like me to do without making at least one mistake. Even with my simplification, this is one of the longest homework problems of Chapter 6.

r 2 d A . 2 .r dr

0

2 r . a0 e a0

Now this is the equation we want to solve (the roots are the values of r for which the function has an extremum): ( r)

0

0

2

r . e a0

d 2 .r dr

2 .r

2 2

r a0

a0

I divided out the A2 because it is never zero. r

.

1 . e a0

a0

2 . 2 .r

2

r . 2 a0

2 .r a0

r

.e

a0

8

r

0

a0

e

2 2

. 1 . 2 .r a0

r a0

2 . 2 .r

2 .r a0

2

r . 2 a0

Because the exponential is nonzero, we once again divide both sides by the leading term, leaving 2 2 2 r 1 r . 2 .r . . . . 0 2r 2 2r 2 a0 a0 a0 a0 2

0

2 .r

r . . 2r a0

0

2 .r

r . a0

2 .r

0

2

2

2

r . 1 a0 a0

2 .r a0

2. 2

4 .r a0

2

r

2

4

a0 6 .r a0

2

r . r a0 a 2 0

2 .r a0

4

The roots of the above equation are found from: 2 .r

0

2

r a0

and

0

6 .r .a 0

2

r

4 .a 0

2

The equation on the left gives r So that r=0 and r=2a 0 are two roots. We will see in a minute 0 r. 2 that these roots correspond to probability minima. a0 The equation on the right is a simple quadratic equation, which has roots

r

r

6 .a 0

36 .a 0

2

16 .a 0

2

2 3 .a 0

1. . a 20 2 0

20 = 4.472

The two maxima thus occur at r

3 .a 0

2.236 .a 0

r

0.764 .a 0

and

r

5.236 .a 0 9

Let's plot again. 1

R 2s ( r )

3

2 . 2 .a 0 r

P 2s ( r ) R 2s ( r )

r . e a0

. 2

2

r 2 .a

0

P 2s ( r )

2 2 r .R 2s ( r )

0 , .05 .. 35

0.2

You can clearly see the two maxima in this plot. Notice also how P(r) is minimum at r=0 and at r=2a0.

0

0.2 0

2

4

6 r

8

10

12

Let's take the derivative and see better where P is zero. d D 2s ( r ) P ( r) d r 2s 0.2

0.2

D 2s ( r ) 0.1

D 2s ( r ) 0.1

0

0

0 0.1

0 0.1

0

1

2 r

4

5 r

6

These two derivative plots show the first maximum, the minimum at r=2a0, and the second maximum. ___________________________________________________________________________

10

Problem 6.19 We simply want to calculate the ratios P a0 a0 P 2

P a0 P 2 .a 0

and

for a 1s electron.

Remember, P ( r)

2 2 r .R ( r )

and for the 1s electron, r

R ( r)

a0

A .e

where

A

2 3

a0

2

2 .r

Thus, P a0 a0 P 2

a0 2 2 r .A .e

where the term in the numerator is evaluated at r=a0 and the term in the denominator is evaluated at r=a0/2

2 .r a0 2 2 r .A .e

The ratio is

2 .a 0

P a0 a0 P 2

2 2 a 0 .A .e

2 .a 0

2

a0 2

a0

.A 2 .e

2 .a 0

2 P a0 4 .e 1 a0 e P 2 The ratio is 4e-1=4/e.

The second part is similar: 2 .r

P a0 P 2 .a 0

a0 2 2 r .A .e 2 .r

where the term in the numerator is evaluated at r=a0 and the term in the denominator is evaluated at r=2a0.

a0 2 2 r .A .e

11

2 .a 0 2 2 a 0 .A .e

P a0 P 2 .a 0

4 .a 0

4 .a 0 P a0 P 2 .a 0 P a0 P 2 .a 0

a0

e

2

.A 2 .e

a0

2

4 .e

4

2

e 4

___________________________________________________________________________ Problem 6.20 I'll let Mathcad do it first. After that, I'll use Mathcad to simulate the paper and pencil solution. a0

1

I already defined a0 like this before, but I'll do it again here for you to see r

2

R 1s ( r )

3

a0

a .e 0

2

∞ expect

2 R 1s ( r ) .r .R 1s ( r ) .r d r

0 expect = 1.5 50 expect

In earlier versions of Mathcad, this was zero. I wonder why? 2 R 1s ( r ) .r .R 1s ( r ) .r d r

0 expect = 1.5

I knew the answer, of course. The integrand peaks up sharply near zero, and goes exponentially to zero for relativly small r (a few Bohr radii). When Mathcad integrates from 0 to ∞, it does the integral numerically, sees an integrand of zero most everywhere, and gets an answer of zero. When I cut the integral off at 50 Bohr radii, Mathcad happily returns the correct answer of 1.5a0. 12

The "paper and pencil" solution: ∞

star .

< r 1s >

R 1s

r .R 1s .r d r 2

0 ∞

2 3 R 1s .r d r

0 ∞

2 .r

4 . a0. 3 e r dr 3

a0 0 ∞

2 .r 3 a0 r .e dr

4 . 3

a0

0

From table of integrals: ∞

1.

n

x

x

Γ ( n)

e dx

0 Where Γ(n)=(n-1)! for n an integer (! is the factorial function). Let x=2r/a0. Then

3

3

r

a0 8

.x3

and

dr

a0 2

.dx

With these substitutions, ∞ < r 1s >

a .x3 .e x . 0 d x 8 2

3

a0

3

a0

4 . 0

4 ∞ 4 .a 0 . 3 a 0 16 0

3 x x .e d x

13

a0

.Γ ( 3 )

4 a0

.( 3 .2 .1 )

4 3 .a 0 2

1.5 .a 0 ___________________________________________________________________________ Problem 6.21 Part a. Calculate the probabilituy of finding a 1s electron at a distance greater than a0 from the nucleus. ∞ 2 2 P r> a 0 R .r d r a0 ∞

2

r

2 3

a0

a .e 0

.r2 d r

2

a0 ∞ 4

a0 2 r .e dr

. 3

a0

2 .r

a0

From table of integrals:

2.

x e

β .x

dx

e

β .x

2

. x β

2 .x β

2

2 β

3

14

2.

r e

2 .r

2 .r

a0

a0

dr

2 .r

e

a0

e

2 r .a 0 . 2

2 r .a 0 . 2

2 2 .r .a 0

2 .a 0

4

8

r .a 0

2 .a 0

2

8

2

3

3

Evaluate the above expression at ∞ and a0 and take the difference to find the value of the integral. Note that lim(e-2r/a0)=0 so that the result is zero at the upper limit, leaving r->∞ 3

P r> a 0

P r> a 0 P r> a 0

4 . 2. a 0 e 3 2 a0

a0 2

3

2 .a 0

3

8

4

. e 2 . 5 .a 3 0 3 4 a0

5 .e

P r> a 0

2

0.68

That is, there is a 68% chance of finding the electron outside a0. Part b. Find the probabilituy that r>2a0. This is identical to part a except that we now have a lower limit of 2a0 on the integral. I will jump directly to the step where we plugged a0 into the expression in the integral, except now I will plug in 2a0. 4 .a 0

P r > 2 .a 0

4 .a 0

2 .a 0

a0

3

2

2

4

4

13 .a 0

4 . e

a0

3

.

a0

3

3

2

P r > 2 .a 0

. e4 .

3

a0

P r > 2 .a 0

13 .e

P r > 2 .a 0

0.24

4

4

There is a 24% chance of finding the 1s electron further away than 2a 0. ___________________________________________________________________________ 15

Problem 6.22 We need to calculate P 2s r < a 0 a0 P 2s r < a 0

and P 2p r < a 0 2 2 R 2s .r d r

0 a0 P 2p r < a 0

2 2 R 2p .r d r

0 For the 2s state, the probability is (plugging in the 2s radial wave function): a0 P 2s r < a 0

r 2 .a

r . e a0

2

2

.r2 d r .

0

0

1 8 .a 0

3

That 1/8a03 term comes in because I almost forgot to include the normalization constant. Let x=r/a0 so that dx=dr/a0. With these substitutions, the above integral becomes P 2s r < a 0

1 1 . 8 0

2 x .( 2

2 x x ) .e d x

We use these two integrals to do the calculation: 1

2 x x .e d x

e

3 x x .e d x

e

x

. x2

2x

x

. x3

3 .x

2

0 1

2

6x

6

0 Skipping a bit through the algebra... P 2s r < a 0

1 . 8 8

20 e

24

64 e

24

65 e 16

P 2s r < a 0

0.0343

A little over a 3% chance.

For the 2p state, the probability is (plugging in the 2p radial wave function): a0 P 2p r < a 0

r a0 4 r .e dr .

0

1 24 .a 0

3

Let x=r/a0 as above. Then P 2p r < a 0

1 1 . 24 0

4 x x .e d x

We use the two integrals from above plus the one below to do the calculation: 1

4 x x .e d x

x

. x4

1 . 24 24

65 e

e

4 .x

3

12 .x

2

24x

24

0 The result is P 2p r < a 0 P 2p r < a 0

0.00366

About a factor of 10 less than for the 2s.

___________________________________________________________________________ Problem 6.24 A hydrogen atom is in the 4p state. To what state or states can it go by radiating a photon in an allowed transition? Solution will be posted before you are tested on this one.

17

18

Problem 7.3 hbar := 1

For convenience, set

The spin angular momentum has a magnitude S :=

3 2

⋅ hbar

The z-component of the spin angular momentum has a magnitude Sz :=

hbar

or

2

Sz :=

−hbar 2

note this second definition of S.z is "turned off in Mathcad"

Now look at Figure 7.2 in Beiser. The angle between the z-axis and the spin angular momentum vector corresponding to ms=1/2 is

 Sz  This is in radians. S

θ := acos

θ := θ ⋅

360 2⋅ π

θ = 54.736

Now look at Figure 7.2 in Beiser. The angle between the z-axis and the spin angular momentum vector corresponding to ms=-1/2 is

 −Sz   S 

θ := acos

θ := θ ⋅

360

θ = 125.264

2⋅ π

_______________________________________________________________________________ Problem 7.9 Remember

s p d f l= 0 1 2 3

so an f subshell corresponds to l=3.

There are 7 possible m.l values for an electron in this subshell: ml := −3 , −2 , −1 , 0 , 1 , 2 , 3 For each m.l, there are two possible values of m.s: ms := −

1 2

ms := or

+

1  2

That gives a total of 7*2=14 electrons in an f subshell. _______________________________________________________________________________ Problem 7.11 See near the end for a simple solution (one that you would want to use on an exam. To see what's going on, let's make a table of possible electron configurations. The table is too big to fit on this page, so I'll continue on the next.

1

n= 6

l= 5 4 3 2 1 0

# of m.l's 11 9 7 5 3 1

# of m.s's 2 2 2 2 2 2

# of elements 11*2=22 9*2=18 7*2=14 5*2=10 3*2=6 1*2=2

5

4 3 2 1 0

9 7 5 3 1

2 2 2 2 2

18 14 10 6 2

4

3 2 1 0

7 5 3 1

2 2 2 2

14 10 6 2

3

2 1 0

5 3 1

2 2 2

10 6 2

2

1 0

3 1

2 2

6 2

1

0

1

2

2

The answer is the sum of all the numbers in the last column: N := 22 + 18 + 14 + 10 + 6 + 2 I do this to keep N from running off the side of page. N := N + 18 + 14 + 10 + 6 + 2 N := N + 14 + 10 + 6 + 2 + 10 + 6 + 2 + 6 + 2 + 2 N = 182 Here's the simpler way. Each shell can contain 2n2 electrons. 2

2

2

2

2

N := 2⋅ 1 + 2⋅ 2 + 2⋅ 3 + 2⋅ 4 + 2⋅ 5 + 2⋅ 6

2

N = 182 _______________________________________________________________________________ Problem 7.16 The effective nuclear charge that acts on the outer electron in the sodium atom is 1.84 e. Use this figure to calculate the ionization energy of sodium. This is just example 7.2 turned around. We are given Also E1 := 13.6

Z := 1.84 This energy has a positive sign because it is the ionization energy.

The outer electron is sodium is a 3s electron,

n := 3

2

En :=

Z ⋅ E1 n

2

En = 5.116 eV

2

_______________________________________________________________________________ Problem 7.40 What element has a Kα x-ray line of wavelength 0.144 nm? 3 ⋅ c⋅ R⋅ ( Z − 1 )

We use f :=

2

4

and c := f ⋅ λ To derive an equation for Z in terms of λ. 2

( Z − 1) :=

Z := 1 +

4⋅ f

Z − 1 :=

3 ⋅ c⋅ R 4⋅ c

3⋅ c⋅ R

Z := 1 +

4⋅ f 3 ⋅ c⋅ R

4

Z := 1 +

3 ⋅ c⋅ R ⋅ λ

4⋅ f

3 ⋅ R⋅ λ

Now define R and turn the equation box into a real equation. 7

R := 1.097⋅ 10 Z( λ ) := 1 +

(

4 3 ⋅ R⋅ λ

) = 30.053

−9

Z 0.144⋅ 10

This corresponds to zinc _______________________________________________________________________________ Problem 7.41 Find the energy and the wavelength of the K α x-rays of aluminum. Aluminum has

Z := 13

E := 10.2⋅ ( Z − 1 ) 8

3

E = 1.469 × 10 eV 7

c := 3 ⋅ 10 f :=

2

R := 1.097⋅ 10

3 ⋅ c⋅ R⋅ ( Z − 1 )

2

4 17

f = 3.554 × 10 Hz Just as a quick double-check: heV := 4.14⋅ 10

− 15

EeV := h eV⋅ f 3

EeV = 1.471 × 10 double-checks OK λ :=

c f

λ = 8.441 × 10

− 10

3

Physics 107

Problem 8.1

O. A. Pringle

The energy needed to detach an electron from a hydrogen atom is 13.6 eV, but the energy needed to detach an electron from a hydrogen molecule is 15.7 eV. Why do you think the latter energy is greater. In H2, each electron experiences an attractive force due to two protons. This attractive force is greater than the attractive force due to the single proton in H. Of course, the electrons also repel each other, but because the electrons are more spread out in space (each occupying a 1s orbital and staying as far away from each other as possible), the net energy increase due to electron-electron repulsion is less than the net energy increase due to electron-proton attraction.

Physics 107

Problem 8.2

O. A. Pringle

The protons in the H2+ molecule are 0.106 nm apart, and the binding energy of H2+ is 2.65 eV. What negative charge must be placed halfway between two protons this distance apart to give the same binding energy? Here is another statement of the problem: −2.65eV := repulsive + attractive where "repulsive" is the Coulomb energy between the two protons and "attractive" is the Coulomb energy between the negative charge and the two protons. So we need to solve −2.65⋅ e :=

e⋅ e 4 ⋅ π ⋅ ε 0⋅ R

q⋅ e

+ 2⋅

4 ⋅ π ⋅ ε 0⋅

Eq. 1 R

2 where e is the charge on the proton and q is the negative charge placed in between, and I am using e to stand for the MAGNITUDE of the charge on the proton. The 2.65*e converts eV to joules − 19

e := 1.6⋅ 10

Define parameters:

ε 0 := 8.85⋅ 10

− 12

−9

R := 0.106⋅ 10

Solve Eq. 1 for q; this is better done on paper.

q :=

π ⋅ ε 0⋅ R e

 

⋅ −2.65⋅ e −

e

2



4 ⋅ π ⋅ ε 0⋅ R 

− 20

q = −4.781 × 10

Coulombs

In terms of e, q e

= −0.299 So the charge to be placed in between is 0.299 times the charge on the electron.

1

Problem 9.1 This is a variation on the example on pages 300-301. n ε2

ε2 ε1 k .T

g ε2 .exp g ε1

n ε1

We are given that n(ε.2)/n(ε.1)=1/1000 and we want to solve for T. Also, we know that g(ε.1)=2 and g(ε.2)=8. We want to solve 1

8.

1000

2

ε2

exp

ε1

for T.

k .T

I'll set this up more generally. n ε2 g ε1 . n ε1 g ε2 ε1

k .T

ε2 ε1 k .T

exp

ε2

ln

k .T

ε2

ε1

T

n ε2 g ε1 . n ε1 g ε2

ln

ε1

k .ln

n ε2 g ε1 . n ε1 g ε2

ε2

n ε2 g ε1 . n ε1 g ε2

Now substitute values and get the answer. ε1 g1

13.6

ε2

13.6 2

g2

8

ε1

T

k .ln

n1

4

k

1000

n2

8.617 .10

5

1

since we are working with eV ε's

ε2

n 2 g1 . n 1 g2

4 T = 1.427 .10

____________________________________________________________________________ Problem 9.2 This is another variation on the example on pages 300-301. n ε2 n ε1

g ε2 .exp g ε1

ε2 ε1 k .T

We are given T=5000 and we want to solve for n(ε.2)/n(ε.1), etc. n

1 .. 4

ε( n )

13.6 n

T

2

g( n )

2 .n

2

k

8.617 .10

5

since we are working with eV ε's

5000

1

g( m ) .

ratio( m , n )

exp

( ε( m) k .T

g( n )

ratio( 1 , 1 ) = 1

ε( n ) )

(just checking)

ratio( 2 , 1 ) = 2.092 .10

10

ratio( 3 , 1 ) = 5.871 .10

12

ratio( 4 , 1 ) = 2.25 .10

12

____________________________________________________________________________ Problem 9.7 1

vbar

3

vbar = 2

2 1

vrms

2

3

2

m/s

vrms = 2.236

2

m/s

____________________________________________________________________________ Problem 9.8 T

273

20

k

8.62 .10

E

3. . kT 2

5

Boltzmann's constant in units of eV/K The average energy per molecule is E = 0.038

eV

This is much less than the 10.2 eV needed to raise a hydrogen atom from its ground state to its first excited state: E 10.2

= 3.71421 .10

3

____________________________________________________________________________ Problem 9.9 k 8.62 .10 Boltzmann's constant in units of eV/K E 13.6 Binding energy of hydrogen in eV 5

We need to solve E T

2 .E 3 .k

T = 1.052 .10

5

3. . kT 2

for T.

or about 10500 K

____________________________________________________________________________

2

Problem 9.10 T

20

273

h

m

2 .16 .1.66 .10

E

3. . kT 2

4 .10

10

k

1.38 .10

= 0.065

23

Boltzmann's constant in mks units

mass of oxygen

λ = 2.612 .10

p λ

34

average oxygen molecule energy

h

λ

27

6.63 .10

11

p

2 . m .E

oxygen momentum

de Broglie wavelength

so the de Broglie wavelength is only about 6.5 percent of the molecular diameter

____________________________________________________________________________ Problem 9.14 The flux is the number of neutrons per square meter per second. If we divide the flux by the average neutron velocity, we get the number of neutrons per cubic meter at any time in the beam port: neutrons

neutrons . seconds_for_neutron_to_travel_1_meter 2 m .s

volume mn

1.675 .10

27

1_meter T

300

k

1.38 .10

23

S

12

10

According to Maxwell-Boltzmann statistics: 8 .k . T π .m n

vbar

ρ

S

8 ρ = 3.986 .10

neutrons/m3 ____________________________________________________________________________ vbar

Problem 9.19 According to the Stefan-Boltzmann law, an object at a temperature T radiates an energy R per second per unit time R

e .σ .T

4

where e is the object's emissivity and σ is a constant. For skin at two temperatures, 4

R2

T2

R1

T1 4 T2

ratio

4

T1

T2

273

35

T1

273

34

4

ratio = 1.013 A 1.3 percent difference; not huge, but measurable. ____________________________________________________________________________

3

Problem 9.20 This is "just like" problem 9.19. T2

5800

ratio

T1

T2

4

T1

4

5000

ratio = 1.811

almost a factor of two different

____________________________________________________________________________ Problem 9.23 e .σ .T 1

R1 R2

4

2

R1 T1

400

T2

4 T 1 .2

R2

e .σ .T 2

R2

T2

4

R1

T1

4

4

273 1

T 2 = 800.336

K or

T2

273 = 527.336

C

____________________________________________________________________________ Problem 9.24 e

0.3

T

R

e .σ .T

400

273

σ

5.67 .10

8

4

R = 3.49 .10

3

This is in units of watts/square meter

To find the rate at which this particular sphere radiates, we need to multiply R by the surface area of the sphere. r

5.

10

2

2 rate

R .4 . π . r

2

rate = 27.407

watts

____________________________________________________________________________ Problem 9.25 In the absence of a given emissivity, let's treat the hole as a blackbody, so e 8 that σ 5.67 .10 Also T 700 273 R

e .σ .T

1

4

4

R = 5.082 .10

4

This is the number of watts/m2 radiated by the hole. To find the rate at which radiation escapes from the hole, multiply R by the area of the hole. Don't forget to convert cm2 to m2. R .10 .10

rate

4

rate = 50.82

watts ____________________________________________________________________________ Problem 9.26 R2

e .σ .T 2

4

R1

e .σ .T 1

4

R1

1.00

R2

R 1.

T1 T2

4

T1

4

R 2 = 3.067

500

273

T2

750

273

kW

____________________________________________________________________________ Problem 9.30 According to Wien's displacement law, 3 λ max .T 2.898 .10 in MKS units T

500

λ max

273 2.898 .10

3

T

λ max = 3.749 .10

6

meters

This wavelength of 3749 nanometers is lower energy than the lowest energy visible photon, which has a wavelength of about 400 nm. Thus, these photons are in the infrared. See figure 2.2 on page 51 for a confirmation of this. ____________________________________________________________________________ Problem 9.31 λ max T

290 .10

2.898 .10 λ max

9

3

equation 9.40, page 316

3 T = 9.993 .10

about 10000 K

____________________________________________________________________________ 5

Problem 9.33 T

34

273 2.898 .10

λ max

3

equation 9.40, page 316

T

λ max = 9.44 .10

6

about 9440 nanometers

This wavelength is in the infrared; see figure 2.2 on page 51. ____________________________________________________________________________ Problem 9.38 The average energy of the free electrons in copper can be found from equation 9.51 εF

7.04 3.

ε avg

5

εF

ε avg = 4.224

eV

The average energy of free electrons at room temperature, according to Maxwell-Boltzmann statistics, is ε bar

3. . kT 2

Beiser says to use kT=0.025 eV, so ε bar

3.

ε bar = 0.038 eV

0.025

2

The actual free electron energy in copper is bigger by a factor of ε avg

= 112.64

ε bar

than the average energy of a gas of free electrons at room temperature. ____________________________________________________________________________ Problem 9.39 Part a. The average energy is given by equation 9.51. εF

5.51

ε avg

3. 5

εF

ε avg = 3.306

eV

Part b. The temperature corresponding to this energy is given by equation 9.13 k

8.617 .10

T

2 .ε avg 3 .k

5

Boltzmann's constant in eV energy units 4 T = 2.558 .10

or about 25600 K 6

Part c. It's safe to do a classical calculation of the velocity of an electron having an energy of 3.306 eV, because the energy is very much less than the electron rest energy of 511 keV. ε avg . 1.6 .10

E

19

first convert energy to joules

From E=mv2/2, we get 2 .E

v

9.11 .10

6 v = 1.078 .10

31

m/s

a fairly high, but still nonrelativistic velocity

____________________________________________________________________________ Problem 9.40 Part a. Use equation 9.50. This is relatively easy in Mathcad, not so easy with paper and pencil. Let's just pick N=1 N εF T

1

1.6 .10

e

19

k

1.38 .10

23

7.04 .e

This is what Beiser uses for a warm "room" temperature.

300 3.

n( ε , T )

N . εF 2 ε

exp

1.5

εF k .T

. ε

See the end of the document for Beiser's solution.

1

Just for kicks, let's plot n(ε) ε F = 1.126 .10 ε

10

19

This gives me a reference for scaling ε

18

, 1.02 .10

19

.. 1.5 .10

18

set a range of energies

This looks reassuringly like figure 9.10. The numbers are big because I haven't multiplied by a dε.

n ε ,T

ε

Electrons in excited states are those with ε>ε.F. Since we defined N=1, we can find the fraction in excited states by integraging n(ε) from ε.F to infinity. 7

∞

I did this on purpose. Mathcad gripes because some numbers involved in calculating n(ε) are getting too big, so I should integrate not to ∞, but to some "big" number, say 3 times ε.F

n( ε , T ) d ε

fraction εF

3 .ε F

If I go any bigger than 3ε.F, I get an overflow.

n( ε , T ) d ε

fraction εF fraction = 3.83 .10

3

about 0.4%

Part b. At T=1083K, T

1083

273

n ε ,T

ε

10 .ε F

n( ε , T ) d ε

fraction εF fraction = 0.017

Now I can go higher in energy without overflowing.

About 1.7%

The above solution is too much for a paper-and-pencil type test problem. All of the excited electrons will have energies just slightly greater than ε.F. Therefore, it is a reasonable approximation to replace ε

in equation 9.50 by ε F Also, for ε greater than εF, the 1 in the denominator of equation 9.50 is much smaller than the exponential. The equation then becomes

n( ε )

3.

εF N . 1 .exp ε ε F .exp . 2 k .T kT

8

The antiderivative of n(ε) is ε . 3 . N .ε 1 .exp F .exp ε F 2 1 k .T k .T . kT 1

When you integrate from ε.F to infinity, the value of the antiderivative is 0 at ∞, and evaluated at ε.F it gives integral( T ) integral( T )

εF εF N 1 .exp ( k .T ) . 3 . . .exp 2 εF k .T k .T N 1 ( k .T ) . 3 . . 2 εF

Evaluating at the two temperatures: integral( 300 ) = 5.5131 .10or

3

about 0.55%

1083 ) = 0.025

or about 2.5% I would expect that if I gave you the appropriate hints as to what approximations to make, that you could do this calculation for an exam. ____________________________________________________________________________ integral( 273

Problem 9.44 The Fermi energy is calculated from εF

or ε F

2 h . 3 .N 2 . m 8 .π . V

3 .n

2

h . 2 . m 8 .π

2 3

2 3

where n=N/V is the density of free electrons. We can get the number of aluminum atoms per volume by multiplying the density by the atomic mass: atoms

n

volume

mass . atom volume mass

Don't forget to convert grams to kilograms and centimeters to meters. According to Table 7.4, aluminum has 2 3s and 1 3p electron. Beiser tells us that the 3s and 3p electrons are close in energy, so we expect all three electrons to become part of the free electron gas. Thus, we need to multiply the number of atoms by 3 to get the number of free electrons in aluminum. h

6.63 .10

34

m

9.11 .10

31 .

0.97 here's

where effective mass enters in

9

n

3 .2.70 .

1 . 2 3. 1 . 1 10 3 26.97 1.66 .10 10

27

The 1/103 converts g-->kg, the (102)3 converts cm3-->m3, and the1/1.66*10-27 converts amu-->kg. εF

2

3 .n

h . 2 . m 8 .π

ε F = 1.929 .10 εF 1.6 .10

19

2 3

18

= 12.055

Joules Fermi energy in eV

10