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Modern Physics for Scientists and Engineers Solutions Joseph N. Burchett after consultations with John C. Morrison August 4, 2010

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2

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Introduction - Solutions 1. We are given a spherical distribution of charge and asked to calculate the potential energy of a point charge due to this distribution. There are two situations to consider which deliver quite different results. The problem does not specify whether the point charge of interest is inside or outside the charge distribution so we must take both possibilities into consideration. Before we get started, let’s set up an appropriate description of our charge distribution so that we may proceed with each of the above situations. It is most useful to express the charge distribution in terms of a charge density ρ: ρ=

Q 4 πR3 3

This is the charge per unit volume as found by merely dividing the total charge Q by the total volume of the sphere. We now handle the “outside the sphere” configuration. All of the charge inside the sphere may be considered as though it is concentrated at a single point. We may then use the Coulomb’s Law result for the force acting on the point charge q: F =

1 Qq 4π0 r2

We integrate to find the potential energy: Z r Qq 1 Qq 1 V =− dx = 02 4π0 ∞ r 4π0 r This result is also identical to the one obtained in the text in the case of the two-point charge configuration. We see that this should in fact be the case since we started with the same quantity of force. Now, to handle the “inside the sphere” situation, we shall use the above stated charge density. We easily see that the distance r from the center of the 3

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4 distribution of charge to the point charge is less than the radius R. Imagine now a sphere of radius r0 that is inside the larger spherical charge distribution but centered at the same point. Only the charge within this inner sphere will act on the point charge. We may now use our charge density to find the charge contained in the smaller sphere: 4 Qin = πr03 ρ 3 The force on a charge q due to the inner sphere is then given by Coulomb’s Law: q 4πr03 ρ qρr0 F = = 4π0 3r02 30 Before we integrate to find the potential energy, we must make an important distinction from the “outside the sphere” procedure above. We were able to integrate directly from infinity to the location of our point charge because the amount of source charge taken into account did not change (Q). However, as we enter the sphere, the amount of source charge begins to decrease until we reach our point at distance r from the center. We split the integration into two parts corresponding to both regions to find the potential energy: Z r Z R qρr0 0 Qq 1 0 dr + dr ] V = −[ 02 R 30 ∞ 4π0 r ρq 2 Qq 1 − (r − R2 ) 4π0 R 60 Note that the first integral above was identical to the one for the outside configuration, just evaluated at r = R. Now, we just substitute our charge density ρ and simplify: =

V =

Qq 1 Qqr2 Qq − + 3 4π0 R 8π0 R 8π0 R =

3Qq Qqr2 − 8π0 R 8π0 R3

3. Looking at the definition of kinetic energy: 1 KE = mv 2 2

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5 Let’s solve for velocity: r v=

2(KE) m

Let KE 0 = 4KE. r v=

2(KE 0 ) = m

r

r 2(4KE) 2(KE) =2 m m = 2v

So, the speed will increase by a factor of 2 if the kinetic energy is increased by a factor of 4. By the definition of momentum: p = mv ⇒ m(2v) = 2p We should also expect the momentum of the particle above to increase by a factor of 2. In fact, as long as the mass does not change, the momentum should increase or decrease by the same factor as the changing speed. 5. The frequency and wavelength of light are related by equation I.23: c f= λ The constant c refers to the speed of light which we know to be 3.00×108 m/s. Taking care to convert our wavelength, which is given in nanometers, to meters: 3.00 × 108 ms = 6 × 1014 Hz f= −9 500 × 10 m 7. Equation I.25 relates the photon energy to the wavelength: hc λ Substituting the value hc = 1240 eV ·nm: E=

E=

1240 eV ·nm = 2.48 eV 500nm

9. The energy of a quantum of light (the photon) must be equal to the difference in energies of the two states: Ephoton = E2 − E1

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6 According to equation I-25, the energy also obeys the following: Ephoton =

hc λ

Equating the two and solving for λ, we get: λ=

hc E2 − E1

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1 The Wave-Particle Duality Solutions 1. The energy of photons in terms of the wavelength of light is given by eq. 1.5. Substituting the given wavelength into eq. 1.5: Ephoton =

hc 1240 eV · nm = = 6.2 eV λ 200nm

3. Considering that we are given the power of the laser in units of milliwatts, the power maybe expressed as: 1 mW = 1 × 10−3 J/s. This gives us a strong hint how to proceed. We are hoping to find the rate of emission of photons, so that if we can find the energy of a single photon, we can use the power as stated above to calculate the number of photons. The energy of a single photon can be calculated as in Problem 1 by substituting the wavelength of the light into equation 1.5: hc 1240 eV · nm = = 1.960 eV λ 632.8 nm We now convert to SI units: 1J 1.960 eV × = 3.14 × 10−19 J −19 1.6 × 10 eV Ephoton =

Now, using the given laser power: J 1 × 10−3 s·photon power photons Rate of emission = = = 3.18×1015 −19 energy per photon 3.14 × 10 J s

7

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8

1. THE WAVE-PARTICLE DUALITY - SOLUTIONS

5. This problem concerns the photoelectric effect. Given the work function of the material and the emitted electron kinetic energy, we wish to calculate the wavelength of the light incident to the surface. Equation 1.6 provides the following: (KE)max =

hc −W λ

where W is the work f unction of the material.

The hc/λ term describes the energy supplied by the incoming photons. By viewing the work function as an energy threshold for producing photoelectric current, we see that the amount by which the photon energy (hc/λ) exceeds the work function is the resultant maximum kinetic energy of the emitted electrons. We may thus write: hc = (KE)max + W λ = 2.3 eV + 0.9 eV = 3.2 eV Using hc = 1240 eV · nm, λ=

1240 eV · nm = 387.5 nm 3.2 eV

6. Here, we are given the stopping potential of a photoelectric experiment and wish to determine the work function of the metal. Since 0.72 eV is the necessary potential energy to cease the flow of electrons, the maximum kinetic energy of the electrons being emitted must equal 0.72 eV. Our problem is then reduced to solving eq. 1.6 for the work function: W =

hc 1240 eV · nm − (KE)max = − 0.72 eV = 1.98 eV λ 460 nm

11. For this atomic transition, the energy of the emitted photon must equal

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9 the difference in energy of the two states of hydrogen (n = 2 and n = 5). Equation 1.22 gives us those energies, thus: Ephoton = E5 − E2 = −

13.6 eV 13.6 eV − (− ) = 2.86 eV 2 5 22

From eq. 1.12: λ=

1240 eV · nm hc = = 434.2 nm ∆E 2.86 eV

15. We must first find the energy of the UV photons, which may be done via eq. 1.5: hc 1240 eV · nm Ephoton = = = 27.6 eV λ 45 nm If electrons are to be emitted, we need to overcome the energy that is keeping them in a bound state. For hydrogen atoms, equation 1.22 gives us that energy. We need only supply the n quantum number for the state of the atom (for ground state, n = 1): E=−

13.6 eV 13.6 eV =− = −13.6 eV 2 n 12

The kinetic energy of these emitted electrons should then be equal to the difference between the energy provided by the incident light and the ground state electron energy: KE = 27.6 eV − 13.6 eV = 14.0 eV For the velocity calculation, it will be useful to convert from eV to J: 14.0 eV ·

1.6 × 10−19 J = 2.24 × 10−18 J 1 eV

(1.1)

We can then find the velocity using the definition of kinetic energy and the mass of the electron: s r 1 2(KE) 2.24 × 10−18 J m KE = mV 2 ⇒ v = = = 2.21 × 106 (1.2) −31 2 m 9.11 × 10 kg s

17. It should be first noted that the wavelength of the emitted light increases

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10

1. THE WAVE-PARTICLE DUALITY - SOLUTIONS

as the photon energy decreases. Thus, the longest acceptable wavelength will correspond to a transition to the nearest excited state. Transitions to higher-lying states will require more energy and thus correspond to a shorter wavelength. We then examine the energy of the transition from the n = 3 state of the hydrogen atom. Equation 1.22 gives: E3 = −

13.6 eV 13.6 eV =− = −1.51 eV 2 n 32

The next available state is the state corresponding to n=4 whose energy may be found by eq. 1.22 again: E4 = −

13.6 eV 13.6 eV =− = −0.85 eV 2 n 42

So, in order for the absorption to take place, the incident photons must contain at least as much energy as the difference in these energies (−0.85 eV − (−1.51 eV ) = .66 eV ). We may now solve eq. 1.5 for the wavelength: λ=

1240 eV · nm hc = = 1878 nm ∆E .66 eV

Thus, any light of wavelength greater than 1878 nm would not possess sufficient energy to be absorbed by the hydrogen atom. 19. Using a procedure such as that of Example 1.5, we can relate the kinetic energy to the de Broglie wavelength by the following equation: h

λ= p

2m(KE)

We need only convert the kinetic energy from eV to J and substitute the appropriate masses. KE = 20 eV × Electron : λ = p

1.6 × 10−19 J = 3.2 × 10−19 J 1 eV

6.626 × 10−34 J·s 2(9.11 ×

10−31 kg)(3.2

×

10−18 J)

6.626 × 10−34 J·s

P roton : λ = p

2(1.67 ×

10−27 kg)(3.2

×

10−18 J)

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= 2.74 × 10−10 m = 6.41 × 10−12 m

11 An α-particle is the most commonly occurring isotope of the helium atom consisting of two protons and two neutrons. We can thus approximate its mass: m = 2(1.672 × 10−27 ) + 2(1.674 × 10−27 ) = 6.692 × 10−27 kg. 6.626 × 10−34 J·s = 3.20 × 10−12 m α − particle : λ = p 2(6.692 × 10−27 kg)(3.2 × 10−18 J)

24. The Davisson-Germer experiment measured the scattering of electrons by a crystal during which the interference patterns characteristic of light were shown to also occur with particles. We can solve Bragg’s Law (eq. 1.24) for the inter-atom spacing of the crystal in terms of the de Broglie wavelength and the scattering angle: 2d sin(θ) = nλ ⇒ d =

nλ 2 sin(θ)

We may use the relation derived in example 1.5 to find the de Broglie wavelength from kinetic energy after first converting our kinetic energy to SI units: 1.6 × 10−19 KE = 54 eV × 1 eV −10 h 6.626 × 10 m λ= p =p = 1.66 × 10−10 m −31 −18 2m(KE) 2(9.11 × 10 kg)(8.64 × 10 J) We let n=1 and solve Bragg’s Law for d, the spacing in the crystal: d=

1.66 × 10−10 m λ = = 1.09 × 10−10 m 2 sin(θ) 2 sin(50◦ )

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12

1. THE WAVE-PARTICLE DUALITY - SOLUTIONS

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2 The Schr¨ odinger Wave Equation - Solutions 1. Equation 2.17 gives the energies for the particles in an infinite potential well: n2 h2 E= 2mL2 Part (a) asks for the size of the region, which we can find by solving eq. 2.17 for L: r n2 h2 L= 8mE We need only supply some information such as the energy (1.0 eV as given), the mass of the electron (9.11 × 10−31 kg from Appendix A), and the appropriate value of n. The lowest energy will be found when the electron is at the ground state (n=1). We must also convert 1.0 ev to 1.6 × 10−19 J. Now we are ready: s 12 (6.63 × 10−34 J·s)2 L= = 6.14 × 10−10 m 8(9.11 × 10−31 kg)(1.6 × 10−19 J) While the derivation of eq. 2.17 is correct, let’s do a little dimensional analysis to better “see through” our length calculation above. The quantum number n is dimensionless as is of course the constant 8, so let’s examine the operations of units of the rest of the quantities: s s r n2 h2 J 2 · s2 J · s2 ⇒ = 8me kg · J kg 13

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14

¨ 2. THE SCHRODINGER WAVE EQUATION - SOLUTIONS 2

Now substitute 1 J = 1 kg ms2 · s2 : s

s2

J· = kg

s

2

kg ms2 · s2 √ 2 = m =m kg

It should be emphasized that the m here is the abbreviation for the unit of length meters, not the quantity of mass. Part (b) can be approached in a couple of different ways. The essential quantity we must obtain now is the energy of the first level above ground state (also called the first excited state). The energy required to excite the electron must be: ∆E = En2 − En1 = E2 − E1 The quantum numbers n=2 and n=1 correspond to the first excited state and ground state respectively. So, for E2 from equation 2.17: E2 =

(6.63 × 10−34 J·s)2 22 h2 = 8mL2 2(9.11 × 10−31 kg)(6.14 × 10−10 m)2 1 eV 1.6 × 10−19 J = 4.00 eV

= 2.14 × 10−18 J ×

Here we used the result of the size of the well from part (a). The other method would involve solving eq. 2.17 for everything except n and using our given quantities as follows: h2 E n2 h2 ⇒ = 2 E= 2 2 2mL 8mL n h2 = (1.0 eV )(12 ) = 1.0 eV 8mL2 En = n2 (1.0 eV ) ⇒ E2 = 4.0 eV Now for our desired result: ∆E = 4.0eV − 1.0eV = 3.0eV

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15 3. As with the atomic transitions we dealt with in chapter 1, the emitted photon energy will equal the difference of the energies of the two states corresponding to n = 3 and n = 2. So our task becomes finding each of these energies via equation 2.17: E3 =

32 h2 9(6.63 × 10−34 J·s)2 = 8mL2 8(9.11 × 10−31 kg)(10 × 10−9 m)2

= 5.43 × 10−21 J · E2 =

1 eV = .034 eV 1.6 × 10−19 J

22 h2 4(6.63 × 10−34 J·s)2 = 8mL2 8(9.11 × 10−31 kg)(10 × 10−9 m)2

= 2.41 × 10−21 J ·

1 eV = .015 eV 1.6 × 10−19 J

∆E = .034 eV − .015 eV = .019 eV This value of ∆E will then be equal to the energy of the photon and we may use eq. 1.5 to calculate the wavelength of the light: KE =

hc hc ⇒λ= λ KE

1240 eV · nm = 65263 nm .019 eV 5. Note the typo, “Draw the wave function...” should be “Find the wave function...” We are given the following information: finite well of depth 0.3 eV and thickness L = 10 nm, we are dealing with conduction electrons if GaAS in their lowest state, and these electrons have an effective mass of 0.067 times the electron mass. From section 2.3, we see the process of solving the solution of tanθ = p p finite square well which culminates in the numerical 2 2 2 (θ0 /θ ) − 1, for the even solutions, and −cotθ = (θ0 /θ2 ) − 1 for the odd case. Our variables θ and θ0 correspond to the following expressions: λ=

θ=

kL mV0 L2 , θ0 2 = 2 2~2

(f rom eq. 2.30)

Notice that the latter of the two is a squared quantity where the former is not, but we’ll deal with that a little later on. We are interested in the ground

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¨ 2. THE SCHRODINGER WAVE EQUATION - SOLUTIONS

16

state of the electron, which will correspond to n=1, leading us to an even solution which may be found by equation 2.29: s θ0 2 tanθ = −1 θ2 In the text near the end of section 2.3, the value of θ0 2 is given to be 13.2. Our equation becomes: r 13.22 tanθ = −1 θ2 By using a computer algebra system or calculator to graph the left-hand and right-hand sides, we find the first intersection at θ ≈ 1.22. Thus, we may solve θ = kL/2 for k: k=

2(1.22) 2θ = = 2.44 × 108 m−1 L 10 × 10−9 m

For the equations outside the well, we must solve for κ as well. Let’s first solve equation 2.24 for E: k 2 ~2 (2.44 × 108 m−1 )2 (1.054 × 10−34 J · s)2 2mE 2 ⇒E= = kg k= ~2 2m 2(6.103 × 10−32 ) E = 5.419/times10−21 J ×

1 eV = .034 eV 1.6 × 10−19 J

Now for κ (from eq. 2.26): s κ=

2(6.103 × 1032 kg)(.3 eV − .034 eV )(1.6 × 10−19 (1.054 × 10−34 J · s)2

J eV

)

= 6.84×10−8 m−1

We know have our necessary information for the wave equations:  : x ≤ − L2  B exp(6.84 × 10−8 m−1 )x A cos(2.44 × 10−8 m−1 )x : − L2 ≤ x ≤ L2 Ψ(x) =  B exp(6.84 × 10−8 m−1 )x : x ≥ L2

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17 9. Outside the finite potential well, V = V0 , thus our time-independent Schrodinger equation is: ~2 d2 ψ − + V0 ψ = Eψ 2m dx2 If we multiply both sides by −2m/~2 : d2 ψ 2mV0 2mE − ψ=− 2 ψ 2 2 dx ~ ~ Moving 2mE/~2 to the LHS: d2 ψ 2m(E − V0 ) + ψ=0 dx2 ~2 We make the following substitution:  k=

2m(E − V0 ) ~2

 12

Since E is greater than V0 , k is in fact a real number and our Schrodinger equation becomes: d2 ψ + k2ψ = 0 2 dx And, we may readily confirm by substitution that the general form of the solution of this equation is a linear combination of the functions, A cos(kx) and B sin(kx), and is oscillatory in nature. 11. The normalized wave function must satisfy the normalization condition (equation 2.18): Z ∞ |ψ(x)|2 dx = 1 −∞

Z

∞

−mωx2 \2~

Ae

Ae

−mωx2 \2~

Z

−∞

2 \~

A2 e−mωx

−∞

Therefore, 1 = A2

∞

dx =

Z

∞

e −∞

−mωx2 \~

Z dx = 2

∞

2 \~

e−mωx

dx

0

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¨ 2. THE SCHRODINGER WAVE EQUATION - SOLUTIONS

18

This looks very similar to the integral we are given if we let a = mω/~. Then: r r 1 π π~ = mω = 2 A mω ~ A=

 mω  14

π~ 13. a) From 0 ≤ x ≤ L, V = 0, and our Schrodinger equation becomes: −

~2 d2 ψ d2 ψ 2mE = Eψ or + 2 ψ=0 2m dx2 dx2 ~

For x ≥ L, V = V0 and: −

d2 ψ 2(V0 − E)m ~2 d2 ψ + V = Eψ or − ψ=0 0 2m dx2 dx2 ~2

b) If we let: r k=

2mE and κ = ~2

r

2m(V0 − E) ~2

Then, d2 ψ + k2ψ = 0 ; 0 ≤ x ≤ L dx2 d2 ψ − k2ψ = 0 ; x ≥ L dx2 Similar to the situation of the finite well discussed in the text, the above equations are thus satisfied by the following: ψ(x) = A cos(kx) and ψ(x) = A sin(kx) ; 0 ≤ x ≤ L ψ(x) = Be−κx ; x ≥ L Once again, the negative argument in the exponential is to insure our wave function decays as x → ∞. c) If the potential is to be infinite at x = 0, then the wave function must go to zero at x = 0. Therefore, we must impose new boundary conditions to insure the continuity of the wave function. Above we had both even and odd solutions for the wave function, so let’s examine their behavior at x = 0 given the new conditions: Even : ψ(0) = A cos(k0) = 0

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19 Odd : ψ(0) = A sin(k0) = 0 We see immediately that the fist of the equations is not physically acceptable, so we must reject the even solutions, thus we are left with: ψ(x) = A sin(kx) for 0 ≤ x ≤ L. We may now proceed with our boundary conditions at x = L: ψ(L) = A sin(kL) = Be−κL Imposing the continuity of the first derivative: Ak cos(kL) = −Bκe−κL Divide eq. 2 by eq. 1: k cot(kL) = −κ κ −cot(kL) = k We insert our κ as defined in part (b): r 2mV0 2mE −cot(kL) = − 2 2 ~2 k 2 ~k Substitute k into the second term of the RHS: r 2mV0 −1 −cot(kL) = ~2 k 2 Let θ = kL:

r

2mV0 L2 −1 ~2 θ2 We can transform this into a form similar to that of eq. 2.31 if: −cot(θ) =

θ0 2 =

2mV0 L2 ~2

Our equation to solve then becomes: s −cot(θ) =

θ0 2 −1 θ2

This can be solved graphically for the values of θ for which the LHS and RHS expressions intersect. From here we obtain the corresponding values of k: k=

θ L

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20

¨ 2. THE SCHRODINGER WAVE EQUATION - SOLUTIONS

And solve eq. 2.24 for E: k 2 ~2 2m 2 Note that the value for θ0 which we derived above is slightly different from that of eq. 2.30. This occurs because our potential well for this problem is actually oriented from x = 0 to x = L, as opposed to x = −L/2 to x = L/2 in the example of section 2.3. This in turn gave us a different argument of the cotangent function and thus necessitated we choose a different θ0 2 in order to keep the same clean form of eq. 2.31 to solve numerically. Procedurally, the process we just followed is identical to that of section 2.3. E=

15. The infinite potential well represents a situation where the potential energy does not evolve with time. The time-dependent solutions must then satisfy eq. 2.45: E Ψ(x, t) = uE (x)e−iωt where w = ~ From eq. 2.20, the normalized spatial wave functions of a particle in the infinte well are given by: r  nπx  2 sin ψ(x) = L L The corresponding energies are given by eq. 2.17: E=

n2 h2 8mL2

We may then calculate ω from eqs. 2.39 and 2.17: ω=

E n2 πh = ~ 4mL2

Our total wave function is then: r    nπx  2 in2 πht sin exp − Ψ(x, t) = L L 4mL2

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3 Operators and Waves Solutions 1. We may obtain the momentum by multiplying the momentum operator as given in equation 3.2 d pˆ = −i~ dx times the wave function given in the problem pˆψ(x) = −i~

d Aeiαx = −i2 α~Aeiαx dx

pˆψ(x) = α~Aeiαx Our wavefunction is therefore seen to be an eigenfunction of the momentum operator corresponding to the eigenvalue α~. Thus, a measurement of the momentum of the electron in this state would yield a value of α~. The kinetic energy may be found via the relation: p2 2m The square of the momentum operator that appears here in the numerator will result in the operator begin applied twice, therefore: KE =

2 d2 2 d = −~ dx2 dx2 We substitute this expression into the expression of the kinetic energy:

pˆ2 = (−i~)2

KE =

pˆ2 ~2 d2 =− 2m 2m dx2 21

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22

3. OPERATORS AND WAVES - SOLUTIONS

and operate on the wavefunction: ~2 d2 ψ ~2 d2 − =− Aeiαx 2 2 2m dx 2m dx =

~2 α2 ~2 Aα2 eiαx ⇒ KE = 2m 2m

3. Our task is to find ψ = Aφ1 (x) + Bφ2 (x) such that: pˆψ = pψ Where p is the eigenvalue of the momentum operator. So, let’s try operating on the following wavefunction: ψ = A cos(kx) + B sin(kx) dψ d d = −i~A cos(kx) − i~B sin(kx) dx dx dx = i~k Asin(kx) − i~k cos(kx)

pˆψ = −i~

This choice of ψ is not generally an eigenfunction of pˆ, as we can not factor out our original ψ = A cos(kx) + B sin(kx). Let us now try the following: ψ = cos(kx) + i sin(kx) pˆψ = −i~k(−sin(kx)) − i2 ~k(cos(kx)) = ~k cos(kx) + i~k sin(kx) = ~k(cos(kx) + i sin(kx)) Thus, we have found an eigenfunction of the momentum operator which corresponds to the eigenvalue ~k. Note, that using the Euler formula: eikx = cos(kx) + i sin(kx) Our wavefunction may be written: ψ = eikx

Using this form of the wavefunction and the result of Problem 1, we may

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23 thus see that the wavefunction is an eigenfunction of the momentum operator corresponding to the eigenvalue ~k. 5. Notice that if the energy is less than the minimum value of V (x), the [V (x) − E] portion of the RHS is always positive. Therefore, if this equation holds, the second derivative of the wavefunction will have the same sign as the wavefunction itself, and the function will increase or decrease in value monotonically. 7. Sections 3.3.1 and 3.3.2 provide the initial setup for this problem. We must solve the appropriate Schrodinger equation for each region, then use boundary conditions to find our desired ratios. In the first region, where V = 0, we have the following from section 3.3.1: d2 ψ1 + k1 2 ψ1 = 0 2 dx and given by eq. 3.32: r k1 =

2mE ~2

This yields solutions: ψ1 = Aeik1 x + Be−ik1 x Region 2 produces equations 3.39 through 3.40: r 2m(V0 − E) d2 ψ2 2 − k ψ = 0 where k = 2 2 2 dx2 ~2 With physically acceptable solutions: ψ2 (x) = De−k2 x Now, we impose continuity of the wave equations at x = 0 and of their first derivatives at x = 0: ψ1 (0) = ψ2 (0) ⇒ A + B = D ψ1 0 (0) = ψ2 0 (0) ⇒ ik1 (A − B) = −k2 D Divide both side of the second equation by ik1 and our two equations become: A+B =D

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(1)

24

3. OPERATORS AND WAVES - SOLUTIONS

k2 k2 D=i D (2) ik1 k1 Here, the fact that −1/i = i was used to bring i out of the denominator and cancel the negative. If we add these two equations, we solve for A in terms of D: k2 2A = D(1 + i ) k1 1 k2 A = D(1 + i ) (3) 2 k1 If we subtract (1) from (2), we can solve for B in terms of D: A−B =−

2B = D(1 − i

k2 ) k1

k2 1 B = D(1 − i ) 2 k1 From here, we divide each (3) and (4) by D:

(4)

A 1 k2 = (1 + i ) D 2 k1 1 k2 B = (1 − i ) D 2 k1 The reflection coefficient R is found by R=

|B|2 v1 |B|2 = |A|2 v1 |A|2

Using our ratios above: R=

1 B ∗ B (1 − i kk21 )(1 + i kk21 ) ) (D) (D 4 = =1 A ∗ A k2 k2 1 (D ) (D) (1 + i )(1 − i ) 4 k1 k1

9. Following figure 3.6, we are given the general solutions for each region: ψ1 (x) = Aeik1 x + Be−ik1 x , k2 x

ψ2 (x) = Ce

+ De

−k2 x

,

ψ3 (x) = Eeik1 x ,

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x≤0 0≤x≤L x≥0

25 We impose the boundary conditions at x = 0 and x = L: ψ1 (0) = ψ2 (0) ψ1 0 (0) = ψ2 0 (0) ψ2 (L) = ψ3 (L) ψ2 0 (L) = ψ3 0 (L) The first two yield the following system of equations: A+B =C +D ik1 (A − B) = k2 (C − D) We divide the second of these by ik1 and add the result to the first to obtain an expression for A in terms of C and D: 2A = C + D − i

k2 (C − D) k1

(1)

Now, let’s turn to the interface between regions 2 and 3. The boundary conditions produce the following system: Cek2 L + De−k2 L = Eek1 L k2 Cek2 L − k2 De−k2 L = ik1 Eek1 L Dividing the second of these equations by k2 reduces to the system to a more manageable one: Cek2 L + De−k2 L = Eek1 L Cek2 L + De−k2 L = i

k1 k1 L Ee k2

Add the two equations and we can solve for C in terms of E: 2Cek2 L = Eek1 L + i E C= 2



k1 1+i k2



k1 k 1 L Ee k2

e(ik1 −k2 )L

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(2)

26

3. OPERATORS AND WAVES - SOLUTIONS

Subtract those same equations to solve for D in terms of E: 2De−k2 L = Eek1 L − i

k1 k1 L Ee k2

  k1 (ik1 +k2 )L 1−i e k2

E D= 2

(3)

Insert (2) and (3) into (1):     k1 (ik1 −k2 )L E k1 (ik1 +k2 )L E 1+i e + 1−i e 2A = 2 k2 2 k    2   k2 E k1 (ik1 −k2 )L k1 (ik1 +k2 )L −i 1+i e + 1−i e k1 2 k2 k2 Divide both sides by A and simplify:         E k1 k1 E k1 k1 ik1 L−k2 L + 4= 1+i − i −1 e 1−i + i + 1 eik1 L+k2 L A k2 k2 A k2 k2      2 2 2 2 E k + k2 k − k1 = 2+i 1 eik1 L−k2 L + 2 + i 2 eik1 L+k2 L A k1 k2 k1 k2 Solve for E/A: E =4 A



k 2 + k22 2+i 1 k1 k2



ik1 L−k2 L

e

  −1 k22 − k12 ik1 L+k2 L + 2+i e k1 k2

11. The Heisenberg uncertainty principle states: ∆E∆t ≥

~ 2

Therefore we solve for ∆E: ~ 2∆t 6.626 × 10−34 J · s = 2(4.0 × 10−10 s) = 8.28 × 10−25 J

∆E =

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27 Now, converting to electron volts: ∆E = 8.28 × 10−25 J ·

1 eV = 5.18 × 10−6 eV 1.6 × 10−19 J

13. Equation 3.48 gives the average value of an observable: Z ˆ < Q >= ψ ∗ (x)Qψ(x) dx ˆ is the corresponding operator, in this case the kinetic energy operWhere Q ator, which is given by: ~2 d2 pˆ2 =− 2m 2m dx2 The integral becomes:   Z 1 ~2 d2 −x Be < KE > = − Be−x dx 2 2m dx 0 Z ~2 2 1 −2x B e dx =− 2m 0  ~2 2  −2 =− B e −1 4m where B is the normalization constant found in problem 12.

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28

3. OPERATORS AND WAVES - SOLUTIONS

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4 The Hydrogen Atom Solutions 2. Per equation 4.9, the radial probability density for the 1s state hydrogen is found by:  r 2 1 r r/a0 2r2 2 |P1s (r)| = 2 e = 3 e−2r/a0 a0 a0 a0 The 1s radial wave function used here is listed in table 4.2 in the row corresponding to n=1 and l=0. Since hydrogen is the atom of interest, we used the appropriate atomic number Z=1. We shall maximize the above expression by taking the first derivative and equating to zero:   4r −2r/a0 2r2 −2 −2r/a0 d 2 |P1s (r)| = 3 e + 3 e =0 dr a0 a0 a0 4r 4r2 = 3 e−2r/a0 − 4 e−2r/a0 = 0 a0 a0 Divide both sides by e−2r/a0 and solve for r: 4r 4r2 − 4 =0 a30 a0   4r r 1− =0 a30 a0 Therefore we have one solution at r = 0 and one at r = a0 . Let’s test the intervals (0, a0 ) and (a0 , ∞) to reveal the nature of these points. Let r = a0 /2 29

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30

4. THE HYDROGEN ATOM - SOLUTIONS

for the former and r = 2a0 for the latter:   d 4  a0  −1 4 a20 −1 2 |P1s (r)| = 3 e − 3 e dr a0 2 a0 4 x=a0 /2   2 1 −1 >0 =e − a20 a20
d 4 4 2 |P1s (r)| = 3 (2a0 ) e−4 − 3 4a20 e−4 dr a0 a0 x=2a0 8 = 2 e−4 (1 − 2) < 0 a0 Since the first derivative of the radial probability density takes a positive values for (0, a0 ) and negative values for (a0 , ∞), we may conclude that r = a0 is in fact the maximum value. This is consistent with the description of the Bohr radius as the innermost orbital radius of the electron in the hydrogen atom. 4. Starting with equations 4.7 and 4.8, dP = |ψ(r)|2 r2 sin(θ) dr dθ dψ We construct our wavefunction by equation 4.4: ψ(r, θ, φ) =

Pnl (r) Θlm (θ) Φm (θ, φ) r

Pnl (r), Θlm (θ), and Φm (θ, φ) for the 1s state may be found in tables 4.1 and 4.2 in rows corresponding to n=1, l=0, and ml =0 (Note that ml =0 is the only possibility for l=0 thus only one row for l=0 appears in table 4.1). Carrying on, 2 ψ(r, θ, φ) = r =

r

1 a0

1

1 e−r/a0 √ π

3/2 a0



r a0



1 1 e−r/a0 √ √ 2 2π

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31 We now integrate to find the probability: Z P = |ψ(r, θ, φ)|2 r2 sin(θ) dr dθ dφ Z 2π Z π Z 1 1 a0 2 −2r/a0 sin(θ) dθ dφ = dr r e π a30 0 0 0 Z a0 4r2 −2r/a0 = dr e a30 0 Grouping 1/a0 with r and dr:   Z a0  2 2r r −2r/a0 P =2 e d a0 a0 0 Substitute ρ =

r : a0

Z P =2

1

ρ2 e−2ρ dρ

0

This integral may be solved analytically using integration by parts: 1 Z 1   Z 1 1 −2ρ 1 −2ρ 2 2 −2ρ 2ρ − e ρe dρ = − e ρ − dρ 2 2 0 0 0 1 Z 1     1 −2 1 −2ρ 1 −2ρ =− e + − e ρ − e dρ − 2 2 2 0 0
1 = 1 − 5e−2 4

5. For a given n-state, the possible values of l are n-1, n-2, . . . 0. Therefore, n=5 allows l=0,1,2,3,4. For any given l value, the possiblities for the magnetic quantum number are ml = -l, -l+1, . . . 0, 1, . . . l. The results are as follows: l=0 l=1 l=2 l=3 l=4

: : : : :

m=0 m = −1, 0, 1 m = −2, −1, 0, 1, 2 m = −3, −2, −1, 0, 1, 2, 3 m = −4, −3, −2, −1, 0, 1, 2, 3, 4

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32

4. THE HYDROGEN ATOM - SOLUTIONS

7. Recall from equation 4.4 our construction of a wave function as a product of radial and angular parts: ψ(r, θ, φ) =

Pnl (r) Θlml (θ)Φ(φ) r

Our strategy will be to express the given wave function in terms of radial and angular parts, then use their properties to check whether the wavefunction is an eigenfunction of the Schrodinger equation and find the corresponding energy. For our given wavefunction, we see that we may separate it in terms of dependence on θ and dependence on r as cos(θ) and r2 e−Zr/2a0 . Note that the presence of r2 is due to fact the radial part is to be divided by r in eq. 4.4 to construct the total wave function. We need not incorporate the constant C as it will not affect whether the wave function is a solution of the Schrodinger equation. For the angular part, note that in table 4.1, cos(θ) appears in the spherical harmonic corresponding to l = 1 and ml = 0. When we distribute the left-hand side of eq. 4.4, we get: ~2 l(l + 1) 1 Ze2 −~2 d2 P (r) + P (r) − Pnl (r) nl nl 2m dr2 2mr2 4π0 r Taking the second derivative of the radial wave function: d2 2rZ −Zr/2a0 Z 2 r2 −Zr/2a0 −Zr/2a0 P (r) = 2e e e − + nl dr2 a0 4a0 Upon substitution and using l = 1, the left-hand side of eq. 4.4 becomes: −~2 d2 ~2 l(l + 1) 1 Ze2 P (r) + P (r) − Pnl (r) nl nl 2m dr2 2mr2 4π0 r ~2 −Zr/2a0 ~2 rZ −Zr/2a0 ~2 Z 2 r2 −Zr/2a0 ~2 −Zr/2a0 Ze2 r −Zr/2a0 e + e + e + e − e m ma0 8ma0 m 4π0 ~2 rZ −Zr/2a0 Ze2 r −Zr/2a0 ~2 Z 2 r2 −Zr/2a0 = e − e + e ma0 4π0 8ma0 =−

Recall from equation 1.20 that the Bohr radius a0 may expressed as follows: a0 =

4π0 ~2 me2

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33 If we substitute this expression for a0 into the first term of the previous equation, the first two terms cancel and equation 4.5 becomes: ~2 Z 2 2 −Zr/2a0 r e = EPnl (r) 8ma0 The radial wave function is in fact an eigenfunction of the Schrodinger equation with corresponding eigenvalue (energy): ~2 Z 2 E= 8ma0

9. We shall start with eq. 4.5:   2 2 ~2 l(l + 1) 1 Ze2 −~ d + − Pnl (r) = EPnl (r) 2m dr2 2mr2 4π0 r And 1.20:

4π0 ~2 me2 The substitution as directed is ρ = r/a0 , so we can directly substitute for F: r = a0 ρ. We let Z=1 as is the case hydrogen and carry out the substitution:   2 ~2 l(l + 1) 1 Ze2 d2 −~ + − Pnl (r) = EPnl (r) 2m d(ρa0 )2 2m(ρa0 )2 4π0 ρa0 a0 =

Since a0 is a constant, we may bring it outside the derivative in the first term on the left-hand side and substitute eq. 1.20 for one a0 in each of the first two terms.   1 ~2 l(l + 1) me2 1 Ze2 1 ~2 me2 d2 + − Pnl (r) = EPnl (r) − a0 2m 4π0 ~2 dρ2 a0 2m(ρa0 )2 4π0 ~2 a0 ρa0 After making the appropriate cancellations and factoring we have:   e2 1 1 d2 1 l(l + 1) 1 − + − Pnl (r) = EPnl (r) a0 4π0 2 dρ2 2 ρ2 ρ   1 d2 l(l + 1) 1 E − + − Pnl (r) = Pnl (r) 2 2 2 dρ 2ρ ρ (1/4π0 )(e2 /a0 )

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34

4. THE HYDROGEN ATOM - SOLUTIONS

12. The transition integral for the general case is given by   Z z ∗ φn2 l2 m2l dV φn1 l1 m1 l a0 We are interested in the transition 3d1 → 2p1 , so the transition integral becomes:   Z z ∗ φ2,1,1 dV φ3,2,1 a0 From here on, we assume that the radial distance will be given in terms relative to a0 , so we take a0 = 1 and constructs the wavefunctions by tables 4.1 and 4.2. P3,2 (r) Y2,1 (θ, φ) r ! r √ 2 2 2 −r/3 15 = √ r e sin(θ)cos(θ)eiφ − 8π 81 15

φ3,2,1 =

P2,1 (r) Y1,1 (θ, φ) r ! r 3 1 = √ re−r/2 − sin(θ)eiφ 8π 2 6

φ2,1,1 =

Now, we take the complex conjugate of φ3,2,1 , substitute z = rcos(θ), and proceed with the integration: Z Z 1 ∗ r4 e−5r/6 sin2 (θ)cos2 (θ) dV φ3,2,1 zφ2,1,1 dV = 648π Z ∞ Z π Z 2π 1 = r6 e−5r/6 sin3 (θ)cos2 (θ) dr dθ dφ 648π 0 0 0 The integration may be separated as follows: 1 648π

Z

∞ 6 −5r/6

r e 0

Z dr

π 3

2

Z

sin (θ)cos (θ) dθ 0

2π

1 40310784 4 2π 648π 15625 15 = 2.123

dφ = 0

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35 13. In order to find the transition coefficient for hydrogen, we will use the results found in Appendix FF (online) along with eqs. 4.20 and 4.21. We According to the selection rules, summarized in table 4.3, ∆ml = ±1 for xand y-polarized light. Therefore, we must calculate the transition integrals for both the 2p1 → 1s0 and 2p−1 → 1s0 cases. As directed in Appendix FF, we use the following in place of the z operator in eq. 4.21: 1 x = (r+ + r− ) 2 1 (r+ − r− ) 2i We first solve the case for x-polarized light. y=

I12 = 2 |φ∗1s0 xφ2p1 |2 + |φ∗1s0 xφ2p−1 |2




Which gives us the following when we substitute the above expression for x: 1 φ∗1s0 xφ2p1 = (φ∗1s0 r+ φ2p1 + φ∗1s0 r− φ2p1 ) 2 1 φ∗1s0 xφ2p−1 = (φ∗1s0 r+ φ2p−1 + φ∗1s0 r− φ2p−1 ) 2 ∗ ∗ Since φ1s0 r+ φ2p1 and φ1s0 r− φ2p−1 will equal zero, we may use the results given in Appendix FF for the remaining terms, giving us:   2 1 2 1 2 Ri + Ri = Ri2 I12 = 2 6 6 3 Using the results of eqs. 4.23 and 4.26 into eq. 4.22: A21 =

6.078 × 1015 1.109 = 6.25 × 108 per atom per second (121.6)3 6

We may follow the same procedure for y-polarized light. First, the transition integral:
I12 = 2 |φ∗1s0 yφ2p1 |2 + |φ∗1s0 yφ2p−1 |2

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36

4. THE HYDROGEN ATOM - SOLUTIONS

Using the operator substitution: φ∗1s0 yφ2p1 =

1 ∗ (φ1s0 r+ φ2p1 − φ∗1s0 r− φ2p1 ) 2i

1 ∗ (φ r+ φ2p−1 − φ∗1s0 r− φ2p−1 ) 2i 1s0 Once again, the nonzero terms lead to:   −2 2 −1 2 −1 2 Ri + Ri = R I12 = 2 6 6 3 i φ∗1s0 yφ2p−1 =

The result for y-polarized light is then: A21 =

6.078 × 1015 −1.109 = −6.25 × 108 per atom per second (121.6)3 6

15. a) Let’s imagine that the angular momentum vector l points from the origin and lies on the surface of a cone, then the radius of the circle at the base of the cone is: r = |l|sinθ And, for a change of the azimuthal angle dφ, the distance that the tip of the angular momentum vector moves is given by: d|l| = |l|sinθ dφ

b) We start with the relationship of the torque to angular to momentum: d|l| = |τ | dt This equation together with eq. 4.40 immediately leads to the equation: d|l| e = |l × B| dt 2m The definition of the vector product then gives: d|l| e = |l|B sin(θ) dt 2m

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37 Substituting d|l| = |l|sinθ dφ, we may then find our Larmor frequencies: |l|sin(θ)

dφ e = |l|B sin(θ) dt 2m

ωL =

dφ eB = dt 2m

17. The total angular momentum of the 4f electron will contain contribution form both the electron’s spin and its orbital angular momentum. Section 4.3.4 shows that the total angular momentum may take on the values: J = j1 + j2 , j1 + j2 − 1, . . . |j2 − j1 | We let j1 be our orbital angular momentum and j2 the spin. The electron occupies an f-state, therefore j1 = 3, and the intrinsic spin of the electron means j2 = 1/2. We may now list the possible values for J: 7 5 J= , 2 2 The spin-orbit coupling energy is then found by: 3ζ~2 ζ~2 l= 2 2 −ζ~2 = (l + 1) = −2ζ~2 2

Es−o = Es−o

1 2 1 f or j = l − 2 f or j = l +

19. Using the two equations following eq. 4.50, the separation in energies due to spin-orbit coupling is:    2 ζ~2 −ζ~2 ζ~ ∆E = l− (l + 1) = (2l + 1) 2 2 2 Since we are using the atomic system of units, ~ = 1 and ζ ∆ = (2l + 1) 2

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(1)

38

4. THE HYDROGEN ATOM - SOLUTIONS

We then let l=1 (due to the occupation of a p-orbital): 3 ∆E = ζ = 7.9 × 10−6 2 ζ = 5.27 × 10−6 The spin-orbit constant ζ for the levels neon may be calculated directly by eq. 4.48, however we may use equation 4.51 to set up a relationship between ζHe and ζN e . Since He+ and N e9+ are both hydrogenic atoms (containing only one electron), the quantity    a0 3 Zr should be equal in both cases as the grouping of a0 and z with r will account for the scaling of the radial distance with changing nuclear charge. We now use the appropriate atomic numbers of helium and neon to generate a ratio between ζHe and ζN e .: 104 ζN e = 4 = 625 ⇒ ζN e = 625ζHe = 3.29 × 10−3 ζHe 2 For the separation between the 3p3/2 and 3p1/2 levels, we use equation (1) above: 3.29 × 10−3 (2l + 1) = 4.94 × 10−3 ∆E = 2 A comparison with the separation of the corresponding helium states reveals the vastly increased impact of spin-orbit coupling in the case of neon.

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5 Many-Electron Atoms Solutions 1. An electron has an intrinsic spin s = 1/2, therefore the possible values of ms are ±1/2. The 4f state of the electron has an orbital angular momentum quantum number l = 3 and therefore may have values ml = −3, −2, −1, 0, 1, 2, 3. In the case of two electrons occupying the 4f state, we first see from our results above, that there are 7 possible ml quantum numbers 2 ms quantum numbers. There are thus 14 possible states in which to put the first electron. Once we assign first electron, there are now 13 possible states. Because the order in which placed them does not matter in our configuration, we then divide the product by two: 14 · 13 = 91 # of distinct states = 2

3. Using the form of eq. 5.6, we attribute each row of the determinant to a particular state and each column to a particular electron occupying a corresponding state (row). Our system contains these electrons, therefore N=3 in our coefficient and r ψ α(1) ψ α(2) ψ α(3) 10 10 1 10 ψ β(1) ψ β(2) ψ β(3) Φ= 10 10 10 3 ψ20 α(1) ψ20 α(2) ψ20 α(3)

39

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40

5. MANY-ELECTRON ATOMS - SOLUTIONS

4. For each of these elements, we shall fill their shells in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 4p, and so on. The number of electrons in a neutral atom of any element will equal the number of protons in its nucleus, therefore we use the atomic number of each element as the number of electrons we use to fill these shells: Flourine(Z = 9) : Magnesium(Z = 12) : Silicon(Z = 14) : Potassium(Z = 19) : Cobalt(Z = 27) :

1s2 1s2 1s2 1s2 1s2

2s2 2s2 2s2 2s2 2s2

2p5 2p6 2p6 2p6 2p6

3s2 3s2 3p2 3s2 3p6 4s1 3s2 3p6 4s2 3d7

[N e] 3s2 [N e] 3s2 3p2 [Ar] 4s1 [Ar] 4s2 3d7

An alternate (shorter) way to notate these, as I’ve done above to the right, is to use the symbol of the inert gas that occurs closest before a particular element to denote the filled shells through complete rows of the periodic table and merely denote the filling of the remaining shells. 7. The nitrogen atom will have the following ground configuration: 1s2 2s2 2p3 . Since the 1s and 2s orbitals have less energy than the 2p orbital, the next higher configurations are 1s2 2s2 2p2 3s and 1s2 2s2 2p2 3p. 8. ns2 : Here are two electrons with l=0, therefore, L=0. Their spins produce S=0,1. Since L+S must be even for two equivalent electrons, the only possibility is 1 S (S=0, L=0). nd2 : Two electrons with l=2 may produce L=4,3,2,1,0. Again, the addition of their spin angular momenta produce S=0,1. Imposing the condition that L+S must be even, we have the following configurations: 1 G (L=4, S=0), 3 F (L=3, S=1), 1 D (L=2, S=0), 3 P (L=1, S=1), and 1 S (S=0, L=0). 4f 2 : These two electrons occupy f-states (l=3), thus may contribute to L=6,5,4,3,2,1,0. As has been the case above, S=0,1. Ipmosing the condition that L+S must be even, we have the following LS terms: 1 I (L=6, S=0), 3 H (L=5, S=1), 1 G (L=4, S=0), 3 F (L=3, S=1), 1 D (L=2, S=0), 3 P (L=1, S=1), and 1 S (S=0, L=0). Let’s apply Hund’s rules to the 4f configuration. Of these terms, the ones

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41 corresponding to the maximum value of S are 3 H, 3 F , and 3 P . Of these, the one with maximum value of L (3 H) should lie lowest. 11. It is important to take note here that we have a case of non-equvalent electrons for the [Xe] 4f 5d configuration and one of equivalent electrons with [Xe] 4f 2 . For [Xe] 4f 5d, one electron has l=3 and the other l=2. Since there is no additional constraint on the values of L and S except for the rules of addition of angular momentum, L may take L=5,4,3,2, or1 and S=0,1. For 4f 2 , we must impose the condition that L+S be even. The two l=3 electrons could form L=6,5,4,3,2,1,0 and S=0,1. The allowed LS terms are then (L=6, S=0), (L=5, S=1), (L=4, S=0), (L=3, S=1), (L=2, S=2), (L=1, S=1), and (L=0,S=0). 13. As described in section 5.5.2, the nuclear charge increases as we go across a row in the periodic table, thus decreasing the average value of the electrons’ distance from the nucleus. This causes an increase in the Coulomb interaction, thus we expect sulfur (S) to possess the greatest Coulomb interaction of these three elements. 15. Recall from equation 4.51 that the strength of the spin-orbit interaction increases with atomic number as Z 4 . Since the spin-orbit interaction is mainly responsible for the breakdown of LS coupling, the O atom has the smallest spin-orbit interaction and the states of O corresponds most nearly correspond to pure LS terms. 17. To find the total energy, we simply click on neon in the periodic table interface, then the red arrow on the bottom row of controls to run the calculation. Then, click on the ”Averages” tab to display the kinetic, potential, and total energies. We see a value of -128.55 for the total energy. We now remove a 2p electron by selecting the 2p shell via the incremental objects in the bottom row of controls, then clicking the (-e) button. Verify you have removed the electron from the correct shell as the configuration in the second to last row of controls should now read 1s2 2s2 2p5 . Now, run the applet again for the neon ion and finding the energy under the ”Averages” tab, you should find the energy to be approximately 127.82. The binding

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42

5. MANY-ELECTRON ATOMS - SOLUTIONS

energy is the difference of these two values: |−128.55 − (−127.82)| ≈ .73 Going back to the neutral neon atom, the third column to the left on the page obtained after striking the Averages tab gives a 2p single electron energy of 0.85. This is very close to the difference in total energies of the neutral Ne atom and the ion. 19. We can calculate the Slater integrals for Si, P, and S by first selecting the appropriate element in the periodic table of the Hartree-Fock applet, clicking the red arrow to run the calculation, then clicking the ”Other” tab. Now, the second box should be changed to ”2”, then continuing across: ”3”, ”p”, ”3”, ”p”. The red arrow on the right runs the calculation. We obtain the following values. F 2 (3p, 3p) Element au cm−1 Si .166 36418 P .197 43180 S .220 48323 We see a definite increase along the row, constitent with the qualitative result obtained in problem 13. 21. Using fig. 5.5, the energy of 1 S of the 1s 3s configuration is approximately 185,000 cm−1 while 1 P has an energy of approx. 170,000 cm−1 . The energy of the emitted photon should equal the difference of the energy levels, thus: ∆E = 25, 000 cm−1 According to equation 5.12, 1 eV of photon energy corresponds to light of wave number 8065.54 cm−1 . The photon energy emitted in this transition must equal: 25, 000 cm−1 = 3.10 eV 8065.54 cm−1 /eV Using the relation E = hc/λ, λ=

hc 1240 eV · nm = = 400 nm E 3.10 eV

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6 The Emergence of Masers and Lasers - Solutions 1. As given in figure 6.7, the range of the 4 F2 band is approximately 17, 000 cm−1 to 19, 000 cm−1 . As these quantities are given in terms of 1/λ, we may quickly obtain the corresponding frequencies in Hz by using the relation I.22: c c = λf ⇒ f = λ Therefore, we may convert c to cm/s and simply multiply:  cm  17 × 103 cm−1 3 × 1010 = 5.1 × 1014 Hz s  cm  19 × 103 cm−1 3 × 1010 = 5.7 × 1014 Hz s Similarly, for 4 F1 :  cm  23 × 103 cm−1 3 × 1010 = 6.9 × 1014 Hz s  cm  27 × 103 cm−1 3 × 1010 = 8.1 × 1014 Hz s

3. The 2p core is given to be in the state 2 P3/2 . Using the following scheme for spectroscopic notation: 2s+1 LJ 43

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44

6. THE EMERGENCE OF MASERS AND LASERS - SOLUTIONS

we take J = 3/2. Since the excited electron (3p) has orbital angular momentum l=1, the possible values for K are: 5 3 1 K = J + l, J + l − 1, . . . |J − l| = , , 2 2 2 The spin orbit interaction spits each level with a value of K into two levels corresponding to the total angular momentum having the values K ± 12 . 5. The situation described in problem 4 gives I = 1/2 and the fact that we are dealing with hydrogen which means we have one electron, for which S = 1/2. The possible values for F are then F = 0, for which MF = 0; and F = 1, for which MF = −1, 0, 1. Since the F=0 level is not split by a magnetic field, we only calculate gF for F = 1.Using eq. 6.17, For F=1, gF =

(1 + 1) − 21 ( 12 + 12 ) + 12 ( 21 + 1) 9 = 1+1 8

Therefore, we consider the F=1 states via equation 6.5. Since gF is positive, the MF = 1 state will have a negative µz value and be drawn to regions of low magnetic field. Consequently, the MF = −1 will have a positive µz value and be drawn to regions of high magnetic field outside the magnetic trap. These atoms will be lost, decreasing the kinetic energy and temperature of the atoms caught in the trap. 7. Given S = 1/2 and I = 5/2, our possible values for F are F = 3 and F = 2. These levels split as follows: F = 3 : MF = −3, −2, −1, 0, 1, 2, 3 F = 2 : MF = −2, −1, 0, 1, 2

As we see, the MF = −3 and MF = 3 states are not mixed with any states where F = 2, thus will be “pure” states with straight lines. All other states of F = 3 and all states where F = 2 will be mixed, thus will have curved lines.

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7 Statistical Physics - Solutions 1. As described in the text, there are sixteen possible outcomes of flipping a coin four times. The possible outcomes can be classified according to the number of heads. We may use equation 7.2 (or 7.3) to calculate the statistical weight associated with each outcome by letting N = 4 and n equal the number of heads obtained in each distribution:

HHHH HHHT HHT T HT T T TTTT

  4 : = 4   4 : = 3   4 : = 2   4 : = 1   4 = : 3

4! 4!0! 4! 3!1! 4! 2!2! 4! 1!3! 4! 0!4!

=1 =4 =6 =4 =1

3. We may set up equation 7.7 for each energy level: n2 n2 N = = e2 /kB T g2 2 Z 45

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(1)

46

7. STATISTICAL PHYSICS - SOLUTIONS n1 n1 N = = e1 /kB T g1 1 Z

(2)

Divide (1) by (2):

n2 = 2e−(2 −1 )/kB T n1 We are given 2 − 1 = 0.025 eV , T = 298K, and we use kB = 8.617 × 10−5 eV /K:   −0.025 eV n2 = 2 exp n1 (8.617 × 10−5 eV /K)(298 K) = .755

5. F1 (u) is defined in conjunction with Fig. 7.2: 4 2 F1 (u) = √ u2 e−u π We maximize F1 (u) by setting the derivative of F1 (u) with respect to u to zero:  4  dF1 2 2 = √ 2ue−u − 2u3 e−u = 0 du π 2

2ue−u (1 − u2 ) = 0 u = 0, ±1 We test the intervals (0, 1) and (1, ∞):   1 0 > 0 and F10 (2) < 0 F1 2 Thus, F1 (u) has a maximum at u = 1. 7. To find the average value of v 2 , we calculate the integral: Z ∞ 2 (v )av = v 2 P (v) dv 0

Use equation 7.22 for P (v)dv:  Z ∞ 2 4 (v )av = 4πv 0

m 2πkB T

3/2

e−mv

2 /2k

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BT

dv

47 √ We may use eq. F.2 in Appendix F to find I4 (1) = 3 π/8, thus (v 2 )av = 3kB T /m. 9. To find rms , we need to first find (2 )av , then take the square root. Analogous to example 7.3, Z ∞ 2 2 P () d ( )av = 0

Using equation 7.26: 2

( )av

2 =√ π

Z 0

∞

5/2 e/kB T d (kB T )3/2

Per equation 7.27, u = /kB T and we may write the above equation in terms of u: Z ∞ 2 2 2 ( )av = √ (kB T ) u5/2 e−u du π 0 At this stage, we can work the integral according to the procedure expressed in eqs. F.8-F.14: 2 (2 )av = √ (kB T )2 Γ(7/2) π 2 = √ (kB T )2 Γ(7/2) π 2 5 = √ (kB T )2 Γ(5/2) 2 π 2 5 3 = √ (kB T )2 Γ(3/2) 2 2 π 2 5 3 1 = √ (kB T )2 Γ(1/2) 2 2 2 π 15 = (kB T )2 4

We now take the square root of this last quantity: √ 15 rms = kB T 2

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48

7. STATISTICAL PHYSICS - SOLUTIONS

11. According to the equipartition of energy (end of section 7.3.1), each particle has an average kinetic energy of kB T /2 for each degree of freedom. For three degrees of freedom: 3 av = kB T 2 Therefore, total kinetic energy of a mole of gas should be: 3 3  = kB T NA = (1.381×10−23 J/K)(298 K)(6.022×1023 particles) = 3717 J 2 2

13. Per equation 7.28, 8πf 2 u(f ) = 3 c



hf



ehf /kB T − 1

We apply the product rule and quotient rule of differentiation: !
2   hf /kB T − 1 − khB fT ehf /kB T du(f ) 16πf 8πf 2 h e hf = 3 + 3 2 df c ehf /kB T − 1 c (ehf /kB T − 1)
2 Let’s set this expression equal to zero and multiply both sides by ehf /kB T − 1 :  
8πf 2 hf /k T
8πf 2 hf hf /k T 16πf 2 hf /kB T B B e −1 + 3 e −1 − 3 e =0 c3 c c kB T Multiply both sides by c3 /8πhr2 and let x = hf /kB T , then simplify: 2 (ex − 1) + (ex − 1) − xex = 0 3ex − xex − 3 = 0 ⇒ (3 − x)ex = 3

15. Equation 7.28 gives the energy density as a function of frequency:   8πf 2 hf u(f ) = 3 c ehf /kB T − 1

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49 We may derive an expression as a function of wavelength by using relation I.23 from the introductory chaper: f = c/λ.     8π(c/λ)2 h(c/λ) 8π h u(λ) = = 3 c3 ehc/λkB T − 1 λ ehc/λkB T − 1

19. According to eq. 7.50, the work done on the gas is given by: dW = −P dV Per the ideal gas law, we solve for P : P =

nRT V

We then integrate to find W and change sign to reflect work done by the gas:   Z V3 V3 dV = nRT ln W = nRT V V2 V2 Eq. 7.48 states dE = dQto + dWon . As found above, the work done by the gas is equal to nRT ln(V3 /V2 ). If the energy remains constant, the amount of heat added to the gas must be equal to the work done by the gas and   V3 dQto = nRT ln V2

21. We first solve eq. 7.66 for Tc : 1 2.612



N V

 Tc =



 =

1 N 2.612 V

2πmkB Tc h2 2/3

3/2

h2 2πmkB

Substitute our given density: N = 5.0 × 1014 atoms/cm3 V

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50

7. STATISTICAL PHYSICS - SOLUTIONS Tc = 1.16 × 109




h2 7.31 × 10−17 = πmkB m

We may supply the appropriate atomic mass to calculate the critical temperature. This equation may appear a bit peculiar due to a quantity of temperature appearing on the left-hand side and one of mass on right. We can take a closer look at the units to more clearly see this relationship. For clarity, let’s use SI units of J, K, m, and kg for the necessary quantities:  Tc =

1 N 2.612 V

2/3

h2 1 J 2 · s2 ⇒ [Tc ] = 2 2πmkB m kg · J/K 1 kg m/s2 · s2 · K = m kg =K

23. Equation 7.76 gives the probability that an energy will fall between  and  + d: 1 3 −3/2 P () d = F 1/2 d (− )/k T F B 2 e +1 Thus, we may substitute our values for F and T then carry out the integration numerically: 3 At T = 295K : 2 3 At T = 3000K : 2

Z

Z

4.1

3−3/2 1/2

3.9

4.1

3.9

3−3/2 1/2

1 e(−F )/295kB ) 1

e(−F )/3000kB )

+1

+1

d = 5.9 × 10−17

d = 5.9×10−17 = 2.41×10−3

It is seen here that the electron is much more likely to be found with an energy between 3.9 ecV and 4.1 eV at 3000K. 25. We substitute our given quantities into the given formula, along with nc = 1 since gold is a monovalent metal: N 19.32 g/cm3 (6.022 × 1023 atoms/mole) = V 197 g/mole

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51 N = 5.90 × 1022 electrons/cm3 V

27. (a) Using the formula from problem 24 and the fact that nc = 1 since sodium is a monovalent metal: nc ρNA (.971 g/cm3 )(6.022 × 1023 atoms/mole) N = = V M 23.0 g/mole 22 = 2.54 × 10 electrons/cm3 (b) To find the Fermi energy, we use equation 7.73 as well as the quantity we just calculated in part (a):  2/3 h2 3 N F = 2m 8π V 2/3  (4.13 × 10−15 )2 3 22 = (2.54 × 10 ) 2(5.48 × 10−4 ) 8π = 3.26 × 10−12 eV (c) The equation after example 7.6 relates the Fermi temperature to the Fermi energy: F F = kB TF ⇒ TF = kB Now, we substitute the Boltzmann constant and the Fermi energy found in part (c): 3 eV TF = = 35, 418K 8.617 × 10− =−5

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52

7. STATISTICAL PHYSICS - SOLUTIONS

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8 Electronic Structure of Solids Solutions 1. The cesium choride lattice is depicted in Figure 8.5. If we take a lightershaded sphere to represent a cesium ion, we see that there are 8 chloride ions that serve as the nearest neighbors. Or, if the cesium ion serves as the center point in a body-centered cubic lattice, the eight nearest neighbors are the choloride ions at the corners. Returning to Figure 8.5, recall that the designation of which shade corresponds to which ion is arbitrary. Therefore, now alllow the darker spheres to be the cesium ions. By extrapolating the figure in one more direction, it can be seen that cesium would have 6 next-nearest neighbors. 3. We refer to Fig. 8.8(b) where we will use vectors to denote the locations of the carbon atoms from a fixed origin, then use vector relationships to determine the bond angle. First, we select our origin to be located at point A in the lattice. The other two point of interest are point B and the atom located in the center of the top face shown. Our primitive ˆi and ˆj will be oriented along the edges of the top face (starting at point A) and the kˆ vector will point directly upward from point A. We may now construct the position vectors to each point. Let a be the position vector from the origin to the top face-centered atom, while b points from point B to the origin (point A). It is our intent to construct a vector c pointing from the top face-centered atom to point B. We will then find the angle between b and c Thus: 1 1 1 1 1 a = ˆi + ˆj, b = − ˆi − ˆj + kˆ 2 2 4 4 4 53

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54

8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS

Our strategy wil be to find the angle between b and c, so let’s first find c. By inspection we see that: c=a+b We substitute the expressions for a and b:     1 1 ˆ 1 1 ˆ 1ˆ − − c=a+b= i+ j− k 2 4 2 4 4 1 1 1 = ˆi + ˆj − kˆ 4 4 4 Now we use the dot (or scalar) product of b and c to find the angle between the two: b · c = |b||c| cos(θ) For the left-hand side: b·c=−

1 1 1 1 − + =− 16 16 16 16

For the right-hand side: r |b||c| cos(θ) =

3 16

r

3 3 cos(θ) = cos(θ) 16 16

Equating the two: 1 3 = cos(θ) 16 16 1 cos(θ) = − 3   −1 θ = cos−1 = 109◦ 280 3 −

5. For the simple cubic lattice, referring to figure 8.2, the nearest neighbor of any point will be located along the length of a side of the cube. Thus, the nearest neighbor distance is simply “a”. In the case of the face-centered cubic lattice, Figure 8.6, for example, we may find the nearest neighbor distance by finding the midpoint distance along the diagonal of a face. We may simply find the diagonal distance via the Pythagorean theorem: √ √ a2 + a2 = 2a

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55 √ The midpoint, thus nearest neighbor distance, is then ( 2/2)a. 7. Our strategy to find the center-to-center distance of the copper ions will be to find the volume of one cell, from which we will find the length of an edge of the cube, then use this to arrive at the face-centered nearest-neighbor distance. First, we calculate the number of atoms per unit volume: mole 8.96 g/cm3 = .141 63.5 g/mole cm3   atoms mole 23 atoms 6.022 × 10 = 8.494 × 1022 .141 3 cm mole cm3 Now, we wish to know the volume per unit cell. A face-centered cubic lattice has four atoms per cell, therefore: 3 4 atoms/cell −23 cm = 4.709 × 10 8.494 × 1022 atoms/cm3 cell

One edge of the cube should now have length: 1.648 × 10−22


1/3

= 3.611 × 10−8 cm

√ As given in problem 5, the nearest-neighbor distance is then ( 2/2)a, where a is the length of an edge of the cube. √ 2 a = 2.553 × 10−8 cm 2

9. We shall begin by constructing the position vectors of each of the corners, starting with the corner directly above the origin indicated in Fig. 8.6 (b) and continuing clockwise: l1 = akˆ l2 = aˆj + akˆ l3 = aˆi + aˆj + akˆ l4 = aˆi + akˆ

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56

8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS

The primitive vectors from eq. 8.4 are: a ˆ ˆ a ˆ ˆ a ˆ ˆ k + i , a3 = a1 = j + k , a2 = i+j 2 2 2 Our strategy will be to solve these primitive vectors for ˆi, ˆj, and kˆ and substitute into the lattice vectors. We first add a1 and a2 : a a1 + a2 = (ˆi + ˆj) + akˆ 2 Then subtract a3 : a1 + a2 − a3 = akˆ ˆ We now solve for k: a1 + a2 − a3 kˆ = a We continue in such manner to find: ˆi = a2 + a3 − a1 a ˆj = a1 − a2 + a3 a We finally carry out the substitution: l1 = a1 + a2 − a3 l2 = a1 − a2 + a3 + a1 + a2 − a3 = 2a1 l3 = a2 + a3 − a1 + a1 − a2 + a3 + a1 + a2 − a3 = a1 + a2 + a3 l4 = a2 + a3 − a1 + a1 + a2 − a3 = 2a2

11. The results from problems 4 & 5 are as follows: a simple cubic : d = a ⇒ R = 2 √ √ 3 3 bcc : d = a⇒R= a 2 √ 4 2 2 fcc : d = a ⇒ R = a 2 4

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57 We now substitute these values of R into the packing fraction equation:
3 4 π a3 π 3 simple cubic : F = 1 = a3 6  √ 3 √ 4 π 43a 3 3π bcc : F = 2 = a3 8  √ 3 √ 4 π 42a 3 2π fcc : F = 4 = a3 6

13. Equation 8.2 gives the following bcc primitive vectors: a ˆ ˆ ˆ a ˆ ˆ ˆ a ˆ ˆ ˆ j + k − i , a2 = k + i − j , a3 = i+j−k a1 = 2 2 2 We must use eqs. 8.18-8.20 to determine the primitive vectors of the corresponding reciprocal lattice. However, it would be of great benefit to calculate the necessary cross-products first. We shall use the determinant method here: ˆ ˆj ˆ k 2 2 i a2 a = ˆj + a kˆ a2 × a3 = 1 −1 1 4 2 2 1 1 −1 ˆ a i a3 × a1 = 1 4 −1 ˆ 2 i a a1 × a2 = −1 4 1 2

ˆj kˆ 1 −1 = 1 1 ˆj kˆ 1 1 = −1 1

a2 ˆ a2 ˆ i+ k 2 2 a2 ˆ a2 ˆ i+ j 2 2

Now we are ready to use eqs. 8.18-8.20 to calculate the primitive vectors of the reciprocal lattice:   a2 ˆ ˆ j + k 3 4π ˆ ˆ   = b1 = 2π 2  j+k a a ˆj + kˆ − ˆi · ˆj + kˆ 4

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58

8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS   ˆi + kˆ 4π ˆ ˆ   = b2 = 2π 2  i+k a a ˆj + kˆ − ˆi · ˆj + kˆ 4   a2 ˆ ˆ i+j 3 4π ˆ ˆ     i+j b3 = 2π 2 = a a ˆj + kˆ − ˆi · ˆj + kˆ 4 a2 3

Compared to eq. 8.4, we see that this reciprocal lattice is nothing but an fcc-lattice with sides of length 8π/a. 15. We shall once again use eqs. 8.18-8.20 to calculate the primitive vectors of the reciprocal lattice. But, again we carry out the cross-products first: ˆ ˆ ˆ 2 i j k a2  ˆ ˆ ˆ  a −i + j + k = a2 × a3 = 1 0 1 4 4 1 1 0 2

a3 × a1 =

a 4

a1 × a2 =

a2 4

ˆi 1 0 ˆi 0 1

kˆ 0 = 1 ˆj kˆ 1 1 = 0 1 ˆj 1 1

a2 ˆ ˆ ˆ i−j+k 4 a2 ˆ ˆ ˆ i+j−k 4

Since the denominator of each of eqs. 8.18-8.20 is the same, we may go ahead and calculate it to substitute into each equation: a ˆ ˆ a2  ˆ ˆ ˆ a3 j+k · −i + j + k = a1 · (a2 × a3 ) = 2 4 4 Now our reciprocal lattice primitive vectors are: 2π  ˆ ˆ ˆ −i + j + k = a 2π ˆ ˆ ˆ b2 = i−j+k = a 2π ˆ ˆ ˆ b3 = i+j−k = a

b1 =

4π 1  ˆ ˆ ˆ −i + j + k a 2 4π 1 ˆ ˆ ˆ i−j+k a 2 4π 1 ˆ ˆ ˆ i+j−k a 2

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59 We factored out 1/2 so as to get us closer to the form of eq. 8.2. Notice that these reciprocal vectors are nothing but the primitive vectors of a bcc-lattice with sides of length 4π/a. 18. Once again, we shall use eqs. 8.18-8.20 extensively as we want to show that: b2 × b3 a1 = 2π (1) b1 · (b2 × b3 ) a2 = 2π

b3 × b1 b1 · (b2 × b3 )

(2)

a1 = 2π

b1 × b2 b1 · (b2 × b3 )

(3)

Where b1 , b2 , and b3 are given by eqs. 8.18-8.20. Let us calculate the cross-product b2 × b3 first since it will be used in each equation.     a3 × a1 a1 × a2 2 × b2 × b3 = 4π a1 · (a2 × a3 ) a1 · (a2 × a3 ) We first make the important observation that the denominators of the terms on each side of the cross-product results in a scalar quantity. Therefore, it can be brought outside the vector product and we may carry out the crossproduct. Let us proceed by writing just the operation in question: (a3 × a1 ) × (a1 × a2 ) The following vector relation, which can be found in most books that discuss vector calculus, can be of immense help here: A × (B × C) = B (A · C) − C (A · B) We relate the vectors that occur in the relation above to the vectors involved in our desired operation as follows: A = a3 × a1 B = a1 C = a2 After the substitution:

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60

8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS

(a3 × a1 ) × (a1 × a2 ) = a1 [(a3 × a1 ) · a2 ] − a2 [(a3 × a1 ) · a1 ] = a1 [a1 · (a2 × a3 )] + a2 [(a1 × a1 ) · a3 ] To arrive at this first term in the second expression of the right-hand side, we used the following vector relation: A · (B × C) = (A × B) · C Notice that the part of this term in brackets is now identical to the expression for the denominator. We also see that the second term is equal to zero since we have a vector in a cross-product with itself. Our result for b2 × b3 is now: b2 × b3 = 4π 2 a1 We can use the same method for the other relevant cross-products to obtain them as well: b3 × b1 = 4π 2 a2 b1 × b2 = 4π 2 a3 The denominators of (1), (2), and (3), which are identical may be also readily solved using the result from the cross-products above: b1 · (b2 × b3 ) = 2π

a2 × a3 · 4π 2 a1 a1 · (a2 × a3 )

= 8π 3 By substituting the results we have obtained into the right-hand side of (1), we see that the equation holds true as can be shown for (2) and (3). 19. Three lattice planes for the simple cubic lattice are depicted in Fig 8.15. Of these, figure 8.15(b) represents a member of the family with Miller indices (110). Therefore, the plane is perpendicular to the reciprocal lattice vector: g = b1 + b2 Equation 8.33 gives the distance between planes: d=

2π |g|

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61 We may find the magnitude of g using the primitive reciprocal vectors found in example 8.1: √ 2π 2π 2π g = ˆi + ˆj ⇒ |g| = 2 a a a Therefore our result: √ a 2a d= √ = 2 2 21. a) First assume g = g 0 : Z Z a −igx igx e e dx =

a

dx = a

0

0

On the other hand, if g 6= g 0 our integral is as follows: Z a Z a 0 −ig 0 x igx e e dx = eix(g−g ) dx 0

0

By using Euler formula, the integrand may be expressed as a combination of sine and cosine functions: Z a cos(x(g − g 0 )) + i sin(x(g − g 0 ))dx 0

Therefore, from equation 8.10 we see that the arguments of the cos and sin functions will be of the form: 2πn x a Therefore, the integral from 0 to a is over a full period. Such integrals of cos and sin functions are equal to zero. b) Eq. 8.14 gives: f (x) =

X

Fg eigx

g 0

Multiply by e−ig x : 0

e−ig x f (x) =

X

0

Fg e−ig x eigx

g

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62

8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS

Integrate from 0 to a: Z a X Z −ig 0 x e f (x) dx = Fg 0

a

0

e−ig x eigx dx

0

g

From part (a), we know that the integral on the right-hand side has a value of zero unless g = g 0 , in which case the value of the integral is ”a”. Thus, the summation disappears as the only term that survives is g = g 0 : Z a 0 e−ig x f (x) dx = aFg0 0

25. We shall first carry out the substitutions then apply orthogonality conditions as directed. Our terms as given in the text are: 1 ψk (r) = 1/2 eik·r V X Vg eig·r V (r) =

Eq. 8.55 : Eq. 8.58 :

g

Our other two needed quantities may be determined from these: 1

ψk0 (r) =

1/2

V Substituting into the integral: Ik∗0 , k

=

0

eik ·r and ψk∗ (r) = X Vg Z g

V

1 V 1/2

e−ik·r

0

ei(g+k −k)·r dV

We separate the exponential terms, so the the integral may be written: X Vg Z 0 ∗ Ik0 , k = ei(g+k )·r e−ik·r dV V g We may also think of this as: X Vg Z g

V

ψk∗ (r)ψ(k0 +g) (r) dV

Using the orthogonality condition of 8.52: k = k 0 + g0 which is equation 8.62.

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10 Semiconductor Lasers Solutions 1. As with other radiative transitions we have dealt with, the photons emitted must have the same energy as the difference in energies between the two states. Here, the two states are the bottom of the conduction band and the top of the valence band, with an energy difference ∆ = 1.519 eV . Using equation I.25: hc E= λ hc 1240 eV · nm λ= = = 816.3nm E 1.519 eV

3. As outlined in section 10.3.1, heterostructures may be formed using semiconductors with similar lattice constants. Thus, the heterostructures grown on a substrate of InP must have the same lattice constant as InP. We’ll use the linear interpolation procedure for each alloy: a(Inx Ga1−x As) = a(InP) = xa(InAs) + (1 − x)a(GaAs) 5.869 = 6.058x + 5.653(1 − x) ⇒ x = .53 This leads to an ally of composition: In.53 Al.48 As p 5. By examining figure 2.6, we see that the graph of (θ02 /θ2 ) − 1 intersects the graph of tan θ and −cot θ three times as it decays. Therefore, if 63

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64

10. SEMICONDUCTOR LASERS - SOLUTIONS

p we insure that (θ02 /θ2 ) − 1 reaches zero after it intersects once, but before it does again, the well will have only one bound state. Per equation 2.30, it θ02 that includes dependence on V0 , the depth of the well. So, we solve for θ0 where the whole expression equals zero: r θ02 − 1 = 0 ⇒ θ02 = θ2 2 θ We kepp only the positive solutions and impose the condition that θ < π/2 as p that is where the graph of −cot θ begins in the positive domain. If (θ02 /θ2 ) − 1 decays completely before this point, then the electron will not have more than one ground state.  π 2 π 2 θ < ⇒ θ0 < 2 2 We now solve eq. 2.30 for V0 , using SI units at first: θ02 =

π π~2 m0 V0 L2 < ⇒ V < 0 2~2 2 m0 L2

π~2 (3.1416)(1.055 × 10−34 J · s)2 = m0 L2 .067(9.109 × 10−31 kg)(10 × 10−9 m)2 = 5.729 × 10−21 J We should now convert to eV: 5.729 × 10−21 J ·

1 eV = .036 eV 1.6 × 10−19 eV /J

The well must be less than 0.036 eV deep to insure that no more than one bound state exists. 7. We will use the ations:  A1 B1 C1

following procedure for these matrix multiplication oper    A2 A3 D1 A1 D1 + A2 D2 + A3 D3 B2 B3  D2  = B1 D1 + B2 D2 + B3 D3  C2 C3 D3 C1 D1 + C2 D2 + C3 D3      1 2 0 1 3 1 1 2 1 = 2 1 3 1 0 4

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65  1 1 1  2 1 1

    0 1 1 2     2 1 0 = 2 1 3 1 4     1 1 1 5 0 1 2 = 2 1 0 1 3

9. We shall proceed left to right, first consolidating the 1/2k1 and 1/2k2 terms as we will reintroduce them after carrying out the matrix operations.     ik L  k1 + k2 k1 − k2 eik2 L 0 e 2 (k1 + k2 ) e−ik2 L (k1 − k2 ) = ik2 L k1 − k2 k1 + k2 0 e−ik2 L e (k1 − k2 ) e−ik2 L (k1 + k2 ) Using this result in eq. 10.30:  ik L   1 e 2 (k1 + k2 ) e−ik2 L (k1 − k2 ) k1 + k2 k2 − k1 T = 4k1 k2 eik2 L (k1 − k2 ) e−ik2 L (k1 + k2 ) k2 − k1 k1 + k2 Carrying this through will give us following:   T11 T12 T = T21 T22 where each element is as follows: T11 =

 1  ik2 L e (k1 + k2 )2 − e−ik2 L (k1 − k2 )2 4k1 k2

T12 =

 −1  ik2 L 2 e (k1 − k22 )2 − e−ik2 L (k12 − k22 ) 4k1 k2

T21 =

 1  ik2 L 2 e (k1 − k22 ) − e−ik2 L (k12 − k22 ) 4k1 k2

T22 =

 −1  ik2 L e (k1 − k2 )2 − e−ik2 L (k1 + k2 )2 4k1 k2

We see that the expression we found for T11 does in fact agree with the first part of eq. 10.31, however the Euler formula was used in the second part to

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66

10. SEMICONDUCTOR LASERS - SOLUTIONS

further simplify:

 1  ik2 L e (k1 + k2 )2 − e−ik2 L (k1 − k2 )2 4k1 k2  1  (cos k2 L + i sin k2 L)(k12 + 2k1 k2 + k22 ) − (cos k2 L − i sin k2 L)(k12 − 2k1 k2 + k22 ) = 4k1 k2 1 = [2k1 k2 cos k2 L + i(k12 + k22 )sin k2 L] 2k1 k2

T11 =

The same may be done for the remaining three elements if desired. 11. Equations10.25 and 10.26 give the following: A1 = T11 A2 + T12 B2 B1 = T21 A2 + T22 B2 The transmission amplitude of light incident from the right will be equal to B1 /B2 and the corresponding reflection amplitude is equal to A2 /B2 . As directed, we set A1 = 0. 0 = A1 = T11 A2 + T12 B2

(1)

B1 = T21 A2 + T22 B2

(2)

Solving (1) for A2 and substituting into (2): B1 =

1 −T21 T12 B2 + T22 B2 = B2 (T11 T22 − T21 T12 ) T11 T11

We recognize the quantity in parentheses as the determinant of the T-matrix, therefore: B1 det T t21 = = B2 T11 This equation 10.38. Now for 10.39, we solve eq. (1) above for A2 /B2 : r21 =

A2 −T12 = B2 T11

15. a) The total rate of change in the amount of water in the reservoir

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67 will equal the rate of water entering minus the rate in which it is leaving. Letting V equal the volume of water in the reservoir: dV = Rf − RD dt Where RD is the total rate at which water drains. Since the reservoir is draining from two ports, we set up the following: RD = Rd1 + Rd2 Therefore, our total rate equation is: dV = Rf − h(C1 + C2 ) dt b) To find the steady state height, we set: dV =0 dt And solve for h: h=

Rf c1 + c2

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68

10. SEMICONDUCTOR LASERS - SOLUTIONS

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11 Relativity I - Solutions 1. Einstein’s second postulate on the special theory of relativity states that the speed of light is independent of the motion of the source, therefore the velocity of the light in the reference frame of the spaceship is c. 3. a) The inverse Lorentz transformations are given by eqs. 11.17. Therefore, we need to first calculate γ: γ=

1 √

1 1−u2 /c2

1 =√ 1 − .252 = 1.033 Now we transform from S to S 0 : x = γ(x0 + ut0 ) = 1.033 (2m + .25(4)m) = 3.10 m y = y0 = 0 m z = z0 = 0 m 1.033 (4 s + .5 s) t = γ(t0 + ux0 /c2 ) = c 4.65 s = c b) We now use eqs. 11.15 and the results from part (a) to transform back 69

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70

11. RELATIVITY I - SOLUTIONS

to S 0 : x0 = γ(x − ut) = 1.033(3.10 m − .25(4.65) m) = 2.0 m 0 y =y=0m z0 = z = 0 m 1.033 t0 = γ(t − ux/c2 ) = (4.39 s − .25(3.10) s) c 4.0 s = c

5. a) We may use equation 11.23 to calculate the length as perceived in our reference frame: r .5c LM = 1 − 2 (1 m) = .86 m c

b) Equation 11.26 will give us the time we record the object passing us. However, we need to calculate ∆tR , which is the time we would record if we were traveling along with meter stick: ∆tR =

11 m d = = 6.67 × 10−9 s 8 v .5(3 × 10 m/s)

Now, to transform to our frame of reference: ∆tM

6.67 × 10−9 s = q = 7.70 × 10−9 s 2 1 − (.5c) c2

7. The percentage of length contraction will be equal to:   LM 1− × 100% LR Therefore, we may calculate LM /LR via eq. 11.23: r r LM u2 (3 × 10−6 c)2 = 1− 2 = 1− LR c c2

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71 LM ≈ 1 ⇒ 0% contraction LR While the high speed of the jet is truly impressive, they still do not travel with enough velocity to display relativistic effects. The time dilation experienced here may be calculated by equation 11.26: ∆tM =

∆tR √ 12 2 1−u /c

1 yr

=q

1−

(3×10−6 c)2 c2

∆tM ≈ 1 yr Once again, the relativistic time dilation has an extremely small effect, well within the intrinsic error of any wristwatch. 9. a) By differentiating both sides of the equation given in the problem for the position of the center of mass with respect to t, we may calculate the velocity of the center of mass: d d 2mxCM = (mxe + mxp ) dt dt dxCM 1 = (ue + up ) dt 2 where ue and up equal the velocities of the electtron and positron, respectively. We now substitute these quantities to obtain the velocity of the center of mass: 1 ux = (.95c + .2c) = .575c 2 b) The lifetime of the produced particle, gives 2.0 × 10−8 s will be taken as ∆tR in eq. 11.26: 2.0 × 10−8 s ∆tM = q = 2.44 × 10−8 s (.575c)2 1 − c2 11. a) The equations of section 11.4.6 will give us the transformed frequencies, however we are given the wavelength of the sodium D2 line. This is a minor obstacle, as we can use equation I.23: f0 =

c 3 × 108 m/s = = 5.09 × 1014 Hz λ0 589.0 × 10−9 m

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11. RELATIVITY I - SOLUTIONS

For an approaching light source: s f=

where β = v/c = 0.3. r f=

1+β f0 1−β

1.3 (5.09 × 1014 Hz) = 6.94 × 1014 Hz 0.7

We then use eq. I.23 to calculate the correspond wavelength: λ=

c 3 × 108 m/s = = 432.2 nm f 6.94 × 1014 Hz

λ − λ0 = 432.2 nm − 589.0 nm = −156.8 nm b) For a receding light source: s r 1−β .7 f= f0 = (5.09 × 1014 Hz) = 3.74 × 1014 Hz 1+β 1.3 3 × 108 m/s c = = 802.1 nm f 3.74 × 1014 Hz λ − λ0 = 802.1 nm − 589.0 nm = 213.1 nm λ=

c) The equation adjacent to figure 11.12 allows us to calculate the Doppler shift due to transverse motion: f=

f0 γ

We calculate γ: 1 1 γ=p =p = 1.05 1 − u2 /c2 1 − (.3)2 And substitute: f=

5.09 × 1014 Hz = 4.86 × 1014 Hz 1.05

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73 λ=

c 3 × 108 m/s = = 617.2 nm f 4.86 × 1014 Hz

λ − λ0 = 617.2 nm − 589.0 nm = 28.2 nm

13. We use eq. 11.41 for the receding source: β= r f=

v = 0.80 c

1 − .80 × 100 M Hz = 33 M Hz 1 + .80

15. Equation 11.16 gives the definition for γ: γ=p

1 1 − u2 /c2

However, we first solve eq. 11.38 for v: v 0 = γ(v − u) · γ(1 + uv 0 /c2 ) We first simplify the right-hand side, then bring the terms containing v 0 to the left, and solve:   vuv 0 u2 v 0 0 2 v =γ v−u+ 2 − 2 c c v0 v 0 vu u2 v 0 − + 2 =v−u γ2 c2 c   uv u2 1 0 v − 2 + 2 =v−u γ2 c c   u2 uv u2 0 v 1− 2 − 2 + 2 =v−u c c c v0 =

v−u 1 − uv c2

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74

11. RELATIVITY I - SOLUTIONS

17. Eq. 11.43 allows us to calculate the proper time interval. We apply this equation at each segment of the observer’s journey. First, the traveler remains stationary for 1 m of time: (proper time)2 = 12 − 02 proper time = 1 m Then, he travels 4 m in 5 m of time: (proper time)2 = 52 − 42 proper time = 3 m Finally he travels 1 m in 2 m of time: (proper time)2 = 22 − 12 √ proper time = 3 m Therefore, the total proper time is 4 +

√

3 m.

22. Per equation 11.60, tσ = g σµ tµ , and eq. 11.58, tµ = gµν tν , the g matrices will serve to raise or lower the indices of a vector. We now use the given expression Λνµ = gµρ Λρσ g σν to operate on vν : g σν vν = v σ Now, the Λρσ will serve to transform to the ρ0 indices: Λρσ g σν vν = Λρσ v σ = v 0ρ We now use gµρ to lower the index: gµρ Λρσ g σν vν = gµρ v 0ρ = vρ0 We have now shown that vµ0 = Λνµ vν , which is similar to the contravariant vector’s transformation in form but requires an alternately defined transformation matrix Λνµ .

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75

23. First recall that gµρ and g σν  1 0  0 0

will have the form:  0 0 0 −1 0 0  0 −1 0  0 0 −1

The first operation, gµρ Λρσ :      1 0 0 0 γ −βγ 0 0 γ −βγ 0 0 0 −1 0   γ 0 0 0 0 0   −βγ  = βγ −γ  0 0 −1 0   0 0 1 0  0 0 −1 0  0 0 0 −1 0 0 0 1 0 0 0 −1 Now, gµρ Λρσ g σν :      γ −βγ 0 0 1 0 0 0 γ βγ 0 0 βγ −γ    0 0 0   0 −1 0  = βγ γ 0 0 0 0 −1 0  0 0 −1 0   0 0 1 0 0 0 0 −1 0 0 0 −1 0 0 0 1

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11. RELATIVITY I - SOLUTIONS

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12 Relativity II - Solutions 1. The components of the four-velocity are given by the relations between eqs. 12.3 and 12.4. Since the electron is traveling in the x-direction, we note that: dz dy = =0 dt dt Therefore, our relevant components are: v 0 = γc and v 1 = γ

dx dt

We then calculate γ: 1

γ=p

1−

u2 /c2

1

=q

1−

(.2c)2 c2

= 1.02

Giving us: v 0 = 1.02c v 1 = 1.02(.2c) = .204c b. The four-momentum as given in eq. 12.6 is simply the product of the electron’s mass and each component of its four-velocity: p0 = 1.02me c p1 = .204me c 77

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78

12. RELATIVITY II - SOLUTIONS

3. We may use equations 12.12 and 12.13 which give expressions for the rest energy and kinetic energy, respectively. Setting R=KE: γmc2 = 2(γ − 1)mc2 γ = 2γ − 2 ⇒ γ = 2 Using eq. 12.3 for γ: γ=p

1 1 − u2 /c2

=2

1 (1 − v 2 /c2 ) = 2   2 v 1 1− 2 = c 4

p

3 v2 = c2 4 √ 3 v= c 2 In order for the kinetic evergy to √ be equal to the rest energy, the particle must be traveling at a speed of ( 3/2)c. 5. We first evaluate the derivatives of f (x) called for in the Taylor series: f 0 (x) =

1 1 2 (1 − x)3/2

f 00 (x) =

1 3 4 (1 − x)5/2

We now evaluate at x=0 and substitute into the given form of the Taylor series: 1 3 f (x) = 1 + x = x2 + . . . 2 8 2 2 Now, let x = v /c : 1 p

1 − u2 /c2

=1+

1 v2 3 v4 = 2 c2 8 c4

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79 This result is identical to eq. 12.10. 7. We shall use eq. 12.13 and let m equal the electron’s mass. However, it will be most convenient to use the mass given in Appendix A in units of M eV /c2 .   M eV c2 100 M eV = (γ − 1) .511 2 c 100 = .511γ − .511 γ = 196.7 We now find the electron’s speed from eq. 12.3: γ = 196.7 = p 38688 = 1−

1 1 − u2 /c2

1 2 1 − vc2

v2 = 2.58 × 10−5 2 c v2 .99997 = 2 c v ≈ .99999c

The electron would have to be traveling toughly the spped of light possess 100 MeV of kinetic energy. 9. a) For the laboratory frame, recall that the target particle is at rest, therefore has zero momentum. We go ahead and carry out the four-vector addition as called for in the numerator of the expression for s: |pA + pB |2 [(EA + EB )2 − (pA + pB )2 ] = c2 c2 Note that we have separated the energy and momentum components of the four-momentum. Since particle B is at rest, we realize that EB must equal the rest energy of the particle and pB = 0: s=

[(EA + mB c2 )2 − |pA |2 ] c2

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12. RELATIVITY II - SOLUTIONS

In the center of mass frame, we take the colliding particles to have equal and opposite momenta. Therefore, we begin as we did for the laboratory frame, however we can no longer take the energy of particle B to be equal to its rest energy and we allow the sum of the momenta to be zero. s=

[(EA + EB )2 − (pA + pB )2 ] (EA + EB )2 = c2 c2

b) We begin by directly substituting the sums of the four-vectors into the sum of the Mandelstam variables:  (EA − EC )2 1 (EA + EB )2 2 − (p + p ) + − (pA − pC )2 s+t+u= 2 A B c c2 c2  (EA − ED )2 2 + − (pA − pD ) c2 By conservation of energy and momentum: EA − ED = E C − E B EA − EC = E D − E B pA − pD = pC − pB pA − pC = pD − pB We substitute these quantities into our equation above:  (ED − EB )2 1 (EA + EB )2 2 − (p + p ) + − (pD − pB )2 s+t+u= 2 A B c c2 c2  (EC − EB )2 2 + − (pC − pB ) c2 We use the fact that in the laboratory frame, pB = 0 and EB = mB c2 and simplify: 2EA mB 2ED mB 2EC mB − − + m2A + 3m2B + m2C + m2D c2 c2 c2 2mB = 2 (EA − ED − EC ) + m2A + 3m2B + m2C + m2D c

s+t+u=

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81 Once again, from the conservation of energy, we know that EB = EA − ED − EC . Making this substitution and using EB = mB c2 : s + t + u = −2m2B + m2A + 3m2B + m2C + m2D = m2A + m2B + m2C + m2D

11. The particle collision here involes two particles with equal mass and speed traveling in the opposite direction. Therefore, using the conservation of momentum: pA + pB = pC = 0 Thus, we are in the center of mass frame. An important result here is that the resulting particle is at rest. We may then use the conservation energy where the energy of the resulting particle is its rest energy and the energy of the colliding particles are given by eq. 12.9. We let m be the mass of the each of the colliding particles and mC be the mass of the resulting particle: EA + EB = EC ⇒ 2γmc2 = mC c2 mC = 2γm Using equation 12.3: m

mC = 2 p

1 − (2/3)2

= 2.68m

We see that since the mass of the resulting particle is greater than the sum of the colliding particles, some of their kinetic energy was converted into the resulting particle’s rest-energy. 13. We consider the center of mass frame where the proton and antiproton have equal and opposite momenta. Since the proton and antiproton have equal masses, their speeds must also be equal. Therefore, when we impose conservation of energy and use eq. 12.9: Ep + Ep¯ = EX where EX is the energy of the resulting particle. As stated above, we are in the center of mass frame, where the resulting particle is at rest and its energy is given by eq. 12.12: 2γmp c2 = mX c2

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82

12. RELATIVITY II - SOLUTIONS 2γ(938 M eV ) = 9700 M eV γ = 5.17

We may then find the kinetic energies of the colliding particles by equation 12.13: KE = (γ − 1)mp c2 = 3912 M eV Their velocities may be calculated by eq. 12.3: 1

γ = 5.17 = p

1 − u2 /c2

v2 = .0374 c2 v = .981c

1−

The proton and antiproton will each have a speed of .981c in the center of mass frame. 15. First, we test the relation: αi αj + αj αi = 2δij I The Dirac-Pauli representation is given in eq. 12.42 and we merely substitute:     i j  0 σi 0 σj σσ 0 i j αα = i = σ 0 σj 0 0 σiσj     j i  0 σj 0 σi σ σ 0 j i αα = j = σ 0 σi 0 0 σj σi   i j σ σ + σj σi 0 i j j i = (σ i σ j + σ j σ i )I α α +α α = 0 σiσj + σj σi It can be shown that the Pauli matrices satisfy the following condition:  I if i = j i j σσ = j i −σ σ if i 6= j Therefore, we see that: i j

j

i

σ σ +σ σ =



2 if i = j 0 if i 6= j

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83 Or, more concisely, σ i σ j + σ j σ i = 2δij I. Thus, αi αj + αj αi = 2δij I . Now, we turn our attention to the relation: β 2 = I. From eq. 12.42, we get the definition for β:   I 0 β= 0 I We carry out the multiplication:     2  I 0 I 0 I 0 = 0 I 0 I 0 I2 Since I 2 = I, β 2 = I. 17. Muons and electrons are both Dirac particles. The Feynman diagrams for muon-electron scattering are equivalent to the Feynman diagram shown in Fig. 12.4 with one of the incoming and outgoing lines corresponding to an electron and the other corresponding to a muon.

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84

12. RELATIVITY II - SOLUTIONS

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13 Particle Physics - Solutions 1. a) This process can not occur as it would violate the conservation of baryon number and lepton number. The baryon number is one of the lefthand side but zero on the right. Likewise, the electron lepton number of the left-hand side is zero while Le = −1 on the right. b) This process can not occur due to violation of energy conservation. The Σ◦ has a rest energy of 1192 MeV while the Λ and the π ◦ have a total rest energy of 1250 MeV. c) Conservations of charge, energy, baryon number, and lepton number are all observed, therefore the process may occur. Furthermore, the process conserves strangeness and proceeds by the strong interaction. d) Here, the conservation of muon lepton number is violated as the total number on the left is 0, but is -1 on the right. e) K − has a strange quantum number S=-1 where the π − π ◦ system has S=0. Since charge, energy, lepton number, and baryon number are all conserved, this process proceeds via the weak interaction. f ) The ρ0 , being a boson, will have a symmetric wave function. The spin of a ρ0 will have a value of 1, therefore its total angular momentum will be J=1. By the conservation of angular momentum, J=1 for the right-hand side, thus the spinless pions would have an antisymmetric total wave function. Also being bosons, this is forbidden. 85

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86

13. PARTICLE PHYSICS - SOLUTIONS

g) Strangeness is not conserved while the other conservation laws are, therefore the process must occur via the weak interaction.

3. The rest energies may be found in tables 13.9 through 13.11. a) Initial state: 140 + 938 = 1078 MeV Final state: 940 + 135 = 1074 Mev b) Initial state: 140 + 938 = 1116 MeV Final state: 938 + 135 = 1073 Mev c) Initial state: 140 + 938 = 1078 MeV Final state: 494 + 1138 = 1632 Mev The kinetic energy of this initial state must be greater than that of the final state to ensure conservation of energy.

5 To calculate the strangeness of these particles, we refer to tables 13.913.11 to find the quark content of each particle, then use eq. 13.26. a) The Ω− has a quark content of sss, therefore using equation 13.26: S = −[3] = −3 We repeat for the particles on the right-hand side to find that S(Λ) = −1 and S(K − ) = −1. Therefore ∆S = 1 and the process must occur via the weak interaction. b) Using the same procedure as part (a), S(Σ◦ ) = −1, S(Λ) = −1, and S(γ) = 0. There is no change in strangeness, therefore the process may occur via the electromagnetic, as we recognize the presence of the photon. c) Using the same procedure as parts (a) and (b) above, S(Λ) = −1, S(p) = 0, and S(π − )=0. ∆S = 1 and the process must occur via the weak interaction.

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87

7. a) This process may occur via the strong interaction as all conservation laws are met. b) The strangeness of the initial state is zero; however, the strangeness has a value of -1 in the final state. The other conservation laws are observed, and therefore this process must occur via the weak interaction. c) This process is strictly forbidden as the baryon number of the initial state is +1 while that of the final is zero. The Λ particle has baryon number ¯ ◦ antiparticle has a baryon number of -1. of +1, while the Σ d) This process is strictly forbiden as the muon lepton number of the antineutrino is -1 which is the total muon lepton number for the initial state. The final state has a muon lepton number zero. Note that the similar conservation of electron lepton number is also broken. e) Here again, lepton number conservation is violated. The electron neutrino on the left-hand side has an electron lepton number of +1 whereas the positron on the right-hand side has an electron lepton number of -1. As a consequence, this process is strictly forbidden. 9. The quark composition may be found in tables 13.9-13.11: a) Ω− : sss Λ : uds

K − : s¯ u

Here we have nothing but strange quarks in the initial state, but the final state contains one fewer s quark but does contain u, u¯, d quarks. b) π − : dˆ u p : uud Λ : uds K ◦ : sd¯ There are no strange quarks in the inital state, but two in the final. Notice that the antiquark in the left-hand meson is a u¯, while the antiquark that appears in the meson that occurs on the right is an antidown quark. c) p : uud K − : s¯ u Ξ− : dss

K + : u¯ s

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13. PARTICLE PHYSICS - SOLUTIONS

There is no change in strangeness during this process. Even though there is an additional strange quark on the right, there is also an s¯ quark which cancels the added strange quantum number. 11. If the wave has periodic boundary conditions at the two ends of the region, then the length L must equal an integer multiple of the wavelength: L = nλ ⇒ λ =

L n

Using the de Broglie relation and this value of the wavelength: p=

h hn = λ L

Recall that ~ = h/2π therefore: p=

2π~n L

We can think of this relation in the following way. In momentum space, each state has a “length” of 2π~/L, therefore, the number of states in a differential length dp will be: number of states =

dp L = dp 2π~/L 2π~

(1)

If we now expand our region to include a cubic volume, extending to x = L, y = L, z = L from the origin, we now have three terms like (1) that form a volume element in a momentum space: L L L dpx dpy dp 2π~ 2π~ 2π~ x V d3 p = (2π~)3

dN =

13. We can use figure 13.29 as a guide to obtain the remaining weights from the highest weight. Notice in figure 13.29(b) the highest weight is 3µ1 , where µ1 is given by eq. 13.84. As we work along a diagonal such as the one with the weights labeled in Fig. 13.29(b), we subtract an additional α1 . As

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89 we work right-to-left along a row, we subtract (α1 + α2 ) for each step. As follows, we shall work from right to left across a row, and completing rows from the top down to 3µ1 − 3α1 . Expressions for α1 and α2 are given in eq. 13.78:     3 1 √ = √ 3µ1 = 3 3/3 3   2 3µ1 − α1 − α2 = √ 3   1 3µ1 − 2α1 − 2α2 = √ 3   0 3µ1 − 3α1 − 3α2 = √ 3   5/2 √ 3µ1 − α1 = 3/2   3/2 √ 3µ1 − 2α1 − α2 = 3/2   5/2 √ 3µ1 − 3α1 − 2α2 = 3/2   2 3µ1 − 2α1 = 0   1 3µ1 − 3α1 − α2 = 0   3/2 √ 3µ1 − 3α1 = − 3/2

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13. PARTICLE PHYSICS - SOLUTIONS

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14 Nuclear Physics - Solutions 1. a) Recall from section 14.2 the nomenclature for various isotopes: A ZX

where A atomic mass number which is the sum of the number of neutrons and number of protons, while z is the atomic number which is the number of protons, and X is the element symbol. Therefore, the number of neutrons is equal to A minus Z: 3 7 Li 63 29 Cu 238 92 U

: 3 protons, 4 neutrons : 29 protons, 34 neutrons : 92 protons, 146 neutrons

3. Equation 14.3 provides an estimate of the radius of a nucleus: R = 1.12A1/3 f m where A is the atomic mass number: 4 2 He 16 8 He 56 26 F e 208 82 F e 237 93 F e

: R = 1.12(4)1/3 f m = 1.78f m : R = 1.12(16)1/3 f m = 2.82f m : R = 1.12(56)1/3 f m = 4.28f m : R = 1.12(208)1/3 f m = 6.64f m : R = 1.12(237)1/3 f m = 6.93f m

91

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14. NUCLEAR PHYSICS - SOLUTIONS

8. The semi-empirical formula without the pairing term is as follows: B(N, Z) = aA − bA2/3 −

dZ 2 (N − Z)2 − s A1/3 A

If N=Z, we see that the fourth term immediately equals zero. Since we want the binding energy per nucleon, we can divide the entire equation by the number of nucleons, which will be equal to A: dZ 2 B(N, Z) = a − ba−1/3 − 4/3 A A Now we let Z=A/2: dA2 4A4/3 d = a − bA−1/3 − A2/3 4

Binding energy per nucleon = a − bA−1/3 −

Now, we differentiate with respect to A and set equal to zero: ¯ dB 1 1 = A−4/3 − dA−1/3 = 0 dA 3 2 Now solve for A:

1 −1/3 2 A (bA−1 − ) = 0 3 d 2 b = A d 2b A= d The critical points are A = 0 and A = 51.3. Per the verification called for in the problem, A/2 ≈ 26. Further testing of the value of the derivative in the intervals (0, 13) and (13, ∞) further identifies A/2 = 26 as a maximum. 9. We use equation 14.13 first which relates the half-life to the proportionality constant λ: t1/2 = 2 min = 120 s = λ=

ln2 λ

ln2 = .006 s−1 120 s

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93 We now use the equation following eq. 14.12 and our initial condition R = 1200 s−1 at t=0 to find N0 : R = 1200s−1 = .006N0 e−.006(0) = (.006s−1 )N0 N0 = 2 × 105 Now we may use this same equation to find the transition rates at the subsequent times: t = 4 min = 240 s :

R = (.006s−1 )(2 × 105 )e−.006(240) R = 284 s−1

t = 6 min = 360 s :

R = (.006s−1 )(2 × 105 )e−.006(360) R = 138 s−1

t = 8 min = 480 s :

R = (.006s−1 )(2 × 105 )e−.006(480) R = 67 s−1

We see, just as expected, that the decay rate itself decays exponentially. 11. Our strategy here should be to ultimately use the rate of decay equation (the one immediately following eq. 14.12) to find the number of 14 6 C atoms that are responsible for the measured radiation in the sample. We may then use the atomic mass of carbon and Avogadro’s number to find the number of total atoms in the one gram sample. Then will we have the necessary information to calculate the proportion. First, since we are given the half-life of 14 6 C, we use eq. 14.13 to obtain the proportionality constant λ: t1/2 =

λ=

ln 2 ln 2 ⇒λ= λ t1/2

ln 2 ln 2 = = 2.30 × 10−10 min−1 9 5730 yr 3.012 × 10 min

Now to bring this into the rate equation: −10

15.3 min−1 = (2.30 × 10−10 min−1 )N0 e−2.30×10 N0 = 6.65 × 1010 atoms

min−1

14 6 C

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94

14. NUCLEAR PHYSICS - SOLUTIONS

We now calculate the total number of carbon atoms in the sample: 1g·

1 mol 6.022 × 1023 atoms · = 5.018 × 1022 atoms C 12 g 1 mol

We now simply find our proportion: 6.65 × 1010 1.32 × 10−12 atoms = 5.018 × 1022 total atoms C

14 6 C

One check from Appendix B shows that this calculation agrees with the fact that the isotope of carbon containingj 14 nucleons is not listed as a percentage. This indicates its rarity and thus reliability for dating ancient relics since these nuclear phenomena are readily measurable. b) We use the value of λ calculated above as well as the value of N0 which would have been relevant at the time that the organism was alive. We convert 20,000 years to minutes and use our rate equation with this as the new time: 20, 000 years = 1.051 × 1010 minutes −10

R = (2.30 × 10−10 min−1 )(6.65 × 1010 min) e−2.30×10 = 1.36 beta rays per minute

min−1 (1.051×1010 min)

We see that there are many fewer counts in the ancient sample than in the current one. 13. We may calculate the release of energy of the 74 Be nucleus by finding the difference in binding energy between 74 Be and 73 He. Since the energy of 74 He is given in table 13.2 to be 39.25 MeV, we need just use the semi-empirical formula (eq. 14.6) for 74 Be: B(3, 4) = 67a − b72/3 −

42 d (−1)2 − s −0 71/3 7

= 97.899 M eV We now subtract the two: ∆E = 97.899 − 39.25 = 58.65 M eV

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95 Thus the energy released is 58.65 MeV. 15. a) We begin by using equation 14.9 to calculate the nuclear masses of 9 22 34U and 9 02 30U : m(92, 142) = 221710 M eV /c2 m(90, 140) = 217913 M eV /c2 Now, using the data from Appendix B, m(2, 2) = 3728.4 M eV /c2 , and: ∆E = 221710 M eV − 217913 M eV − 3728.4 M eV = 68.44 M eV

b) The energy carried off by the alpha-particle will equal the difference in binding energies of 9 22 34U and 9 02 30U . Using equation 14.6, B(92, 142) = 1923.88 M eV B(90, 140) = 1883.46 M eV Thus, the alpha-particle carries off 40.42 MeV of energy. 17. We follow a very similar procedure to that used in section 14.5. However, we will use the following subscripts to represent the various particles in the process: D represents the deuteron, X ∗ represents the excited nucleus, and p represents the proton. We begin with our statements of conservation of momentum: pD = pp cosθ + Px 0 = pp sinθ + Py Solving for Px and Py : Px = pD − pp cosθ Px = −pp sinθ The total kinetic energy before the collision is equal to the kinetic energy of the proton: p2 Ei = D 2mD

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96

14. NUCLEAR PHYSICS - SOLUTIONS

The total kinetic energy after the collision is: Ef =

p2p 1 1 + (pD − pp cosθ)2 + (pp sinθ)2 2mp mX ∗ 2mX ∗

Due to the conservation of energy, the energy absorbed by the nucleus will be equal to the loss of kinetic energy: E = Ei − E f =

p2p 1 p2D − − (p2 + p2p − 2pD pp cosθ) 2mD 2mp mX ∗ D

We now substitute the corresponding energies for the first two terms as well as the following: p2D = E D mD 2 p2p = Ep mp 2 pD pD p = ED Ep mD mp 2 Our expression for the energy of the excited nucleus due to deuteron stripping is now: p     2 ED Ep mD mp mp mD − Ep 1 − − cosθ E = ED 1 − mX ∗ mX ∗ mX ∗

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