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PHYS 560: Assignment 6 : SOLUTIONS Martin J. Savage December 18, 2009 Abstract The sixth assignment for Nuclear Physics, PHY560, Autumn 2009.

December 2009

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Assignment 6 : PHYS 560 Due : Dec 14 1. Wong : 6.1: We wish to construct the allowed values of total angular momentum for a system composed of two phonons each with angular momentum λ. The wavefunction must be totally symmetric under interchange. If we start by considering the stretch state, M = 2λ, then (a) There is only one way to obtain this : |λi ⊗ |λi. This state is allowed as it is symmetric under interchange, and is the stretch state associated with J = 2λ. (b) For the state with M = 2λ − 1, we have states |λi ⊗ |λ − 1i and |λ − 1i ⊗ |λi. The sum of these states gives the M = 2λ − 1 state of the J = 2λ multiplet. (c) For the state with M = 2λ − 2, we have states |λi ⊗ |λ − 2i, |λ − 2i ⊗ |λi, |λ − 1i ⊗ |λ − 1i. The linear combination ∝ |λi ⊗ |λ − 2i + |λ − 2i ⊗ |λi + 2|λ − 1i ⊗ |λ − 1i is the M = 2λ − 2 member of the J = 2λ multiplet as is obvious from acting with the lowering operator on the M = 2λ−1 state), while the combination ∝ |λi ⊗ |λ − 2i + |λ − 2i ⊗ |λi − |λ − 1i ⊗ |λ − 1i is the M = 2λ − 2 member of the J = 2λ − 2. (d) This pattern continues all the way through the spectrum. Therefore there are no states with odd J in the state formed from two identical bosons each with λ. For the state composed of three λ = 2 bosons, we simply construct all possible states. I will shorthand this, and write states with symmetrization implicit. The number of states with different values of the mi corresponds to the number of symmetrized, linearly independent states–hence states of the systems: M = 6 : Denoting the tensor product of states by |m1 , m2 , m3 i, we have |2, 2, 2i, and hence only one state with J = 6. M = 5 : |2, 2, 1i, hence only one state, there is NO J = 5 state. M = 4 : |2, 2, 0i, |2, 1, 1i, hence there are two states, there is J = 4 state. M = 3 : |2, 2, −1i, |2, 1, 0i,|1, 1, 1i, hence there are three states, there is J = 3 state. 2

M = 2 : |2, 2, −2i, |2, 1, −1i,|2, 0, 0i,|1, 1, 0i, hence there are four states, there is J = 2 state. M = 1 : |2, 1, −2i, |2, 0, −1i,|1, 1, −1i,|1, 0, 0i, hence there are four states, there is NO J = 1 state. M = 0 : |2, 0, −2i, |2, −1, −1i, |2, −2, 0i,|1, 0, −1i,|0, 0, 0i, hence there are five states, there is a J = 0 state. Therefore, we have that three identical spin-2 bosons will form states with J = 0 ⊕ 2 ⊕ 3 ⊕ 4 ⊕ 6. 2. Wong : 6.2: This problem involves taking the measured values of the energy levels in 25 Mg, and their identification as members of rotational bands (via their electromagnetic transition strengths) and determining the momentum of inertia of the intrinsics state that the rotational state + is built on, and the decoupling parameters in the K π = 21 bands. The energy of states in a rotational band have the form   1 1 J+ 21 (J + ) + EK , (1) EJ,K = J(J + 1) + a δK, 1 (−) 2 2I 2 where I is the moment of inertia and a is the decoupling parameter. The easiest way to determine the parameters is to form the differences in energies, eliminating EK , and then performing a fit to all the available data. I used the Nonlinearregression package in mathematica that produces fit values and standard errors, along with the confidence intervals and error-ellipses. I will only quote the standard errors on the parameters, but one should be aware of the correlation between the extracted parameters. Fitting for I and a is highly correlated, and so 1 I extracted the coefficient of the J(J + 1) and the δK, 1 (−)J+ 2 (J + 21 ) 2 terms. I find that for the K =

5+ 2

band

I = 2.52 ± 0.16 MeV−1 ,

(2)

which gives fit energy-levels at 0, 1.39, 3.8, 5.36MeV, c/w 0, 1.64, 3.41, 5.45MeV with uncertainties that I have not propagated (but could and should!). I find that for the lowest K =

1+ 2

band

I = 3.18 ± 0.12 MeV−1 and a = −0.356 ± 0.036 , 3

(3)

which gives fit energy-levels at 0.585(input), 0.89, 1.96, 2.66, 4.58, 5.69MeV, c/w 0.585, 0.98, 1.96, 2.74, 4.70, 5.74MeV. I find that for the second K =

1+ 2

band

I = 2.73 ± 0.12 MeV−1 and a = −0.21 ± 0.14 ,

(4)

which gives fit energy-levels at 2.56(input), 2.997, 4.10, 5.12MeV, c/w 2.56, 2.80, 3.905, 5.00MeV. 3. Wong : 6.5: We are extracting the moment of inertia and the intrinsic quadrupole moment of 20 N e from the B(E2) matrix elements and the energy eigenvalues. Fitting the energy levels (j π , E(M eV )) = (0, 0+ ), (2+ , 1.22), (4+ , 4.08), (6+ , 8.57), (8+ , 14.7), using the same procedure as described earlier gives I = 2.45 ± 0.15 MeV−1 . The inband B(E2)-values are related to the intrinsic quadrupole moment, Q0 , by B(E2; Ji → Jf )

5 2 2 e Q0 |hJi K20|Jf 0i|2 16π 5 2 2 3J(J − 1) e Q0 16π 2(2J + 1)(2J − 1)

= K=0

→

,

(5)

and the transition rate is given by W (λ; Ji → Jf ) = α

8π(λ + 1) E 2λ+1 B(Eλ) , λ[(2λ + 1)!!]2 γ

(6)

where λ is the transition multipolarity, and I have tossed all factors of ~ and c. Fitting the lowest three transition strengths, Γ(2+ , 1.63MeV → 0+ , g.s.) = 6.3 × 10−4 eV Γ(4+ , 4.25MeV → 2+ , 1.63 MeV) = 7.1 × 10−3 eV Γ(6+ , 8.78MeV → 4+ , 4.25 MeV) = 0.10 eV ,

(7)

gives eQ0 = 49.8 ± 3.4 e fm2 . The fit values of the transition rates are 4.5 × 10−4 , 7.0 × 10−3 , 0.12eV. Using the relation QJK =

3K 2 − J(J + 1) Q0 (J + 1)(2J + 3) 4

,

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gives the quadrupole moment of the J π = 2+ state to be Q20 = −14.2 e fm2 . 4. Wong : 9.5: We need to calculate the quadrupole moment and moment of inertia about the symmetry axis of an axially-symmetric uniform charge distribution, bounded by r < R0 (1 + βY20 (Ω)). This is straightforward: r Z 16π d3 r r2 Y20 (Ω) ρ(r) Qzz = 5 r Z Z R0 (1+βY20 (Ω)) 16π dr r4 Y20 (Ω) dΩ = 5 0 r Z   16π 5 = R 0 ρ0 dΩ β (Y20 (Ω))2 + 2β 2 (Y20 (Ω))3 + O(β 3 ) 5 " # r r 16π 5 2 5 2 R ρ0 β + 2β = 5 0 7 4π =

  3ZR02 √ β 1 + 0.3604β + O(β 3 ) 5π

where we have used Z =

4π ρ R3 3 0 0

,

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+ O(β 2 ).

The moment of inertia about the z-axis is given by Z Izz = ρ0

R0 (1+βY20 (Ω))

Z dΩ Z

dr r4 sin2 θ

0

1 = ρ0 R05 5

dΩ sin2 θ 1 + 5β Y20 (Ω) + O(β 2 ) # " √ 5 8π = ρ0 R05 1 − √ β + O(β 2 ) 15 2 π   2 M R02 1 − 0.63β + O(β 2 ) , = 5



(10)

which differs from the desired result given in Wong. Wongs answer is incorrect, as one can show that Ixx = Iyy =

  2 M R02 1 + 0.31β + O(β 2 ) 5 5

.

(11)

5. Wong : 10.6: We wish to calculate the production of 37 Ar via νe +37 Cl → e− +37 Ar. From the solar constant of 1350 W/m2 we have 1 135 J/s/m2 of neutrinos impacting the earth. With an average 0.9 neutrino energy of Eν = 1 MeV, this corresponds to a flux of φν = 8.4 × 1014 m−2 s−1 . This is further reduced by a factor of 100 due to the energy required to produce 37 Ar. The number of 37 Cl atoms in the tank is NCl = 2.42×1030 . The rate of production of 37 Ar is the product of the cross-section, the flux and NCl . When converted into events per day, I find a rate of 1.76 events/day. 6. Wong : 10.8: Neutron Stars (NS): Using the density of nuclear matter (NM) to be ρ = 0.17 N/fm3 and the mass of the NS to be MN S = 3 × 1030 Kg I find a radius of RN S = 13.68 km. If th core is all 56 Ni, then there are NN i = 3.2 × 1055 atoms, and if al the protons transmute into neutrons, this produces (Z = 28), 56 The integrated (in time) flux at the earth is RNν ∼ 9 × 10 neutrinos. 13 −2 dtΦν = 3.14 × 10 m , where we used the distance of d = 160000 light years to the earth. Assuming al the 3000 ton detector are free nucleons, we find a total number of nucleons to be Ntank = 1.83 × 1033 . Thus the number of events seen in this tank should be events = (3.14 × 1013 )(10−48 )(1.83 × 1033 ) = 0.057. The total energy emitted corresponds to about 1% of the rest mass of the sun. 7. YN Interactions: In order to determine, or constrain, the scattering parameters describing YN processes, we need to get the limited data from the Nijmegen web-page. One cannot use the phase-shifts that they give as they have already been processed once-must use the experimental measurements..including their uncertainties. Isolating the s-wave cross-section requires a measurement of the isotropy of the scattering, and we restrict ourselves to low-energies only. Ideally we would use on p < 70 MeV (in CoM frame) in order to be below the pion cut, where the effective range expansion is valid, but this is not available. So to solve this problem

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we do something that is not justified, and write the cross section as σ =

4π |k|2

+

( −1 a

+

1 r |k|2 2 0

+ 21 r1 |k|4 + ..)2

.

(12)

The cross section data is on units of millibarns, and is multiplied by 0.1 to convert to fm2 . The actual fitting is done with mathematica, using the Nonlinearregression fitting package. As we are dealing with fitting intervals and data with pretty large errors, I chose to Monte-Carlo the error analysis. I ran many fits to data where the momentum is uniformly sampled from each interval, and a Gaussian distribution is used for the cross section. The extracted values result from the mean and standard deviation of the fit values. I only did the Λ-proton scattering amplitude as I was pressed for time, but al can be done the same way. The data here is perhaps the best set. I could not extract the effective range with any meaningful errors, and so only fit the scattering length. I found aΛ−p to lie between 2.2 and 4.3 fm, using the data below p ∼ 320 MeV. In fitting, I actually fit a−1 as fitting a was not stable for obvious reason. I found a−1 = 67 ± 20 MeV. This makes clear the problem of determining the range for a as at 3σ the range is huge. Keep in mind that we are doing something illegal here by using the ER expansion at such large momentum, and so the results must be regarded as a guide at best.

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Figure 1: Fit to the Λ − p cross section. The three curves correspond to the mean and mean± 1-σ for the scattering length.

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