3 - PHYS 560: Solution 5 George Newman December 8, 2006...

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Unformatted text preview: PHYS 560: Solution 5 George Newman December 8, 2006 Abstract The solutions to the fifth assignment for Nuclear Physics, PHY560, Autumn 2006. 1 Question 1: Wong 4.11 Obtain the masses of members of the A = 135 isobar from a table of binding energies and plot the results as a function of Z. From the results deduce the value of the symmetry binding energy parameter 0:4 in the Weizacker mass formula.Carry out the same calcualtion for members of the A = 136 isobar and estimate the value of the pairing parameter 6 from the results. Solution 1. The included Mathematica code has been used to perform a fit to data obtained from the NIST website. The tables provided there were of masses, and binding energies were obtained from these - that’s the reverse of what the problem asks for, but mass data is more readily available. The fit returns the coefficients in Eb = 0.22 + bZ + c. When we write out the Weizacker formula in terms of Z and A, Z Z —1 A — ZZ 2 Eb(A, = 05114 *‘ 0(2A2/3 — 01% — 0%, and isolate the coefficients of Z and Z 2, we find that _ A(a+ b) “ 4(A—1) The values for both isobars agree at this level of accuracy. (14 = 19.9MeV. (2) 2. We can estimate the value of 6 from the residuals of the fit to the A = 136 isobar, since all of these nuclei are either even-even or odd- odd. Averaging these residuals for our data gave 6 m 1.9MeV. 2 Question 2: Wong 4.12 The Weizacker mass formula is useful to obtain global distribution of binding energies as a function of A, Z, and N. Use this formula to show that the Q-value, i.e., kinetic energy released, in fission is positive only for heavy nuclei. Solution In order for positive Q to be relaeased, we must have Eb(products) > Eb(parent). The Weizacker mass formula divides the binding energy into contribu- tions that behave differently as Z, N, and A vary. Below, I’ll refer to these components of Eb as the bulk (B), surface (S), coulomb (C), and symmetry (H), components. (There is a pairing component as well, but it will depend periodically on A, so we should ignore it in answering this question.) The binding energy for a species is, Eb(A,Z) = B—S—C—H (4) Z(Z—1)_ (A-2Z)2 A1/3 a A ’ respectively. We will consider only a 1 —> 2 fission process, since any m ore complicated process could be described as a progression of 1 -—> 2 events. The first two terms depend only on A. For A = A1 + A2, .43” + A3” > (A1 + A2)“. (5) Thus, we know (B —S)1+2 < (B—S)1+(B—S)2, and we always loose kinetic energy to surface tension when fission occurrs. If we want to get Q > 0 the other terms must compensate for this loss. Specifically, [CH2 — Cl — 02] + [H1+2 — 1‘11 — H2] > [51 + 5'2 — 51+2l (5) = 01A — (12/42/3 — 01 must be true for fission to be exothermic. Lets start by considering H. We can parametrize the proton and neutron content of the daughter species in terms of the parent by writing, Z1 = 2:2, Z2 = (1 — m)Z and N1 = yN, N2 = (1 — y)N. In the simple case where m = y we can plug these values into the equation for H in each species, and find that the symmetry energy is unchanged, no matter what the values of Z, N, and A are. Consider next the case where one of the daughter species is perfectly symmetric. In that case, 1 1 [H1+2 — H1 — H2] = 014(N *‘ Z)2 — . We can imagine going farther than this as well. If the parent has an excess of neutrons, consider putting all that difference and—then-some into one of 3 the daughters, thus causing the other daughter to have a proton excess. This would give a still more negative symmetry energy to the products than in the case with one symmetric daughter. No matter what we do, we can at best leave the symmetry energy unchanged. When a: 96 y, in addition to surface tension, we would also have spent kinetic energy on increasing asymmetry. However, we can at least tell that this possible loss decreases with A, (unlike the loss to surface tension.) Our hopes now rest on sufficient coulomb energy being released to pay the bill for surface tension and possible symmetry losses, while still leaving some positive Q. In the valley of stability, we can expect that for A > 1 we have, Z or A” with m < 1. This means that, C' 0: a3A2”_%. (8) and correspondingly, __l_ $-l [C1+2 — CI — 02] cc 013K241 + A2)2”“i — A?“ 3 + A: 3]. (9) So long as a: > 1/2 (which is true in the valley of stabillity) this will grow faster with A than either of the negative Q contributions (surface tension being most important), and we can find some minimum A for the parent such that fission will always be exothermic. Below this value, (which turns out to be about A = 56) the other terms will win out, and it will be endothermic. 3 Question 3: Wong 4.13 Use the Kelson—Garvey mass relation to find the binding energy of 120 from the values of nearby nuclei. Estimate the uncertainties in the value deduced, Using the value of the binding energy obtained, and a minimum amount of additional experimental data, find the locations of isobaric analogue states of 120 and 12N relative to the ground state in 120. (Note to the reader: I reworded the last sentence since I could not be certain of the meaning of the question in Wong. The question written above is the one I answer below — though I can’t be certain it is the one intended by Wong.) Solution l. The Kelson—Garvey mass formula may be expressed as, 0 = Eb(Z+1,N-1)+Eb(Z—1,N)+Eb(Z,N+1) —Eb(Z,N—1)—Eb(Z+l,N)—Eb(Z—1,N+1). We need data on five neighboring nuclei to make use of this. (There is a simpler version which completely ignores all two body interactions, and only requires 3 inputs. It has a much higher uncertainty.) We’ll use, Eb(l20)+Eb(11C)+Eb(13N)— Eb(uN) — Eb(130) — Eb(120) = 0. (10) (There is no data for any isotope of oxygen lighter than 120, so we must use the formula with (N — 1) = 4.) Grabbing some numbers off the NIST website gives us, 2Eb(120) = 54.4313MeV. (Data from the same source have, Eb(120) = 54.4593MeV.) This is off for our isotope by about 0.03MeV. In general, one can input six values into the equation and check the magnitude of the deviation from zero, then take the rms of this deviation for many sets of input. This amounts to treating the difference from zero as a random variable - whose standard deviation we will take to be the expected error in the Kelson—Garvey formula. Doing this with measured masses for many species has shown the error to be about 0.1MeV on the average (Wong, p.143). 2. An isobaric analogue state is one related to the original through the action of an isospin raising or lowering operator. That is, nothing has been changed except the third component of isospin — transforming some number of neutrons to protons, or vice versa. It will thus have the same angular momentum (component by component) and total isospin as the original state, but a different nucleon content. I think this question is very poorly worded, but I believe what is being asked for is the excitation energy of the “12O—isobaric-analogue-to—12C” relative to the ground state of 120 (and the same for 12N). There are three contributions to be considered here. One is the bind- ing energy, which may be looked up, or obtained through the Kelson— Garvey formula. Another, is the mass difference from transmuting be- tween proton and neutron. (The binding energy of 12O is determined 5 w.r. to the mass of 8 free protons and 4 free neutrons, while 12C is compared to 6 of each.) The last is the coulomb energy change due to changing the electric charge in the nucleus (which must still be the same volume, since the 120 configuration space wave functions aren’t changed in building the isobaric analogues.) The excitation energy will thus be E(l20gs) '_ E(I2OIAS) = E02015)-{E(”CJ+[EC(8,12)— Ec(6112)l+ 2(mp — ms}. The coulomb energy difference is found by means of a uniform sphere approximation, with radius (1.2fm)A1/3: [13,,(2 = s, A = 12) — Ec(Z = 6, A =12)] = 8.177MeV. (11) Plugging in this, and the nucleon masses gives, E(120g,)— ECZOIAS) = —29.08MeV. * (12) Why I chose to calculate the negative of the excitation energy is a mystery that will no doubt follow me to the grave. 4 Question 4: Wong 7.3 The energies of 17O with respect to the 160 core are —4.15MeV for the g+ state, —3.28MeV for the g state, and +0.93MeV for the %+ state. Assuming these values are the single—particle energies of the ds—orbits, use an indepen— dant particle model to find the relative energies of the lowest, y, %+, and %+ states in 390a with respect to the 40Ca core. Solution The three states of 17O we have information on all involve adding a single neutron to the one of the three distinct N = 2 orbitals. To make 400a we would have to fill all three of these levels with two neutrons and two protons each. 390a would be the same, but with one neutron removed from the state with the given angular momentum — the angular momentum of that “hole” would be the spin of the nucleus in that state. 6 In an independant particle model, we would ignore any potential interac- tion between the nucleons in 390a that aren’t in 17O and each of these states would lose a neutron single particle energy with respect to 4E’Ca. €1d5/2 = 6231/2 = and Eldg/g == for 3903.. 5 Question 5: Wong 7.4 The ds—shell single-particle energies with respect to 160 core are 61d5/2 = —4.15MeV, 6251/2 = —3.28MeV, and Ema/2 = +0.93MeV. A particular ef- fective interaction has the following set of two-body matrix elements for (J,T) = (0,1): (1d5/2, ld5/2; J = 0, T = 1|V|1d5/2,1d5,2; J = 0, T = 1) (1d5/2,1d5/2; J = 0, T = 1|V|1d3/2,1d3/2; J = 0, T = 1) (1d5/2,1d5/2; J = 0 , T =1|V|281/2,251/2; J = 0, T = 1) (1d3/2, 1d3/2; J = 0, T = llVIld3/2, Ida/2; J = 0, T = 1) (Ida/2, 1d3/2; J = 0, T =1|V|281/2,281/2; J = 0, T = (251/2,281/2; J = 0, T =1lV|2Sl/2,231/2; J = 0, T = 1) ll —2.0094MeV —3.8935MeV —- 1.3225MeV —O.8119MeV -—0.8385MeV —2.3068MeV 1. Calculate the ground state binding energy of 18O with respect to 16O and compare to the result obtained from a table of binding energies. What are the excitation energies of the two other 0+ states in this space? 2. Obtain the ground state wave function of 180 and use it to calculate the relative probability of finding a neutron in the 1d5/2, 231/2, and 1d3/2 single—particle states in 18O. The results are essentially the spec— troscopic factors for one neutron pickup reactions. Solution 1. There are two neutrons paired in some linear combination of the three available single-particle states, in addition to the core: llSO) = l160)(al1d5/2,1d5/2) +fil2d1/2, 2sup) + ’Yllds/za 1d3/2)) (13) 6 The Hamiltonian (assuming 0"” states only) is a three by three matrix, —4.15 0 0 —2.0094 —1.3225 —3.8935 'H = 2 0 —3.28 O + —1.3225 —0.8119 —0.8385 . 0 0 —0.93 —3.8935 —0.8385 —2.3068 (14) the ground state eiganvalues and eigenvectors (found in Mathematica) are, — a 0.866 139, = —12.20MeV , 119, = fi = 0.414 7 0.281 The other eigenvalues give the excited state energies: E1 = —8.20MeV for the first excited state, and 132 = 2.27MeV for the second. (We don’t need the eigenvectors here, but we will need the ground state for the second part.) The ground state is at ~12.20MeV relative to 16O, and the excitation energies are 4.01MeV and 14.47MeV relative to the ground state. . The ground state wave function is, £180)” = I16O)[(0.866)|1d5/2,1d5/2)+(0.414)|2d1/2,2s1/2)+(0.281)|1d3/2,Ida/2)]. (15) The probability for finding a netron in the 1d5/2, (231/2), [1d5/2] state is 2 x (0.866)2 = 1.5, (2 x (0.414)2 = 0.34), [2 x (0.281)2 = 0.16]. (I took this last part to mean “If I look in each state, what is the expected number of neutrons I will find there?” Hence, the “2x”. Some took it to mean “Given that a single neutron is in one of these states, what is its likelyhood of being in each?” Either is acceptable, I think.) Question 6: Wong 7.5 If the wave function of the lowest 1"' state in 18O is |J" = 1+, T =1) =l1d3/22s1/2; J” = 1+, T = 1) (16) find the magnetic dipole moment of this state. 8 Solution All the nucleons in this isotope are paired, except for two neutrons which the wave function tells us are in the Ida/2 and 251/2 states. Hence, the magnetic moment will be _ 11 all §§ 3% In general, 1 .. 3 .. _ . . _ (JJlS I33) — 2j+1lJ(J+1l+5(5+1) L(L+1)]- (18) For the d we have, 3' = 3/2, L = 2, s = 1/2, and for the s we have, 3' = 1/2, L = 0, s = 1/2. This leads to M = (2/5)I$n = —0.765/J.N. 7 Question 2: Shell-Model Assuming that 25Mg is a spherical nucleus described by the single—particle shell-model, what is its spin and parity? Does this agree with data? Solution We have 12 protons and 13 neutrons to accomodate. The Is and 1p states will be filled for both, accounting for 8 of each. The next lowest state is the Ids/2 state, which can hold up to 6 of each particle. This is the highest energy state that should see any occupation in 25Mg. The four protons in these states can pair off in “up-down” duos, so they do. Four of the neutrons do the same, leaving us with a spin zero contribution, plus one neutron in an aligned L = 2, .S' = 1/2 state. Orbital angular momentum L = 2 states have even parity, so the state is J7r = 3+. The data graciously agree with this prediction. This does not agree with the “spherical nucleus” assumption, however. For that, we would need L = 0 for the 13th neutron, and we would expect 1r_l+ J _2. soEKKL5_auainb 111(41):: £[z_, L] := 2 (938.272) + (A-Z) (939.566) mm}.- b[u_, z_, A_] := Hz, A] —u (931.494) Inl43]:= 14333135 := {134.92517, 134.91645, 134.910050, 134.907207, 134.905972, 134.905683, 134.906971, 134.909146, 134.91314, 134.91824, 134.92462, 134.93235); MasslBG := {135.93066 , 135.92010 , 135.921.4660, 135.907220, 135.907306 , 135.904570 , 135.907650 , 135. 907140 , 135.912650 , 135. 915020 , 135. 92345 , 135.92830 }; In[121]:= M135 := Tab1e[{i+50, Mass135[[1]]}, {1, 12)],- 14136:: Table [{i + 50, Massl36 [ [i]]}, {1, 12)]; ListPlot [14135, AxesLabel a {2, man}, PlotLabel -v 135 Isobar] ListPlot [11136, AxesLabel a {2, amu}, PlotLabel -> 136 Isobar] amu 135Isobar 134 . 93 134.925 . 134. 92 134.915 Out11231= - Graphics - amu l36Isobar 135293 135. 925 135 . 92 135.915 . . 135.905 . Out [124]= - Graphics - Data135 :=Tab1e[{i+50, b[Massl35[[i]], use, 135]}, {1, 1, 11, 2}]; Data136:=‘1'ab1e[(i+50, b[Ma55136[[i]], 1+5o, 136]}, {1, 2, 12, 2}]; Unpair[x_] =Fit[Dat3135, {1, 3:, x2}, x] Paired [x__] =E‘it[Dat3136, {1, x, x2}, x] Out{601= -1085.39 + 79.6403 x — 0.723521 x2 Out[61}= —1098.95+ 79.79 x— 0.71908 x2 501560_5_au06.nb 2 _—___—__—_—__*—___— Plot [{Unpair [y] , Paired[y] }, (y, 51, 62}, PlotStyle -» {RGBColor [1, O, 0] , RGBColor [0, 0, 1]}, AxesLabel —> (z, Mev}] MeV “www (13'3") mm? (130 Out[92}= - Graphics - 1 6 = —Sum[ 12 Abs[b[Mass136[[i]], 1+5o, 136] - (4093.95 +79.79(1+50)-o.72(1+50)2)], {1,1, 12]] Outf2181= l . 97725 In{125}:= 135 [79.64 —O.72] 4 134 0ut(125]= 19. 8772 In[126}:= 136 [79.79 - 0.72 J 4 135 Out{126}= 19.9139 In[132]:= Mass-413 := {11.01143, 11.02680, 12, 13.00574, 13.02481} In[l33}:= MassAlB[[2]] Out{133]= 11.0268 In[139]:= b413:={b[Ma33413[[1]], 6, 11], b[Mass413[[2]], '7, 11], b[Mass413[[3]], 6, 12], b[Mass413 [[4]], 7, 13], b[Mass413[[5]], 8, 13]} In[140]:= b413 Out[140}= (70.381, 54.77, 89.1, 90.5312, 71.4736) .- (O / d, / J / s01560’_5_au06.nb In[144}:= -b413[[1]]+b413[[2]]+b413[[3]]—b413[['4]]+b413[[5]] Out(l44}= 54.4313 In[143]:= b[12.03441, 8, 12] Out{1431= 54.4593 In[145]:= 0ut[144] -0ut[143] Outfl451= —0.0279448 ...
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3 - PHYS 560: Solution 5 George Newman December 8, 2006...

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