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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 eveneven 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 Qvalue, 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)_ (A2Z)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—thensome 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,N1)+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—isobaricanalogueto—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 singleparticle 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 twobody matrix elements for (J,T) = (0,1): (1d5/2, ld5/2; J = 0, T = 1V1d5/2,1d5,2; J = 0, T = 1)
(1d5/2,1d5/2; J = 0, T = 1V1d3/2,1d3/2; J = 0, T = 1)
(1d5/2,1d5/2; J = 0 , T =1V281/2,251/2; 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 =1V281/2,281/2; J = 0, T = (251/2,281/2; J = 0, T =1lV2Sl/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 singleparticle 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: ShellModel Assuming that 25Mg is a spherical nucleus described by the single—particle
shellmodel, 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 “updown” 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) + (AZ) (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]:=
Mass413 := {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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