Physics 249 Homework 9
Due Nov 30th
1) Considering the model where the deuteron consists of two protons and an electron.
Compute the spin and magnetic moment of the ground state. Compare the values with
those given in the table in chapter 11.
In this model the deuteron is composed of two identical spin ½ protons and an electron.
Since the protons are identical they must have spins that are anti-aligned in the group
state. Therefore the proton-proton spin is s=1/2-1/2=0. The total spin is
s=0+1/2 = ½ℏ.
The proton magnetic momentum would also be anti-aligned and the magnetic momentum
would be due to the electron with a magnitude of 1 Bohr magniton, 𝜇𝐵 .
The correct model of the deuteron consists of one neutron and one proton. Since they are
not identical the you can add the spins as s = ½ + ½ or s = ½ - ½. The second is true
since it allows the magnetic momentum to be anti-aligned. Then the magnetic
momentum is approximately 2.79𝜇𝑁 − 1.91𝜇𝑁 ≅ 0.86𝜇𝑁
2) Compute the total binding energy per nucleon of 3He, 4He, 236U. Express your answer
in MeV. Masses are given in the appendix.
Using
𝐵𝑛𝑢𝑐𝑙𝑒𝑎𝑟 = 𝑍𝑀𝐻 𝑐 2 + 𝑁𝑚𝑛 𝑐 2 − 𝑀𝐴 𝑐 2
𝐵𝑛𝑢𝑐𝑙𝑒𝑎𝑟,3𝐻𝑒 = 2 ∗ 1.007825u + 1 ∗ 1.008665u − 3.016029u
931.5𝑀𝑒𝑉
= 8.286x10−3 𝑢 ∗ (
) = 7.718𝑀𝑒𝑣, 2.573𝑀𝑒𝑉
𝑢
𝐵𝑛𝑢𝑐𝑙𝑒𝑎𝑟,4𝐻𝑒 = 2 ∗ 1.007825u + 2 ∗ 1.008665u − 4.002602u
931.5𝑀𝑒𝑉
= 3.0378x10−2 𝑢 ∗ (
) = 28.30𝑀𝑒𝑣, 7.074𝑀𝑒𝑉
𝑢
𝐵𝑛𝑢𝑐𝑙𝑒𝑎𝑟,236𝑈 = 92 ∗ 1.007825u + 144 ∗ 1.008665u − 236.046559u
931.5𝑀𝑒𝑉
= 1.921101x100 𝑢 ∗ (
) = 1790𝑀𝑒𝑣, 7.583𝑀𝑒𝑉
𝑢
3) Find the energy necessary to remove a neutron from 4He and to add a neutron to 235U.
Express your answer in MeV.
3.016029u + 1.008665u − 4.002602𝑢 = 0.022093 ∗ (
931.5𝑀𝑒𝑉
) u = 20.58Mev
𝑢
931.5𝑀𝑒𝑉
235.043924u + 1.008665u − 236.045562u = 0.007023 ∗ (
)u
𝑢
= 6.542Mev
4) Use the shell model to analyze the following nuclei. The number of nucleons is given.
29Si, 33S, 38Cl, 40Ar, 53Cr
a) List the outermost shell type and number of nucleons in the outermost shell for protons
and neutrons.
b) What are the orbital, total angular momentum, and nuclear magnetic moment for the
first two nuclei.
c) Which nuclei are unstable and why?
d) From the information above what are the first two cases where a neutron orbital is
filled before the remaining positions in proton orbital due to proton repulsion.
29 Si A = 29, Z = 14, N = 15
Z, 6/6 1d5/2
N 1/2 2s1/2
Both proton and neutrons have a full subshells up to 1d5/2. These shells will have total
angular momentum zero.
The total angular momentum is set by the unfilled shells. l=0, s=1/2 j=1/2
Also the nuclear magnetic moment of the filled shells will be zero. Therefore, the
nuclear magnetic moment will come completely from the one neutron in the unfilled
shell.
-1.91𝜇𝑁
33 S A = 33, Z = 16, A = 17
Z, 2/2 2s1/2
N, 1/4 1d3/2
l = 2, s=1/2, j=3/2,
** Again the nuclear magnetic moment is set by the final nucleon. In this case j=3/2
considering axis where the total angular momentum (and it’s contribution to the magnetic
moment) is maximum, mj=3/2, then ml=2 and ms=-1/2 or ml=1 and ms=1/2. We have to
added the contributions from the intrinsic nuclear magnetic moment which is oriented
relative to the spin and the orbital angular momentum component. No points were
deducted for this problems
38 Cl A= 38, Z = 17, A = 21
Z, 1/4 1d3/2
N, 1/8 1f7/2
A neutron has filled the 1f7/2 orbital before the lower energy proton 1d3/2 orbit was
completed. However, this atom is unstable because the neutron orbit has been filled first,
so this is not the first stable reversal of normal ordering.
40 Ar, Z = 18, N = 22
Z, 2/4 1d3/2
N, 2/8 1f7/2
Two neutrons have filled the 1f7/2 orbital before the lower energy proton 1d3/2 orbit was
completed. This is the first stable reversal of the normal ordering due to proton
repulsion.
53 Cr A= 53, Z = 24, A = 29
Z, 4/8 1f7/2
N, 1/4 2p3/2
A neutron has filled the 2p3/2 orbital before the lower energy proton 1f7/2 orbit was
completed.
5) If a nuclei has a lifetime of 23.5min (239U). What is the probability that it has not
decayed after 1 year? Note, please evaluate the exponential and express the answer as a
final numerical value. (hint: use the definition of the natural logarithm and logarithm)
a) 239U lifetime 23.5min, 4.4711x10-5 years
** Note the above number is actually the half-life. Full credit was given for either
treating this as the lifetime or the half-life. Below we treat it as the lifetime as the
problem was written. Treating it as the half-file would add an extra factor on ln(2).
Prob for a 239U to exist after 1year
P = e-1.0/4.4711x10-5
ln(P) = -2.2365x104
log(P)ln(10) = -2.2365x104
log(P) = -2.2365x104/ln(10)
then using the definition of log if x=10y then y=log(x)
P = 10^(-9712.996)= 10^(-9712)* 10^(-0.996)=1.01x10^(-9713)!