# Single particle shell model for nucleus

1. Apr 1, 2014

### BOYLANATOR

1. The problem statement, all variables and given/known data

Predict the ground state spin and parity for $^{7}_{3}Li$, $^{15}_{7}N$. $^{31}_{15}P$.

2. Relevant equations

Fill up shells for neutrons and protons separately.
Shells in increasing energy are: $1S_{1/2}$, $1P_{3/2}$, $1P_{1/2}$, $1S_{1/2}$, $1D_{5/2}$, $1D_{3/2}$
and have occupancies: 2, 4, 2, 6, 4, 8, 6

A missing proton is a hole with parity -1.
An unpaired neutron has parity +1.

3. The attempt at a solution

$^{7}_{3}Li$: We have odd protons so only interested in this. $1S_{1/2}$ fills up. Then only one proton out of four sits in $1P_{3/2}$. It has j = 3/2 and parity -1.

$^{15}_{7}N$: We are interested in the 7 protons. We will be left with one hole in $1P_{1/2}$ so it has j = 1/2 and parity -1.

$^{31}_{15}P$: We are interested in the 15 protons. We fill up 2, 4, 2, 6 and then are left with 1 out of 4 in $1D_{3/2}$. This unpaired proton should have spin 3/2 and parity -1 if I use my same method.
However the answer given is that the unpaired proton sits in the $2S_{1/2}$ so has spin
1/2 and parity +1. How am I to know that jumping up to 2S is more favourable than carrying on through the 1X's? Why is the parity now +1. Does it flip when you jump from 1X to 2X?

2. Apr 3, 2014

### BvU

Can you check your list ? there are 6 shells with $1S_{1/2}$ appearing twice (I take it the second should be $2S_{1/2}$ ?) and there are 7 occupancies, with unlikely values. I would expect something like 2,4,2,2,6,4 but that still brings you to $1D_{5/2}$ with P, so it looks as if the $2S_{1/2}$ sits between $1D_{5/2}$ and $1D_{3/2}$ , which isn't impossible. See http://jol.liljenzin.se/KAPITEL/CH11NY3.PDF 310-313

3. Apr 3, 2014

### BOYLANATOR

Sorry for sloppy post, and also you are right about the ordering. Let me write it out correctly.

4. Apr 3, 2014

### BOYLANATOR

$^{31}_{15}P$: We are interested in the 15 protons. We fill up 2, 4, 2, 6 and then are left with 1 out of 2 in $2S_{1/2}$. The hole should have spin 1/2 as given in solution.

Regarding the parity, does it flip when we jump up an energy level. (Going from $1S_{1/2}$ to $2S_{1/2}$ is jumping up an energy level right? Increasing the quantum number n.)

5. Apr 4, 2014

### BvU

I don't really know much about parity and I don't understand the "a missing proton is a hole with parity -1".

As far as I know parity depends on angular momentum L as (-1)L and the primary quantum number n doesn't appear. S states are L=0 states, so definitely even parity: there is no $\theta,\, \phi$ dependence.

6. Apr 4, 2014

### BOYLANATOR

I found this on wikipedia http://en.wikipedia.org/wiki/Nuclear_shell_model#Other_properties_of_nuclei.

"All protons in the same level (n) have the same parity (either +1 or −1), and since the parity of a pair of particles is the product of their parities, an even number of protons from the same level (n) will have +1 parity"

Although it doesn't state it explicitly this suggests that the parity does switch as you move up n.

7. Apr 5, 2014

### BvU

I am not buying it. The whole S, P, D thing comes from splitting the r dependency in the wavefunction from the $\theta, \phi$ dependence. n has to do with the r ($|\vec r|$) dependence -- even parity, period. Proton itself has intrinsic party +1 (definition). Odd parity can come from l and m and it turns out p = (-1)l.

Wikipedia isn't always right. I'm not always right either, so if someone could set this straight, I'd be grateful.

8. Apr 5, 2014

### BOYLANATOR

Yes, the missing proton is a hole with parity -1 is the problem statement here I think. I read this out of context and it doesn't help.

I've learned one should just think in terms of unpaired protons in terms of holes. So the parity is simply +1 as it is in an l = 0 shell and the unpaired proton has +1 parity.

I get it now. Thanks!