Total Angular Momentum of an odd-parity shell-model state

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The discussion focuses on determining the total angular momentum (J) and orbital angular momentum (L) for odd-parity shell-model states with a maximum of 4 and 6 nucleons. For 4 nucleons, L is found to be 1, resulting in J = 3/2, while for 6 nucleons, L remains 1, also yielding J = 3/2. The parity of the state is odd, indicating that L must be an odd number. The example of titanium-47 illustrates how unpaired nucleons contribute to the total angular momentum and parity. Additionally, odd-odd nuclei are discussed, where the total spin is derived from the vector sum of the angular momenta of unpaired protons and neutrons.
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Homework Statement


A certain odd-parity shell-model state can hold up to a maximum of 4 nucleons. What are its values of J and L? What about an odd-parity shell-model state with a maximum of 6 nucleons?

Homework Equations


Parity = (-1)L
J = L+S
Total angular momentum, J, is equal to orbital angular momentum, L, plus spin, S.
Nucleons are both fermions and have spin = 1/2
Therefore, J = L + 1/2

The Attempt at a Solution


Parity is odd ∴ L has to be an odd number (Parity = (-1)L)
maximum of 4 nucleons: 1s1/2 shell is full for either protons or neutrons. 1p3/2 shell is not full.

∴ J= 3/2

3n + 1p OR 1n + 3p

J = L ± 1/2 = 1 + 1/2 → J = 3/2, L=1

---------------------------------------------------------------

maximum of 6 nucleons
Parity is odd ∴ L has to be an odd number (Parity = (-1)L)
1s1/2+1p3/2 shells have maximum of 6 nucleons. ∴ L = 1
With 6 nucleons: 1s1/2 shell is full for both protons and neutrons. 1p3/2 shell is not full.

L=1
J = L ± 1/2 = 1 + 1/2 → J = 3/2, L=1
 
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Odd A nuclei have one unpaired nucleon.

The spin of the nucleus is equal to the j value of that unpaired nucleon and the parity is (−1)l , where l is the orbital angular momentum of the unpaired nucleon.

For example, 47 Ti22 (titanium) has an even number of protons and 25 neutrons. 20 of the neutrons fill the shells up to magic number 20 and there are 5 in the 1f 7/ 2 state (l = 3, j = 7/ 2 ) Four of these form pairs and the remaining one leads to a nuclear spin of 7 /2 and parity (−1)3 = −1. •

Odd-odd nuclei. In this case there is an unpaired proton whose total angular momentum is j1 and an unpaired neutron whose total angular momentum is j2.

The total spin of the nucleus is the (vector) sum of these angular momenta and can take values between |j1 − j2| and |j1 + j2| (in unit steps).

The parity is given by (−1)(l1+l2) , where l1 and l2 are the orbital angular momenta of the unpaired proton and neutron respectively.

For example, 6 Li 3(lithium) has 3 neutrons and 3 protons. The first two of each fill the 1s level and the third is in the 1p 3 /2 level.

The orbital angular momentum of each is l = 1 so the parity is (−1)x(-1) = +1 (even), but the spin can be anywhere between 0 and 3.

see ref.http://www.personal.soton.ac.uk/ab1u06/teaching/phys3002/course/05_shell.pdf
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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