Nuclear shell model of double magic nucleus 132Sn

[Marioweee]

Homework Statement

See below

Relevant Equations

See below

The independent particle energies for protons and neutrons around the
exotic doubly magic core 132Sn are shown in the figure below, where
π refers to protons and ν to neutrons.
Using the nuclear shell model and using this figure as a guide, answer
to the following questions:
a)Estimate Jπ for the lowest energy states of 132Sb. From graph determine the ground state energy of 132Sb with respect to
to 132Sn.
b)In 132Sb the analogous isobaric state (IAS) $|Sb_{IAS}>$ of the ground state of 132Sn $|Sn_{g.s}>$ is given by the decay $|Sb_{IAS}>=T^{+}|Sn_{g.s}>$
where the operator $T^{+}=\sum_{i}t^{+}_{i}$ and $t_{i}$ exchange a neutron for a proton. Write the quantum numbers of spin, parity, isospin and third component of isospin for the IAS in 132Sb and try to estimate its energy with respect to the ground state of 132Sn using some macroscopic model.

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a) 132Sb has p=51 and n=81. According to the order of filling the layers, all of them will be complete until $\pi g_{9/2}$ and $\nu h_{11/2}$. We have an unpaired neutron and proton in $\pi g_{7/2}$ and $\nu d_{3/2}$. It is this unpaired proton and neutron that will give us the possible spins in the ground state of 132Sb. The possible states are: $$J^{\pi}=(2)^+, (3)^+, (4)^+, (5)^+$$. Would this reasoning be correct?
As for the energy of the ground state of 132Sb from the graph, the truth is that I do not know how it would be done.

b)I am also a bit lost on this question. All I know is that the third isospin component of 132Sb is $T_z=\dfrac{Z-N}{2}=-15$ and therefore the isospin has to be between $15<= T <= A/2=66$.
Any suggestions on how to continue? The truth is that I do not understand very well what is being asked of me either, since I am not very familiar with the concept of analogous isobaric state and I do not find much information on the matter either.

 

[member38644]

a) Your reasoning for the possible spins in the ground state of 132Sb is correct. The unpaired proton and neutron in the πg7/2 and νd3/2 shells will give rise to Jπ=(2)+, (3)+, (4)+, and (5)+ states. To estimate the ground state energy of 132Sb with respect to 132Sn, we can use the fact that the ground state energy of 132Sn is at 0 MeV on the graph. Looking at the graph, we can see that the energy for the πg7/2 proton and the νd3/2 neutron is at approximately 2.5 MeV. Therefore, we can estimate the ground state energy of 132Sb to be around 2.5 MeV higher than that of 132Sn.

b) The analogous isobaric state (IAS) in 132Sb is formed by exchanging a neutron for a proton in the ground state of 132Sn. This means that the IAS will have the same spin and parity as the ground state of 132Sn, but with a different isospin. The quantum numbers for the IAS in 132Sb will be Jπ=(0)+, T=15, and Tz=-15. To estimate its energy with respect to the ground state of 132Sn, we can use a macroscopic model such as the liquid drop model. This model predicts that the IAS will have a higher binding energy than the ground state of 132Sn, as the exchange of a neutron for a proton will result in a more stable nucleus. However, the exact energy cannot be determined without more information about the specific model being used.