- #1
JD_PM
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- TL;DR Summary
- Issue explaining why sometimes the table just gives one of the possible values for the total AM of an O/O nucleus.
What I know is the following:
The total angular momentum of the nucleus is just the total sum of the angular momentum of each nucleon.
If the nucleons are even the total angular momentum in the ground state will simply be ##0+##.
If the odd number of nucleons is close to one of the magic numbers then the shell model can explain what total angular momentum we'll get.
There are 3 cases to consider:
1) E/E nucleus (40-Ca for instance): Both proton and neutron numbers are even so the total angular momentum of the nucleus (I have read in other sources they call it Isospin) in the ground state will be ##0+##.
2) E/O nucleus (41-Ca for instance): The proton number is even. Thus the contribution by protons to the final angular momentum is none because ##j_p = 0##. What about neutrons? The unpaired neutron will be located in the 1f7/2 shell, then ##j_n = 7/2-##. Then the final angular momentum of 41-Ca in its GS will be ##I = 7/2-##.
Here comes my issue:
3) O/O nucleus (14-N for instance): here both proton and neutron numbers are odd so we expect that both contribute to the final angular momentum of 14-N.
Proton contribution: The unpaired proton will be located in the 1p1/2 shell, then ##j_p = 1/2-##.
Neutron contribution: The unpaired neutron will be located in the 1p1/2 shell, then ##j_n = 1/2-##.
Now it is about adding both AMs up to get ##I##:
$$I = 1+, 0+$$
But in the table just 1+ is given. Why?
The total angular momentum of the nucleus is just the total sum of the angular momentum of each nucleon.
If the nucleons are even the total angular momentum in the ground state will simply be ##0+##.
If the odd number of nucleons is close to one of the magic numbers then the shell model can explain what total angular momentum we'll get.
There are 3 cases to consider:
1) E/E nucleus (40-Ca for instance): Both proton and neutron numbers are even so the total angular momentum of the nucleus (I have read in other sources they call it Isospin) in the ground state will be ##0+##.
2) E/O nucleus (41-Ca for instance): The proton number is even. Thus the contribution by protons to the final angular momentum is none because ##j_p = 0##. What about neutrons? The unpaired neutron will be located in the 1f7/2 shell, then ##j_n = 7/2-##. Then the final angular momentum of 41-Ca in its GS will be ##I = 7/2-##.
Here comes my issue:
3) O/O nucleus (14-N for instance): here both proton and neutron numbers are odd so we expect that both contribute to the final angular momentum of 14-N.
Proton contribution: The unpaired proton will be located in the 1p1/2 shell, then ##j_p = 1/2-##.
Neutron contribution: The unpaired neutron will be located in the 1p1/2 shell, then ##j_n = 1/2-##.
Now it is about adding both AMs up to get ##I##:
$$I = 1+, 0+$$
But in the table just 1+ is given. Why?