Spin and parity for nuclear shell model

[dingo_d]

Homework Statement

I have an element: ${}^{207}Pb$, with Z=82, and A=125, and I need to find spin and parity of it.

Homework Equations

The angular momentum is determined by the angular momentum of the last nucleon that is odd, in this case it will be one unpaired neutron.

Now what confuses me is that the last neutron is in $3p_{1/2}$ shell, but when I look at the picture:

and start to fill the shells, how do I know that it must be $3p_{1/2}$ and not $2f_{5/2}$? Is it because of the energy difference? That is I take the one that is the closest to the filled shell?

That is basically what bothers me a bit. The spin and parity is easy once I know this. Since the protons are all paired I only need to look at this unpaired neutron, and the answer is $J^{\pi}=\frac{1}{2}^-$

EDIT: I totally misunderstood this magic stuff xD I need to fill each shell so that in the end I have all paired up, and if I have unpaired it needs to be in the next shell (if it is energy efficient ofc)! That's why I have it in $3p_{1/2}$!

:D

 

[mark726]

Dear poster,

Thank you for your question. You are correct in your understanding that the spin and parity of an element can be determined by looking at the last unpaired nucleon. In this case, since we have a neutron in the 3p_{1/2} shell, we can determine that the spin and parity of this element is J^{\pi}=\frac{1}{2}^-.

As for your confusion about choosing between the 3p_{1/2} and 2f_{5/2} shells, it is indeed based on energy difference. In general, the energy of a shell increases as the principal quantum number (n) increases, and within a given shell, the energy increases as the orbital angular momentum quantum number (l) increases. In this case, the 3p_{1/2} shell is closer to being filled than the 2f_{5/2} shell, so it is more energetically favorable for the unpaired neutron to be in the 3p_{1/2} shell.

I hope this clears up your confusion and helps you in your further studies of nuclear physics. Best of luck!