- #1

Warr

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In my textbook, one of the questions states:

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The ground state of the radioisotope 17-F-9 has spin-parity j_P = (5/2)+ and the first excited state has j_P=(1/2)-. Suggest two possible configurations for the latter state.

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Here is the answer in the back:

The configuration of the ground state is:

protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2(1d_\frac{5}{2})[/tex]

neutrons:[tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2[/tex]

To get j_P= (1/2)-, one could promote a p_1/2 proton to the d_5/2 shell giving

protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^{-1}(1d_\frac{5}{2})^2[/tex]

Then by the pairing hypothesis, the two d_5/2 protons could give j_P = 0+ so that the total spin-parity would be determined by the unpaired p_1/2 neutron (j_P=(1/2)-).

Alternatively, one of the p_3/2 protons could be promoted to the d_5/2 shell, giving

protons: protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^{-1}(1p_\frac{1}{2})^2(1d_\frac{5}{2})^2[/tex]

and the two d_5/2 protons could combine to give j_P = 2+, so that when this combines with the single unpaired j_P = 3/2- proton, the overall spin is j_P = 1/2-

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So here are two things I am confused about:

Firstly, how can the two d_5/2 protons combine to have j_P = 0+ in the first case and j_P = 2+ in the second case?

Secondly, how is it that in the second case, the spin-parity ends up being j_P = 1/2-. Is it that the parities of the two are multiplied (ie the parity of the two d_5/2 protons is 1+ and the parity of the unpaired p_3/2 proton is 1-, giving an overall parity of 1-, and then the spin is 2 - 3/2 = 1/2? I don't really get how that works).

If I can understand this I may be able to even get started on the homework.