Proton and neutron configurations (shell model)

[SUDOnym]

Question is:

Determine the neutron and proton configurations of the first and second excited states of:

[tex]_{20}^{41}Ca[/tex]

which have spin-parities of [tex]3/2^{-}[/tex] and [tex]3/2^{+}[/tex] respectively.My attempt at solution:

(I haven't encountered a problem quite of this type before so am a little unsure as to how to go about it...)

But my reasoning was that the proton states would stay the same as if in ground state as there are no unpaired protons, ie.:

protons:
[tex](1s_{\frac{1}{2}})^{2}(1p_{\frac{3}{2}})^{4}(1p_{\frac{1}{2}})^{2}(1d_{\frac{5}{2}})^{6}(2s_{\frac{1}{2}})^{2}(1d_{\frac{3}{2}})^{4}[/tex]

so the difference would lie with the neutrons, first off to say neutron ground state config is:

[tex](1s_{\frac{1}{2}})^{2}(1p_{\frac{3}{2}})^{4}(1p_{\frac{1}{2}})^{2}(1d_{\frac{5}{2}})^{6}(2s_{\frac{1}{2}})^{2}(1d_{\frac{3}{2}})^{4}(1f_{\frac{7}{2}})^{1}[/tex]

and since the first excited state is [tex]3/2^{-}[/tex] I made a guess this means that is only the outer most neutron that jumps up to the nearest level with this spin-parity:

implies first excited state neutron configuration is:

[tex](1s_{\frac{1}{2}})^{2}(1p_{\frac{3}{2}})^{4}(1p_{\frac{1}{2}})^{2}(1d_{\frac{5}{2}})^{6}(2s_{\frac{1}{2}})^{2}(1d_{\frac{3}{2}})^{4}(2p_{\frac{3}{2}})^{1}[/tex]

and the second excited state configuration for neutrons with spin parity given by:
[tex]3/2^{+}[/tex]
is

[tex](1s_{\frac{1}{2}})^{2}(1p_{\frac{3}{2}})^{4}(1p_{\frac{1}{2}})^{2}(1d_{\frac{5}{2}})^{6}(2s_{\frac{1}{2}})^{2}(1d_{\frac{3}{2}})^{4}(2d_{\frac{3}{2}})^{1}[/tex]

So is this the correct method/reasoning I used? if not what is the correct approach to take?

 

[jbsteele]

Yes, this is the correct method and reasoning you used. The way to approach this type of problem is to look at the ground state configuration and determine which particle in the ground state would have to be excited to a higher energy level with the given spin-parity in order to reach the first and second excited states. In this case, the outermost neutron was needed to be excited to the next higher energy level with the given spin-parity in order to reach the first and second excited states.