# Parity of $$^{9}_{5}B^{5}$$

1. Aug 19, 2010

### tirrel

Hy... I'm reviewing some nuclear physics but don't remember certain topics.

This exercise for example:

Knowing that $$^{9}_{4}Be^{5}$$ and $$^{9}_{5}B^{4}$$ have both as quantum numbers $$\frac{3}{2}^{-}$$ and their properties are determined by the odd nucleon, show that the ground state of $$^{9}_{5}B^{5}$$ can be $$\frac{3}^{+}$$.

My thinkings, which are correct?:

First of all we see that $$^{9}_{5}B^{5}$$ has as outermost nucleons the odd nucleon of $$^{9}_{4}Be^{5}$$ and the odd nucleon of $$^{9}_{5}B^{4}$$. So $$I=I_1+I_2$$, and thanks to the rules of q.m. $$I \in \{0,1,2,3\}$$. 3 is present and we are happy.
So for both $$^{9}_{4}Be^{5}$$ and $$^{9}_{5}B^{4}$$ we have for the odd nucleon $$I=\frac{3}{2}$$. but thanks to the rules of angular momentum I=L+S and so since $$I=\frac{1}{2}$$, $$I \in \{l+\frac{1}{2},l-\frac{1}{2}\}$$ and this forces $$l \in \{1,2\}$$ , Now the parity is (-1)^l, so we are forced to have l=1 for both particles. I don't know is this is useful but I think is correct.
Now maybe we can say that the state in $$^{9}_{5}B^{5}$$ will be a superposition of vectors of the type $$v \otimes w$$, where the vector $$v$$ refers to one nucleon and the vector $$w$$ to the other. The parity on each member acts on each subspace: $$\Pi v \otimes w= \Pi_1 v \otimes \Pi w_2=(-1)^2v \otimes w$$. and so the parity is positive. This reasoning seem to tell that to find the parity I simply have to multplicate the parities of the individual nucleons. is it right? sounds strange...

Last edited: Aug 19, 2010
2. Aug 19, 2010

### tirrel

ehm... the firs messag behaves strange and I'm not able to modify it... can you delete and leave only the second?:

Hy... I'm reviewing some nuclear physics but don't remember certain topics.

This exercise for example:

Knowing that $$^{9}_{4}Be^{5}$$ and $$^{9}_{5}B^{4}$$ have both as quantum numbers $$\frac{3}{2}^{-}$$ and their properties are determined by the odd nucleon, show that the ground state of $$^{9}_{5}B^{5}$$ can be $$3^{+}$$.

My thinkings, which are correct?:

First of all we see that $$^{9}_{5}B^{5}$$ has as outermost nucleons the odd nucleon of $$^{9}_{4}Be^{5}$$ and the odd nucleon of $$^{9}_{5}B^{4}$$. So $$I=I_1+I_2$$, and thanks to the rules of q.m. $$I \in \{0,1,2,3\}$$. 3 is present and we are happy.
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For both $$^{9}_{4}Be^{5}$$ and $$^{9}_{5}B^{4}$$ we have for the odd nucleon $$I=\frac{3}{2}$$. but thanks to the rules of angular momentum I=L+S and so since $$S=\frac{1}{2}$$, $$I \in \{l+\frac{1}{2},l-\frac{1}{2}\}$$ and this forces $$l \in \{1,2\}$$ , Now the parity is (-1)^l, so we are forced to have l=1 for both particles. I don't know is this is useful but I think is correct.
\\
\\
\\
\\
Now maybe we can say that the state in $$^{9}_{5}B^{5}$$ will be a superposition of vectors of the type $$v \otimes w$$, where the vector $$v$$ refers to one nucleon and the vector $$w$$ to the other. The parity on each member acts on each subspace: $$\Pi v \otimes w= \Pi_1 v \otimes \Pi_2 w=(-1)^2v \otimes w$$. and so the parity is positive. This reasoning seem to tell that to find the parity I simply have to multplicate the parities of the individual nucleons. is it right? sounds strange...