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Parity of [tex]^{9}_{5}B^{5}[/tex]

  1. Aug 19, 2010 #1
    Hy... I'm reviewing some nuclear physics but don't remember certain topics.

    This exercise for example:

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

    My thinkings, which are correct?:

    First of all we see that [tex]^{9}_{5}B^{5}[/tex] has as outermost nucleons the odd nucleon of [tex]^{9}_{4}Be^{5}[/tex] and the odd nucleon of [tex]^{9}_{5}B^{4}[/tex]. So [tex]I=I_1+I_2[/tex], and thanks to the rules of q.m. [tex]I \in \{0,1,2,3\}[/tex]. 3 is present and we are happy.
    So for both [tex]^{9}_{4}Be^{5}[/tex] and [tex]^{9}_{5}B^{4}[/tex] we have for the odd nucleon [tex]I=\frac{3}{2}[/tex]. but thanks to the rules of angular momentum I=L+S and so since [tex]I=\frac{1}{2}[/tex], [tex]I \in \{l+\frac{1}{2},l-\frac{1}{2}\}[/tex] and this forces [tex]l \in \{1,2\}[/tex] , 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 [tex]^{9}_{5}B^{5}[/tex] will be a superposition of vectors of the type [tex]v \otimes w[/tex], where the vector [tex]v[/tex] refers to one nucleon and the vector [tex]w[/tex] to the other. The parity on each member acts on each subspace: [tex]\Pi v \otimes w= \Pi_1 v \otimes \Pi w_2=(-1)^2v \otimes w[/tex]. 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. jcsd
  3. Aug 19, 2010 #2
    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 [tex]^{9}_{4}Be^{5}[/tex] and [tex]^{9}_{5}B^{4}[/tex] have both as quantum numbers [tex]\frac{3}{2}^{-}[/tex] and their properties are determined by the odd nucleon, show that the ground state of [tex]^{9}_{5}B^{5}[/tex] can be [tex]3^{+}[/tex].

    My thinkings, which are correct?:

    First of all we see that [tex]^{9}_{5}B^{5}[/tex] has as outermost nucleons the odd nucleon of [tex]^{9}_{4}Be^{5}[/tex] and the odd nucleon of [tex]^{9}_{5}B^{4}[/tex]. So [tex]I=I_1+I_2[/tex], and thanks to the rules of q.m. [tex]I \in \{0,1,2,3\}[/tex]. 3 is present and we are happy.
    \\
    \\
    \\
    \\
    For both [tex]^{9}_{4}Be^{5}[/tex] and [tex]^{9}_{5}B^{4}[/tex] we have for the odd nucleon [tex]I=\frac{3}{2}[/tex]. but thanks to the rules of angular momentum I=L+S and so since [tex]S=\frac{1}{2}[/tex], [tex]I \in \{l+\frac{1}{2},l-\frac{1}{2}\}[/tex] and this forces [tex]l \in \{1,2\}[/tex] , 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 [tex]^{9}_{5}B^{5}[/tex] will be a superposition of vectors of the type [tex]v \otimes w[/tex], where the vector [tex]v[/tex] refers to one nucleon and the vector [tex]w[/tex] to the other. The parity on each member acts on each subspace: [tex]\Pi v \otimes w= \Pi_1 v \otimes \Pi_2 w=(-1)^2v \otimes w[/tex]. 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...
     
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