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Statement about Forbidden Alpha decay transitions.

  1. Mar 6, 2013 #1
    Hi, I'm confused about a statement about the change in final/initial states of the daughter/parent atom in an alpha decay. It is the following;

    "The spin between the parent ([itex]I_i[/itex]) and daughter ([itex]I_f[/itex]) can change by [itex] lh[/itex] (h being h-bar, l is the orbital angular quantum number of the alpha particle), where;

    [itex]\vec{I_i} = \vec{I_f} + \vec{l}[/itex]

    and the parity changes by [itex] (-1)^l [/itex]
    "

    I'm confused because if, for example, we take the initial state of the parent to be [itex] 0^+ [/itex], then there are the following cases;

    [itex]\vec{I_i} = \vec{I_f} + \vec{l}[/itex] means

    [itex]|\vec{I_i}| = |\vec{I_f} + \vec{l}|[/itex]

    Coupling angular momentum together would surely mean that the total orbital quantum number would be of multiple values;

    [itex] L = I_f + l, I_f + l - 1, I_f + l - 2, ... I_f - l [/itex]

    So if [itex] I_i = 0^+ [/itex] we could have multiple [itex] I_f [/itex] for a given [itex] l [/itex]. For example [itex] l = 1 [/itex], then

    [itex] L = I_f + 1, I_f , I_f - 1 = I_i = 0 [/itex]

    Meaning [itex] I_f [/itex] could take on values [itex] 0 [/itex] or [itex] 1 [/itex]. My notes seem to suggest only the [itex] I_f = 1 [/itex] state is possible. Am I looking at this in completely the wrong way? I think I don't fully understand what it means by "can change by [itex] lh [/itex]", what is the signficance of the "can" change?

    I looked on the internet and in my textbook, not much to a detail that I can understand. Beta decay forbidden decays seemed to be the closest I could find which might explain it but I don't know how applicable it is to this situation.
     
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 6, 2013 #2

    mfb

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    ##I_i=0## gives ##|\vec{I_f}+\vec{l}|=0##, which has the solution ##\vec{I_f}=-\vec{l}## only.
     
  4. Mar 6, 2013 #3
    Ah yes clearly. I should've realised that. Here's a better example which might help me understand it for general [itex] l [/itex] and [itex] I_f [/itex].

    Say the initial state was [itex] I_i = 1^+ [/itex]. We must have the coupled orbital angular momentum quantum number [itex] L [/itex] equal to [itex] 1 [/itex]. Then say the alpha particle had [itex] l = 2 [/itex]

    The coupled orbital angular momentum quantum number of

    [itex]\vec{I_i} = \vec{I_f} + \vec{l}[/itex]

    Would be

    [itex] L = I_f + l, |I_f + l - 1|, |I_f + l - 2|, ... |I_f - l| [/itex]

    So if we had [itex] I_f = 3 [/itex]

    [itex] L = 5,4,3,2,1 [/itex]

    Corresponding to a possibility to have [itex] L = 1 [/itex]

    If we had [itex] I_f = 2 [/itex]

    [itex] L = 4,3,2,1,0 [/itex]

    Corresponding to another possibility to have [itex] L = 1 [/itex]

    And lastly; If we had [itex] I_f = 1 [/itex]

    [itex] L = 3,2,1 [/itex]

    Corresponding to another possibility to have [itex] L = 1 [/itex]


    So in a transition from a state [itex] 1^+ [/itex] with the emission of an alpha with [itex] l = 2 [/itex], it's possible to have 3 final states for the daughter ([itex] 3^-, 2^-, 1^- [/itex])? Is this correct?
     
  5. Mar 7, 2013 #4

    mfb

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    Looks correct.

    ##|l-I_i| \leq I_f \leq |I+I_i|##
     
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