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Cobalt decays to Nickel - Why is this suppressed?

  1. Apr 17, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Show maximum energy transferred in compton scattering is as such.
    (b) Identify what the peaks are and why some decays are suppressed.
    (c) How do you distinguish between these 2 decays

    2014_B4_Q4.png

    2. Relevant equations


    3. The attempt at a solution

    Part(a)
    Bookwork. For completeness, I simply state
    [tex]\Delta E = \frac{E_\gamma}{1 + \frac{m_e}{(1-cos\theta)E_\gamma}}[/tex]
    where ##c=1##.

    Other processes important in ##\gamma## ray detection are observation of galaxies and X-ray spectroscopy.

    Part(b)
    First peak is ##4^+ \rightarrow 2^+##. Second peak is ##2^+ \rightarrow 0^+##.

    Why is the decay ##4^+ \rightarrow 0^+## forbidden?

    Part(c)
    Not sure how to determine the branching ratios?
     
  2. jcsd
  3. Apr 17, 2015 #2

    mfb

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    I think the question asks for other processes the photons can make in the material, not other applications of gamma detection.

    Hint: this quoted question alone is sufficient to find the answer, even without knowing which states are meant.

    Can you count how frequent some things are happening?
     
  4. Apr 18, 2015 #3
    Apart from scattering off the electrons, not sure what it can do. Can it be absorbed?

    I think since the maximum energy that can be transferred to an electron is ##~0.91E_\gamma##, so ##4^+ \rightarrow 0^+## is forbidden.
     
  5. Apr 18, 2015 #4

    mfb

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    It can create new particles.
    Huh?
    How can the interaction of photons with matter long after their emission influence which photon production is allowed?
    Also, where does that 0.91 come from? It is a not a universal limit.
     
  6. Apr 18, 2015 #5
    Ok, so it can undergo EM interaction to produce particles such as pair creation ##\gamma + Co \rightarrow Co + e^+ + e^-##.

    Ok, I must be confused as I simply let ##E_\gamma = 2.505 MeV## which is wrong, as ##E_\gamma## is the energy of the photon and not the energy of the excited state.

    I'm guessing there are some form of selection rules?
     
  7. Apr 18, 2015 #6

    mfb

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    Pair creation, right.

    2.502 MeV is the energy of the excitation and also the energy of an emitted photon from this transition to a good approximation.
    Right. And ##4^+ \rightarrow 0^+## is sufficient to identify the selection rule.
     
  8. Apr 18, 2015 #7
    Ok, I just looked up Cottingham and the selection rule is:

    If the nucleus changes spin from ##J_I## to ##J_f## through ##\gamma##-decay, then
    [tex]|J_I + J_f| \geq J \geq |J_I-J_f| [/tex]
    The decay ##4^+ \rightarrow 2^+## represents ##6 \geq J \geq 2##.
    The decay ##4^+ \rightarrow 0^+## represents ## 4 \geq J \geq 4##.
    The decay ##2^+ \rightarrow 0^+## represents ## 2 \geq J \geq 2##.

    Why is the decay ##4^+ \rightarrow 0^+## suppresed? Is it because the most likely angular momentum for the photon is ##J=2## from the first and third reactions?
     
  9. Apr 18, 2015 #8

    mfb

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    Right.
    Large changes in angular momentum make the transition unlikely. As extreme example 180Ta needs a change of 8 and the decay has not been observed yet.
     
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