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Minimum Energy Of A Beta Particle

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data
    When Ne (Z=10) decays to Na (Z=11), what is the maximum kinetic energy of the emitted electron? What is the minimum energy? What is the energy of the neutrino in each case? Ignore recoil of the daughter nucleus.


    2. Relevant equations
    [tex]E = mc^2[/tex]

    3. The attempt at a solution
    The following is the decay:
    [tex]^{23}_{10} Ne \rightarrow ^{23}_{11} Na + \beta ^- + n + Q[/tex]

    I know that there will be extra energy, Q, emitted in the decay and I know I can find it using the difference in mass between the products and initial particles. Now I'm assuming that this energy will go towards the kinetic energy of the neutron and the electron. But first, the momentum should be conserved. So:
    [tex]m_{\beta} v_{\beta} = m_n v_n[/tex] or [tex]v_n = \frac{m_{\beta}}{m_n} v_{\beta}[/tex]

    Then I can use this together with the energy equation:
    [tex]\frac{m_{\beta} v_{\beta}^2}{2} + \frac{m_n v_n^2}{2} = Q[/tex]
    [tex]\frac{m_{\beta} v_{\beta}^2}{2} + \frac{m_{\beta}^2 v_{\beta}^2}{ 2 m_n } = Q[/tex]
    [tex]\frac{m_{\beta} v_{\beta}^2}{2} = \frac{Q}{( 1 + \frac{m_{\beta}}{m_n} )}[/tex]

    So as the mass of the neutrino approaches 0, then the fraction m_\beta / m_n approaches infinity (so 1+ that approaches infinity) and Q divided by a very big number would be about 0.

    That is what am I unsure of, is the kinetic energy of the electron really 0 or did I make a mistake somewhere in my assumptions?

    My second question concerns the minimum kinetic energy of the electron. Would the minimum KE be when all of Q is heat, so in that case KE=0?
     
  2. jcsd
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