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Kinetic energies of particle decay

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data
    A lambda hyperon [tex]\Lambda^{0}[/tex] (mass = 1115.7 MeV/c^2) at rest in the lab frame decays into a neutron [tex]n[/tex] (mass = 939.6 MeV/c^2) and a pion [tex]\pi^{0}[/tex]: [tex]\Lambda^{0} \rightarrow n + \pi^{0}[/tex]

    What are the kinetic energies (in the lab frame) of the neutron and pion after the decay?


    2. Relevant equations
    [tex] KE = m c^{2} - m_{0} c^{2} = m_{0} c^{2} (\gamma - 1)[/tex]
    Conservation of Momentum: [tex] p_{n} = - p_{\pi} [/tex]
    [tex] p = \gamma m v [/tex]


    3. The attempt at a solution

    Using the relevant equations, there are two equations and two unknowns ([tex]v_{n}[/tex] and [tex]v_{\pi}[/tex]). However, all my algebra attempts failed here. Any ideas? Is the above what I should be starting with?
     
  2. jcsd
  3. Jul 7, 2008 #2

    Vanadium 50

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    Your first "relevant equation" isn't particularly relevant.

    Note that the pion and proton masses aren't anywhere in your equations. Does this seem right to you?
     
    Last edited: Jul 7, 2008
  4. Jul 7, 2008 #3
    I suppose I should've filled in more what I implied by the first equation.
    [tex] E_{i}= E_{f} [/tex]

    [tex] E_{i}= m_{\Lambda} c^{2} [/tex]

    [tex] E_{f}= \gamma_{n} m_{n} c^{2} + \gamma_{\pi} m_{\pi} c^{2} [/tex]

    for the first equation, conservation of energy.

    And

    [tex] \gamma_{n} m_{n} v_{n} = - \gamma_{\pi} m_{\pi} v_{\pi} [/tex]

    for the conservation of momentum. Those are the two equations and two unknowns I used, where

    [tex] \gamma_{i} = \left( \sqrt{1- (\frac{v_{i}}{c})^{2} } \right) ^{-1} [/tex]
     
  5. Jul 7, 2008 #4

    Vanadium 50

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    The gammas aren't helping. You might want to find an equation without them.
     
  6. Jul 7, 2008 #5

    Vanadium 50

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    Or rather...the gammas aren't helping right away.

    You need to start by getting one equation in one unknown. What would be the most useful thing to solve for? Hint: it's common to both decay particles.
     
  7. Jul 8, 2008 #6
    Ah, right. Good ol' [tex] E^2 = p^2 c^2 + m_{i}^{2} c^4 [/tex] The momentums are equal so can set both p's to be the same, we know the initial energy, we know the masses, solve for momentum, etc...

    Not pretty math, but it works out.

    Thanks Vanadium 50.
     
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