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Homework Help: Special Relativity and Particle Decay

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle with rest mass [tex]M_0[/tex] can decay at rest into a pair of particles each with rest
    mass [tex]m_0[/tex]. Calculate the following in the rest inertial frame of the original particle
    using [tex]M_0c^2=600[/tex]MeV and [tex]m_0c^2=150[/tex] MeV .
    (i) The total energy of each particle produced in the decay.
    (ii) The magnitude of the relativistic linear momentum of each particle produced in
    the decay.
    (iii) The speed of the particles produced in the decay.
    2. Relevant equations
    [tex]E^2=p^2c^2 + E_0^2[/tex] [1]
    [tex]E=\gamma m_0c^2[/tex] [2]

    3. The attempt at a solution
    i)Ok, first I'm assuming that the new particles will have equal energy, so each E=300MeV
    ii) I think I'm meant to use equation [1], where E=300 and E0=150. This gives [tex]p=\frac{259.8}{c}=0.866[/tex]. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?
    iii) I use the fact that [tex]E=\gamma m_0c^2[/tex], giving [tex]\gamma=2 \Rightarrow v=\frac{\sqrt{3}}{2}c[/tex]

    Thanks very much for helping!!
  2. jcsd
  3. Apr 25, 2010 #2


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    That's correct. You can work out the units: E^2-E0^2 has units of MeV^2, so p has the unit MeV*s/m. To get the result in the familiar kg*m/s, just convert MeV to J before doing the calculation.
  4. Apr 25, 2010 #3


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    Good. You're not really assuming the answer though; you're deducing it from symmetry.
    The method is correct. Just write the units in explicitly, and you'll see how they turn out.

    [tex](pc)^2 = E^2 - E_0^2 \rightarrow pc = \sqrt{(300~\textrm{MeV})^2 - (150~\textrm{MeV})^2} = 259.8~\textrm{MeV}[/tex]

    So your answer is in units of MeV/c.
    Since you have the momentum, you can also find the velocity using β=v/c=pc/E. Either method is fine, though.
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