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A proton emitting and re-absorbing a pi meson.

  1. Oct 28, 2007 #1
    A proton or a neutron can sometimes "violate" conservation of energy by emitting and then re-absorbing a pi meson, which as a mass of 135 MeV/c^2. This is possible as long as the pi meson is re-absorbed within a shoart enough time [tex]\Delta T[/tex] consistent with the Uncertainty Principle. Consider therefore the following:

    A) A proton p undergoes the following process: [tex]p \rightarrow p + \pi[/tex]. By what amount is the energy violated (ignore any kinetic energies and assume all particles are at rest).

    B) For how long a time [tex]\Delta T[/tex] can the pi meson exist?

    C) Assuming now that the pi meson is traveling at very nearly the speed of light, how far from the proton can it go?

    2. Relevant equations
    [tex]\Delta P \Delta x = \frac{\hbar}{2}[/tex]
    [tex]\Delta E \Delta t = \frac{\hbar}{2}[/tex]
    [tex]E \approx pc[/tex] when [tex]v \approx c[/tex]

    3. The attempt at a solution
    A) 938.27 MeV -> 938.27 Mev + 135 MeV. A violation of 135 MeV.
    B) t = 2.4376 E -24 s (Using [tex]\Delta E \Delta t = \frac{\hbar}{2}[/tex].)
    C) I have no idea where to start here... I assume the E is going to be equal to pc, but how do I get the pc?
  2. jcsd

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