A proton emitting and re-absorbing a pi meson

[lylos]

Homework Statement

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 \Delta T consistent with the Uncertainty Principle. Consider therefore the following:

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

B) For how long a time \Delta T 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?

Homework Equations

\Delta P \Delta x = \frac{\hbar}{2}
\Delta E \Delta t = \frac{\hbar}{2}
E \approx pc when v \approx c

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 \Delta E \Delta t = \frac{\hbar}{2}.)
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?

 

[owensomething]

E=mc^2 ignoring kinetic energies. From there find delta E, and plug back into the uncertainty equation to find delta t. For part c, assume the particle is traveling at c with a lifetime equal to delta t and find the distance it can travel.

 

[Steve4Physics]

E=mc^2 ignoring kinetic energies. From there find delta E, and plug back into the uncertainty equation to find delta t. For part c, assume the particle is traveling at c with a lifetime equal to delta t and find the distance it can travel.

Hi @owensomething. For information, note that the Post you replied to is from October 2007!