One of the possible decay modes of the neutral kaon is K-> pion + pion The rest energies of the K0 and pion are 498 MeV and 135 MeV, respectively. The kaon is initially at rest when it decays.(adsbygoogle = window.adsbygoogle || []).push({});

a) How much energy is released in the decay?

b) What are the momentum and relative direction of the two neutral pions

Possible equations:

E0=mc^2

E=(gamma)mc^2

p=(gamma)mv

I got logged out of my last attempt at a post, so let me see if I can replicate it ;x

I figured at first that the energy released would deal with the loss of energy from the 498 MeV = 135MeV+135MeV +???, but these are all rest energies. I'm not sure if there is a difference, but if the energy released by this was not 228 (498-170) MeV then I would have to delve into the wonderful world of relativity. However, I am not sure of what options I have here. Multiplying it by gamma seems too easy. :/

For momentum we take p=mv(gamma). We need to find gamma and v. E0=mc^2 gives us the mass (m=E0/c^2). E=(gamma)mc^2 throws us another nice equation using these variables. We can plop E0=mc^2 in there to target the gamma. We then get E=(gamma)E0. or (gamma)=E/E0. But now we need to find E also.

Any help would be appreciated

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# Homework Help: Energy released in particle decay w/ relativity

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