Gamma Ray Energy from Decay of Boron

AI Thread Summary
In the decay of 1000 atoms of (12/5) Boron, 1000 beta particles are released, corresponding to the decay into (12/6) Carbon and an electron. The energy released as gamma rays is calculated to be 88 MeV, derived from the fact that 2% of the atoms emit energy at 4.4 MeV. The calculation involves multiplying the number of atoms by the percentage that emits gamma rays and the energy per emission. The solution process highlights the application of E = mc^2 and the importance of understanding decay processes. The individual successfully resolves the problem independently, emphasizing the satisfaction of solving complex physics questions.
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Homework Statement



I have to figure how many beta particles are released and how much energy is released as gamma rays in the complete decay of 1000 atoms of (12/5) Boron.

Homework Equations



Mass of one boron atom is 10.811 g. I think we may be able to use E = mc^2 somehow but I'm not sure.

The Attempt at a Solution



The answer the book gives is that 1000 beta particles are released and 88 MeV of energy is released as gamma rays. The first seems obvious because (12/5) boron decays into (12/6) C + an electron. So if you have 1000 atoms you will have 1000 electrons / beta particles. The second I'm not so sure of. How can you calculate the energy released as gamma rays from just this information?
 
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Never mind, I figured it out :)

There were 1000 initial atoms and 2% of them emit energy in the form of gamma rays at 4.4 MeV.

So 1000 x 0.02 x 4.4 MeV = 88 MeV
Huzzah, huzzah u_u it feels good to solve your own problems.
 
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