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- Thread starter Ryan.L
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the mass defect of X is 2MeV since the K.E. released(energy of the products) must be equal to the mass defect.

in 2nd eqn, the mass defect is sum of Mass defect of X and the electron

therefore we add 0.5MeV to the X's 2.0MeV to get 2.5MeV

proton007007

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Just so you learn from your 'mistakes': "I am having difficulty on solving the Problem 5

I also have a question for proton: in the sescond reaction you also don't have to create a positron. Doesn't that yield another .5 MeV ?

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