i think its 2.5MeV
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 2.5MeVproton007007
Hello Ryan, welcome to PF :)
Just so you learn from your 'mistakes': "I am having difficulty on solving the Problem 5" does not count as a relevant equation. Neither does " The answer I obtained is 1.5MeV, is it correct ? " count as an attempt at solution. PF has this template for very good reasons (see guidelines). We're not here to stamp-approve homework, but to help people along. So we need to know where you are and you can show by properly formulating the problem (i.e. not linking to a rotated test page) and listing off variables & known/given data, line up the equations you think you need and show what you have done.
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 ?