Question

Consider the following nuclear reaction 158O → 157N + 01e + ν and determine the Q value or total energy released (in MeV) in this process. (Let the atomic mass of 158O be 15.0031 u, the atomic mass of 157N be 15.0001 u and the mass of 01e be 5.4858 ✕ 10−4 u.)

Answer 1

Q value of nuclear reaction is given by:

Q = dm*c^2

dm = mass difference = Mass of O-15 - Mass of N-15 + Mass of e

dm = 15.0031 u - 15.0001 u - 5.4858*10^-4 u

dm = 0.00245142 u

c = speed of light

So

Q_value = Q = dm*c^2

Q = 0.00245142 u*c^2*((931.5 MeV/c^2)/1 u)

Q = 0.00245142*931.5

Q = 2.2835 MeV

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