A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A= 120 with BE/A = 8.5 MeV. Calculate the released energy.

Explanation

Since the nucleus has a mass number A = 240 and binding energy per nucleon of 7.6 MeV, its total binding energy `E_(1) = 240 xx 7.6 = 1824 MeV`. As both fragments of mass number A = 120 has a binding energy per nucleon of 8.5 MeV, total binding energy of the fragments `E_(2)=2 xx 120 xx 8.5 = 2040 MeV`
`therefore` Energy released during fission reaction `E = E_(2)-E_(1) = 2040 - 1824 = 216 MeV`