Question

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.

Answer 1

Gain in binding energy for nucleon is about 0.9 MeV.
Binding energy of the nucleus, B₁= 7.6 x 240 = 1824 MeV
Binding energy of each product nucleus, B₂ = 8.5 x 120 = 1020 MeV. Then, energy released as the nucleus breaks.
E = 2B₂-B₁= 2 x 1020 – 1824 = 216 MeV