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.
OR
Calculate the energy in the fusion reaction:
Solution
The B.E. of the nucleus of mass number 240, B1 = 7.6 x 240 = 1824 MeV
The B.E of each product nucleus, B2 = 8.5 x 120 - 1020 MeV
Then, the energy released as the nucleus breaks is given by,
E = 2B2 - B1 = 2 x 1020 - 1824 = 216 MeV
OR
Given:
B.E of 
B.E of 
Energy in the fusion reaction is given by, 
