A nucleus with mass number $242$ and binding energy per nucleon as $7.6 \mathrm{MeV}$ breaks into two fragment each with mass number $121$. If each fragment nucleus has binding energy per nucleon as $8.1 \mathrm{MeV},$ the total gain in binding energy is $\_\_\_\_\_\_\_ \mathrm{MeV}$.

Binding energy is given by $E=∆m{c}^2$

where $∆m$ is the mass defect.

The energy per nucleon of the nucleus having mass number $242$ is $7.6 \mathrm{MeV}$.

The initial binding energy is,

$BE=242\times 7.6 \mathrm{MeV}$

The energy per nucleon of the nucleus with mass number $121$ is $8.1 \mathrm{MeV}$.

Therefore, binding energy is

$BE\text{'}=2\left(121\times 8.1 \mathrm{MeV}\right)$

The gain in the binding energy is 

$BE\text{'}-BE = (8.1 – 7.6) \times 242 \mathrm{MeV}=121 \mathrm{MeV}$

$= 121 \mathrm{MeV}$