$A$ nucleus with mass number $A = 240$ and binding energy per nucleon $= 7.6 \,MeV$ breaks into two fragments each of $A = 120$ with binding energy per nucleon $= 8.5 \,MeV$. Calculate the released energy.

(216 MEV) The binding energy of the parent nucleus with $A = 240$ is given by:
$E_{b1} = 240 \times 7.6 \,MeV = 1824 \,MeV$
The binding energy of the two daughter fragments,each with $A = 120$,is given by:
$E_{b2} = 2 \times (120 \times 8.5 \,MeV) = 2 \times 1020 \,MeV = 2040 \,MeV$
The energy released in the process is the difference between the total binding energy of the products and the binding energy of the parent nucleus:
$Q = E_{b2} - E_{b1}$
$Q = 2040 \,MeV - 1824 \,MeV = 216 \,MeV$