Question

A nuclear fission is given below
$A^{240}\to B^{100}+C^{140}+Q\left(\text{energy}\right)$.
Let binding energy per nucleon of nucleus $A,B$ and $C$ is $7.6\text{MeV},8.1\text{MeV}$ and $8.1\text{MeV}$ respectively. Value of $Q$ is : (Approximately)

• A. $20\text{MeV}$
• B. $220\text{MeV}$
• C. $120\text{MeV}$
• D. $240\text{MeV}$

Answer 1

The correct option is C $120\text{MeV}$

Binding Energy of $A=240\times 7.6=1824\text{MeV}$
Binding Energy of $B=100\times 8.1=810\text{MeV}$
Binding Energy of $C=140\times 8.1=1134\text{MeV}$
So $Q=(810+1134)\text{MeV}-1824\text{MeV}=120\text{MeV}$