Question

Consider the nuclear reaction
$X^{200}\to A^{110}+B^{90}$
If the binding energy per nucleon for $X,A$ and $B$ are $7.4\text{MeV},8.2\text{MeV}$ and $8.2\text{MeV}$ respectively, the energy released in the reaction is

• A. $160\text{MeV}$
• B. $110\text{MeV}$
• C. $200\text{MeV}$
• D. $90\text{MeV}$

Answer 1

The correct option is A $160\text{MeV}$

Binding energy of $X$
$(B.E{)}_X=200\times 7.4\text{MeV}$

Binding energy of $A$
$(B.E{)}_A=100\times 8.2\text{MeV}$

Binding energy of $B$
$(B.E{)}_B=90\times 8.2\text{MeV}$

If $Q$ is the energy released in the reaction,

By conservation of energy,

$Q=B.E\text{of products}-B.E\text{of reactants}$

$Q=(B.E{)}_A+(B.E{)}_B-(B.E{)}_X$

$\Rightarrow Q=(110\times 8.2)+(90\times 8.2)-200\times 7.4$

$\Rightarrow Q=200\times 8.2-200\times 7.4$

$\Rightarrow Q=200[8.2-7.4]$

$\Rightarrow Q=200\times 0.8$

$\Rightarrow Q=160\text{MeV}$

Hence, option $\left(A\right)$ is correct.