Question

A nuclear fission is represented by the following reaction

${\text{U}}^{236}\to {\text{X}}^{111}+{\text{Y}}^{122}+3_0{\text{n}}^1$

If the binding energies per nucleon of ${\text{X}}^{111},$ ${\text{Y}}^{122}$ and ${\text{U}}^{236}$ are $8.6\text{MeV},8.5\text{MeV}$$\text{and}7.6\text{MeV}$ respectively, then the energy released in the reaction will be

• A. $498\text{MeV}$
• B. $398\text{MeV}$
• C. $298\text{MeV}$
• D. $198\text{MeV}$

Answer 1

The correct option is D $198\text{MeV}$

The binding energy of ${\text{X}}^{111}$,
$B_X=111\times 8.6=954.6\text{MeV}$

The binding energy of ${\text{Y}}^{122}$,
$B_{\text{Y}}=122\times 8.5=1037\text{MeV}$

The binding enegy of ${\text{U}}^{236}$,
$B_{\text{U}}=236\times 7.6=1793.6\text{MeV}$

Total B.E of products $=B_{\text{X}}+B_{\text{Y}}$
$B.E_{products}=954.6+1037=1991.6\text{MeV}$

As, $B.E_{reactants}<B.E_{products}$, So exothermic reaction.

The energy released in the fission reaction is
$E=B.E_{products}-B.E_{reactants}$

$\Rightarrow E=1991.6-1793.6=198\text{MeV}$

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

Why this question? This question gives an idea how to calculate the energy in fission process.