Question

The mass of a certain nucleus $X^4$ is less than that of its constituent particles by $0.03u$. The binding energy per nucleon of $X^4$ nucleus will be

• A. $7\text{MeV}$
• B. $3.5\text{MeV}$
• C. $14\text{MeV}$
• D. $21\text{MeV}$

Answer 1

The correct option is A $7\text{MeV}$

Given that, the mass defect,
$\Delta m=0.03u$

Binding Energy,
$B.E=\Delta m\times c^2$

$\Rightarrow B.E=0.03u\times c^2$

We know that, $1u{c}^2=931.5\text{MeV}$

$\Rightarrow B.E=0.03\times 931.5\text{MeV}$

$\Rightarrow B.E=27.945\text{MeV}$

Binding energy per nucleon,

$\frac{B.E}{A}=\frac{27.945}{4}=6.986\text{MeV}$

$\Rightarrow \frac{B.E}{A}=7\text{MeV}$

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