Question

The mass of a ${}_3{\text{Li}}^7$ nucleus is $0.042u$ less than the sum of the masses of all its nucleons. The binding energy per nucleon of ${}_3{\text{Li}}^7$ nucleus is nearly,

• A. $3.9\text{MeV}$
• B. $5.6\text{MeV}$
• C. $23\text{MeV}$
• D. $46\text{MeV}$

Answer 1

The correct option is B $5.6\text{MeV}$

Given:
Mass defect $\Delta m=0.042u$

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

$\Rightarrow B.E=0.042u\times c^2$

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

$\Rightarrow B.E=0.042\times 931.5\text{MeV}$

$\Rightarrow B.E=39.123\text{MeV}$

The number of nucleons $\left(A\right)$ in ${}_3{\text{Li}}^7=7$

The binding energy per nucleon of

${}_3{\text{Li}}^7=\frac{B.E}{A}=\frac{39.123}{7}$

$\Rightarrow \frac{B.E}{A}=5.589\approx 5.6\text{MeV}$

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