Question

The mass of a $_{3}^{7}Li$ nucleus is $0.042 \, u$ less than the sum of the masses of all its nucleons. The binding energy per nucleon of $_{3}^{7}Li$ nucleus is nearly

• A. 46 MeV
• B. 5.6 MeV
• C. 3.9 MeV
• D. 23 MeV

Answer 1

Answer: (B)

For $_{3}^{7}Li$ nucleus,
Mass defect, $\Delta M=0.042u$
∵ $1u=931.5MeVc^{- 2}$
∴ $\Delta M=0.042\times 931.5MeVc^{- 2}$
$=39.1MeVc^{- 2}$
Binding energy, $E_{b}=\Delta Mc^{2}$
$=\left(3 9 \cdot 1 MeV c^{- 2}\right)c^{2}$
$=39.1MeV$
Binding energy per nucleon
$E_{b n}=\frac{E_{b}}{A}=\frac{39 \cdot 1 MeV}{7}\approx5\cdot 6MeV$