Question

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

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

Answer 1

Answer: (B)

5.6MeV

Answer 2

For $^7_3$Li nucleus , Mass defect, $\Delta M=0.042 u$ $\because \, 1 u=931.5 MeV/c^2$ $\therefore \, \, \Delta M=0.042 \times 931.5 MeV/c^2$ $=39.1 \,MeV/c^2$ Binding energy, $E_b =\Delta Mc^2$ $=\bigg(39.1 \frac{MeV}{c^2}\bigg)c^2$ $=39.1 \,MeV$ Binding energy per nucleon, $E_{bn}=\frac{E_b}{A}=\frac{39.1 \,MeV}{7}$ $=5.6 \,MeV$