Question

If the binding energy per nucleon of deuterium is $1.115 \,MeV$, its mass defect in atomic mass unit is

• A. $0.0048$
• B. $0.0024$
• C. $0.0012$
• D. $0.0006$

Answer 1

Answer: (B)

$0.0024$

Answer 2

Binding energy per nucleon is $E_{bn}=\frac{E_{b}}{A}$ For deuterium, $A = 2$ $\therefore 1.115\, MeV =\frac{E_{b}}{2}$ or $E_{b}=2\times1.115\,MeV$ or $E_{b}=\Delta Mc^{2}$ Mass defect, $\Delta M=\frac{2\times1.115}{931.5}u \left[\because1\,u=931.5\,MeV/ c^{2}\right]$ $=0.0024\,u$