Question

The mass defect of ${}_2^{4}He$ is $0.03u$. The binding energy per nucleon of helium (in $MeV$) is

• A. $69.825$
• B. $6.9825$
• C. $2.793$
• D. $27.93$

Answer 1

The correct option is B $6.9825$

Mass defect of ${}_2^{4}He$, $\Delta M=0.03u$

Binding energy of helium $E=\Delta M\times 931$ $MeV$

$E=0.03\times 931.5=27.93MeV$

Number of nucleons in helium $n=4$

Binding energy per nucleon $\frac{E}{n}=\frac{27.93}{4}=6.9825$