Question

The mass defect for the nucleus of helium is $0.0303\text{amu}$. The binding energy per nucleon in $\text{MeV}$ is nearly

• A. $28$
• B. $7$
• C. $4$
• D. $1$

Answer 1

The correct option is B $7$

Given, $\Delta m=0.0303\text{amu}$

Number of Nucleon, $A=4\left(\text{Helium}\right)$

$B.E=\Delta m{c}^2=0.0303\text{(amu)}\times 931.5\text{(MeV/amu)}=28.2\text{MeV}$

Binding energy per nucleon will be

$\frac{\text{B.E.}}{A}=\frac{28.2}{4}=7.05\text{MeV}$

Hence, $\left(\text{B}\right)$ is the correct answer.