Question

Mass defect for the helium nucleus is $0.0303amu$. Binding energy per nucleon for this (in $MeV$) will be_______

• A. $28$
• B. $7$
• C. $4$
• D. $1$

Answer 1

The correct option is B $7$

Mass defect of helium nucleus $\Delta M=0.0303amu$
Binding energy $E=\Delta M{c}^2=\Delta M\times 931.5MeV$
$⟹E=0.0303\times 931.5=28MeV$
Number of nucleon in helium nucleus $=4$
Thus binding energy per nucleon $=\frac{28}{4}=7MeV$