Question

The mass defect for helium nucleus is $0.0304$ $a.m.u$. The binding energy per nucleon of helium nucleus is ________

• A. $28.3$ $MeV$
• B. $7.075$ $MeV$
• C. $9.31$ $MeV$
• D. $200$ $MeV$

Answer 1

The correct option is B $7.075$ $MeV$

$1amu=1.67\times {10}^{-27}kg$
Using this and Einsteins relation, we get the energy equivalent to the rest mass as:
$E=m{c}^2=(1.67\times {10}^{-27}\times 0.0304)\times (3\times {10}^8{)}^2=4.569\times {10}^{-12}J$

Converting this energy in electron volts.
$E=\frac{4.569\times {10}^{-12}}{1.6\times {10}^{-19}}=28MeV$

This is per nucleon of Helium. One Helium nucleus has four nucleons.

Hence, binding energy per nucleon is $7MeV$.