Question

The binding energy of a deuterium nucleus is about $1.115\text{MeV}$ per nucleon. Then the mass defect of the nucleus is about

• A. $2077u$
• B. $2.23u$
• C. $0.0024u$
• D. $0.024u$

Answer 1

The correct option is C $0.0024u$

Given:
For ${}_1{\text{H}}^2,\frac{B.E}{A}=1.115\text{MeV}$

$\Rightarrow B.E=1.115\times A=1.115\times 2$

$\Rightarrow B.E=2.23\text{MeV}$

We know that, $1u\times c^2=931.5\text{MeV}$

$\Rightarrow \frac{1}{c^2}=\frac{u}{931.5\text{MeV}}$

Also, We know that,

$B.E=\Delta m\times c^2$

$\Rightarrow \Delta m=\frac{2.23\text{MeV}}{c^2}$

$\Rightarrow \Delta m=\frac{2.23\text{MeV}u}{931.5\text{MeV}}$

$\Rightarrow \begin{array}{c}\Delta m=0.0024u\end{array}$

Hence, option $\left(C\right)$ is correct.