Q135 / 200 · NEET 2010 (Screening) · Physics

The mass of a ${ }_3 \mathrm{Li}^7$ nucleus is $0.042 \mathrm{u}$ less than the sum of the masses of all its nucleons. The binding energy per nucleon of ${ }_3 \mathrm{Li}^7$ nucleus is nearly

A
$46 \mathrm{MeV}$
Correct answer
B
$5.6 \mathrm{MeV}$
C
$3.9 \mathrm{MeV}$
D
$23 \mathrm{MeV}$

Solution

If $\mathrm{m}=1 \mathrm{u}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$, then $\mathrm{E}=931 \mathrm{MeV}$ ie, $1 \mathrm{u}=931 \mathrm{MeV}$ Binding energy $=0.042 \times 931=39.10 \mathrm{MeV}$ $\therefore$ Binding energy per nucleon $$ =\frac{39.10}{7}=5.58 \approx 5.6 \mathrm{MeV} $$

Hence, the correct answer is option: 'B'.

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