The mass defect for the nucleus of Helium is $0.0303 a m u$ . The binding energy per nucleon in $M e V$ is

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Solution

The correct option is B 7
$B . E = (931.478 \times .0303 \frac{M e V}{c^{2}}) c^{2}$
$= 28.22 M e V$
$B . E p e r n u c l e o n = \frac{B E}{A} = \frac{28.22}{4} = 7.05 M e V$

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