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Q.

The mass defect for the nucleus of helium is 0.0303 amu. The binding energy per nucleon in MeV is nearly

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a

7

b

28

c

1

d

4

answer is B.

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Detailed Solution

$\begin{array}{l} BE = Δm \times 931 MeV = 0 .0303 \times 931 MeV \\ \frac{BE}{A} = \frac{0 .0303 \times 931}{4} = 7 MeV \end{array}$

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The mass defect for the nucleus of helium is 0.0303 amu. The binding energy per nucleon in MeV is nearly