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(B) The mass defect $(\Delta m)$ for the helium nucleus is $0.0303 \, a.m.u.$We know that $1 \, a.m.u. = 931 \, MeV$ of energy.Total binding energy $(

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$7$

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(B) The mass defect $(\Delta m)$ for the helium nucleus is $0.0303 \, a.m.u.$We know that $1 \, a.m.u. = 931 \, MeV$ of energy.Total binding energy $(BE)$ = $\Delta m 	imes 931 \, MeV = 0.0303 	imes 931 \approx 28.21 \, MeV$.Helium nucleus $(_{2}^{4}He)$ has $4$ nucleons ($2$ protons and $2$ neutrons).Binding energy per nucleon = $\frac{BE}{	ext{Number of nucleons}} = \frac{28.21}{4} \approx 7.05 \, MeV$.Thus, the binding energy per nucleon is approximately $7 \, MeV$.

The mass defect for the nucleus of helium is $0.0303 \, a.m.u.$ What is the binding energy per nucleon for helium in $MeV$?

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