If the binding energy per nucleon of deuteron is 1.115 MeV , its mass defect in atomic mass unit is\

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Professor PC Thomas Classes · Answer

Binding energy per nucleus is\ $E_{b n}=\frac{E_{b}}{A}$\ For deuteron, $\mathrm{A}=2$\ $	herefore 1.116 \mathrm{MeV}=\frac{\mathrm{E}_{\mathrm{b}}}{2}$\ $\mathrm{E}_{\mathrm{b}}=2 	imes 1.115 \mathrm{MeV}$\ $\mathrm{E}_{\mathrm{b}}=\Delta \mathrm{Mc}^{2}$\ Mass defect,\ $\Delta \mathrm{M}=\frac{2 	imes 1.115}{931.5} \mathrm{u}\left[\because 1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}ight]$\ $=0.0024 \mathrm{u}$\