Calculate the binding energy per nucleon (in MeV ) of a helium nucleus ( _2^4 He ). Given: m (_2^4 He )=4 · 002603 u m_n =1.008665 u; m_H =1.007825…

\(\displaystyle \Delta \mathrm{m}=\left(2 \mathrm{m}_{\mathrm{n}}+2 \mathrm{m}_{\mathrm{H}}\right)-\mathrm{m}\left({ }_{2}^{4} \mathrm{He}\right) =2 \times 1.008665+2 \times 1.007825-4.002603 =0.0303774 \mathrm{u}\) \(\displaystyle B E=\Delta \mathrm{m} \mathrm{c}^{2} =0.030377 \times 931 =28.2962 \mathrm{MeV}\) \(\displaystyle \frac{\mathrm{BE}}{\text { nucleon }}=\frac{28.2962}{4}=7.07 \mathrm{MeV}\)