(a)OR(b)

Calculate the binding energy of an alpha particle in MeV . Given mass of a proton $\displaystyle =1.007825 \mathrm{u}$ mass of a neutron $\displaystyle =1.008665 \mathrm{u}$ mass of He nucleus $\displaystyle =4.002800 \mathrm{u}$ $\displaystyle 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2}$

(a) \begin{tabular}[t]{|l|l|}

Number of Protons \(\displaystyle =2\) Number of neutrons \(\displaystyle =2\) \[\text { B.E. }=\left[\left(Z M_{\mathrm{P}}+(A-Z) \mathrm{m}_{n}\right)-\mathrm{m}_{N}\right] C^{2} \] \(\displaystyle \Delta \mathrm{m}=[(2 \times 1.007825+2 \times 1.008665)-(4.002800)] \mathrm{u}\) B. \(\displaystyle \mathrm{E} .=0.03018 \times 931 \mathrm{MeV}\) B \(\displaystyle \mathrm{E} .=28 \mathrm{MeV}\)(b) Binding energy of nucleus \(\displaystyle \mathrm{P}=7.6 \times 240=1824 \mathrm{MeV}\) Binding energy of nucleus \(\displaystyle \mathrm{Q}=8.5 \times 110=935 \mathrm{MeV}\) Binding energy of nucleus \(\displaystyle \mathrm{R}=8.4 \times 130=1092 \mathrm{MeV}\) Binding energy of \(\displaystyle (\mathrm{Q}+\mathrm{R})=935+1092=2027 \mathrm{MeV}\) Energy released = Binding Energy of \(\displaystyle (\mathrm{Q}+\mathrm{R})-\) Binding energy of P Energy released \(\displaystyle =(2027-1824) \mathrm{MeV}\) \(\displaystyle =203 \mathrm{MeV}\) \end{tabular}