question_answer

Calculate the binding energy per nucleon (in \[\mathbf{MeV}\]) for \[_{2}^{4}\] He and \[_{2}^{3}\] He. Comment on the difference of these binding energies and its significance in relation to \[\alpha \] -decay of the nuclei. [Given : mass of \[_{1}^{1}\] \[H=\mathbf{1}.\mathbf{00783}\text{ }\mathbf{u},\] mass of \[_{0}^{1}n=1.00867u,\] mass of \[_{2}^{3}\] He  \[=3.01664u,\] mass of \[_{2}^{4}\] He \[=4.00387u]\]

Answer:

                B.E. of \[_{2}^{4}\]He \[=[2{{m}_{p}}+2{{m}_{n}}-m(_{2}^{4}\text{He})]\times {{c}^{2}}\] \[=[2\times 1.00783+2\times 1.00867-4.00387]\times 931\text{MeV}\]\[=[4.03390-4.00387]\times 931=0.02933\times 931\text{MeV}\]\[=27.30623\text{MeV}\] B.E. per nucleon of \[_{2}^{4}He\] \[=\frac{27.30623}{4}=6.83\,MeV\] B.E. of \[_{2}^{3}He\] \[=[2{{m}_{p}}+{{m}_{n}}-m(_{2}^{3}He)]{{c}^{2}}\] \[=[2\times 1.00783+1.00867-3.01664]\times 931\,MeV\] \[=0.00769\times 931\,MeV=7.16\,MeV\] B.E. per nucleon of \[_{2}^{3}He\] \[=\frac{7.16}{3}=2.39\,MeV\] As the binding energy per nucleon of \[_{2}^{4}\]He is larger than that of \[_{2}^{3}\]He, so unstable heavy nuclei prefer to get stabilised through \[\alpha \]-decay.