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.
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