The binding energy of deuteron is 2.2 , MeV a | vedclass

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(B) The nuclear fusion reaction is given by: ${}_1^2H + {}_1^2H \rightarrow {}_2^4He + Q$.Binding energy of one deuteron is $2.2 \, MeV$. Therefore,th

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$23.6$

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(B) The nuclear fusion reaction is given by: ${}_1^2H + {}_1^2H ightarrow {}_2^4He + Q$.Binding energy of one deuteron is $2.2 \, MeV$. Therefore,the total binding energy of two deuterons is $2 	imes 2.2 \, MeV = 4.4 \, MeV$.The binding energy of one ${}_2^4He$ nucleus is $28 \, MeV$.The energy released $(Q)$ in the fusion process is equal to the difference between the binding energy of the product nucleus and the total binding energy of the reactant nuclei.$Q = BE({}_2^4He) - 2 	imes BE({}_1^2H)$$Q = 28 \, MeV - 4.4 \, MeV = 23.6 \, MeV$.

The binding energy of deuteron is $2.2 \, MeV$ and that of ${}_2^4He$ is $28 \, MeV$. If two deuterons are fused to form one ${}_2^4He$ nucleus,then the energy released is ......... $MeV$.

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