If the binding energy per nucleon of deuteron | vedclass

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(C) The reaction is ${ }_1 H^2 + { }_1 H^2 \rightarrow { }_2 He^4 + Q$.Total binding energy of reactants: Each deuteron has $2$ nucleons,and there are

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$5.95$

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(C) The reaction is ${ }_1 H^2 + { }_1 H^2 ightarrow { }_2 He^4 + Q$.Total binding energy of reactants: Each deuteron has $2$ nucleons,and there are $2$ deuterons. Total nucleons = $4$. Binding energy per nucleon = $1.15 	ext{ MeV}$. Total binding energy = $4 	imes 1.15 	ext{ MeV} = 4.6 	ext{ MeV}$.Total binding energy of products: The $\alpha$-particle $({ }_2 He^4)$ has $4$ nucleons. Binding energy per nucleon = $7.1 	ext{ MeV}$. Total binding energy = $4 	imes 7.1 	ext{ MeV} = 28.4 	ext{ MeV}$.Energy released $(Q)$ = (Total binding energy of products) - (Total binding energy of reactants) = $28.4 	ext{ MeV} - 4.6 	ext{ MeV} = 23.8 	ext{ MeV}$.The energy released per nucleon is calculated by dividing the total energy released by the total number of nucleons in the product nucleus $({ }_2 He^4)$,which is $4$.Energy released per nucleon = $23.8 	ext{ MeV} / 4 = 5.95 	ext{ MeV}$.

If the binding energy per nucleon of deuteron $({ }_1 H^2)$ is $1.15 \text{ MeV}$ and an $\alpha$-particle has a binding energy of $7.1 \text{ MeV}$ per nucleon,then the energy released per nucleon in the given reaction is ${ }_1 H^2 + { }_1 H^2 \rightarrow { }_2 He^4 + Q$. (in $\text{ MeV}$)

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