Binding energy per nucleon of { }_1^2 H and { | vedclass

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Question

(D) The energy released in a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy

Vedclass · Accepted Answer

$23.6$

Vedclass · Answer

(D) The energy released in a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactants.Total binding energy of the product ${ }_2^4 He = 	ext{Number of nucleons} 	imes 	ext{Binding energy per nucleon} = 4 	imes 7.0 \ MeV = 28.0 \ MeV$.Total binding energy of the reactants $2 	imes ({ }_1^2 H) = 2 	imes (2 	imes 1.1 \ MeV) = 2 	imes 2.2 \ MeV = 4.4 \ MeV$.Energy released $\Delta E = (28.0 \ MeV) - (4.4 \ MeV) = 23.6 \ MeV$.

Binding energy per nucleon of ${ }_1^2 H$ and ${ }_2^4 He$ are $1.1 \ MeV$ and $7.0 \ MeV$ respectively. Energy released in the process ${ }_1^2 H + { }_1^2 H \rightarrow { }_2^4 He$ is: (in $MeV$)

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