The binding energy per nucleon of deuteron ({ | vedclass

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(C) Given:Binding energy per nucleon for ${ }_{1} H^{2} = 1.1 \ MeV$.Binding energy per nucleon for ${ }_{2} He^{4} = 7 \ MeV$.The nuclear fusion reac

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$23.6$

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(C) Given:Binding energy per nucleon for ${ }_{1} H^{2} = 1.1 \ MeV$.Binding energy per nucleon for ${ }_{2} He^{4} = 7 \ MeV$.The nuclear fusion reaction is: ${ }_{1} H^{2} + { }_{1} H^{2} \longrightarrow { }_{2} He^{4} + Q$.Total binding energy of reactants (two deuterons) = $2 	imes (2 	imes 1.1 \ MeV) = 4.4 \ MeV$.Total binding energy of the product (one helium nucleus) = $4 	imes 7 \ MeV = 28 \ MeV$.The energy released $Q$ is the difference between the total binding energy of the product and the reactants:$Q = BE_{	ext{product}} - BE_{	ext{reactants}}$$Q = 28 \ MeV - 4.4 \ MeV = 23.6 \ MeV$.

The binding energy per nucleon of deuteron $({ }_{1} H^{2})$ and the helium atom $({ }_{2} He^{4})$ are $1.1 \ MeV$ and $7 \ MeV$ respectively. If two deuteron nuclei fuse to form a single helium nucleus,then the energy released is: (in $MeV$)

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