Energy released when two deuterons left({ }_1 | vedclass

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(C) The nuclear fusion reaction is given by: ${ }_1 H^2 + { }_1 H^2 ightarrow { }_2 He^4$.Total binding energy of reactants: $2 	imes (2 	imes 1.1

Vedclass · Accepted Answer

$23.6$

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(C) The nuclear fusion reaction is given by: ${ }_1 H^2 + { }_1 H^2 ightarrow { }_2 He^4$.Total binding energy of reactants: $2 	imes (2 	imes 1.1 \ 	ext{MeV}) = 4.4 \ 	ext{MeV}$.Total binding energy of the product: $4 	imes 7.0 \ 	ext{MeV} = 28.0 \ 	ext{MeV}$.The energy released ($Q$-value) is the difference between the total binding energy of the product and the total binding energy of the reactants: $Q = BE_{	ext{product}} - BE_{	ext{reactants}}$.$Q = 28.0 \ 	ext{MeV} - 4.4 \ 	ext{MeV} = 23.6 \ 	ext{MeV}$.

Energy released when two deuterons $\left({ }_1 H ^2\right)$ fuse to form a helium nucleus $\left({ }_2 He ^4\right)$ is $:$
(Given $:$ Binding energy per nucleon of ${ }_1 H ^2=1.1 \ \text{MeV}$ and binding energy per nucleon of ${ }_2 He ^4=7.0 \ \text{MeV}$) (in $\text{MeV}$)

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Energy released when two deuterons $\left({ }_1 H ^2\right)$ fuse to form a helium nucleus $\left({ }_2 He ^4\right)$ is $:$(Given $:$ Binding energy per nucleon of ${ }_1 H ^2=1.1 \ \text{MeV}$ and binding energy per nucleon of ${ }_2 He ^4=7.0 \ \text{MeV}$) (in $\text{MeV}$)