The binding energy of deuteron {}_1^2H is 1.1 | vedclass

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Question

(C) The total binding energy of the initial reactants (two deuterons) is: $2 	imes (2 	imes 1.112 	ext{ MeV}) = 4.448 	ext{ MeV}$.The total bindin

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$23.8$

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(C) The total binding energy of the initial reactants (two deuterons) is: $2 	imes (2 	imes 1.112 	ext{ MeV}) = 4.448 	ext{ MeV}$.The total binding energy of the final product (one $\alpha$-particle) is: $4 	imes 7.047 	ext{ MeV} = 28.188 	ext{ MeV}$.The energy $Q$ released in the reaction is the difference between the total binding energy of the products and the total binding energy of the reactants:$Q = [	ext{Total B.E. of products}] - [	ext{Total B.E. of reactants}]$$Q = 28.188 	ext{ MeV} - 4.448 	ext{ MeV} = 23.74 	ext{ MeV}$.Rounding to the nearest provided option,the energy released is $23.8 	ext{ MeV}$.

The binding energy of deuteron ${}_1^2H$ is $1.112 \text{ MeV}$ per nucleon and an $\alpha$-particle ${}_2^4He$ has a binding energy of $7.047 \text{ MeV}$ per nucleon. Then in the fusion reaction ${}_1^2H + {}_1^2H \to {}_2^4He + Q$,the energy $Q$ released is ........... $\text{MeV}$.

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