The binding energies per nucleon for a deuter | vedclass

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(B) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.For the reactant side, we hav

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$4(x_2 - x_1)$

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(B) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.For the reactant side, we have two deuterons $(_{1}H^{2})$. Each deuteron has $2$ nucleons. The binding energy of one deuteron is $2x_1$. Since there are two deuterons, the total binding energy of the reactants is $2 	imes (2x_1) = 4x_1$.For the product side, we have one $\alpha$-particle $(_{2}He^{4})$. An $\alpha$-particle has $4$ nucleons. The binding energy of the $\alpha$-particle is $4x_2$.The energy released $Q$ in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants.Therefore, $Q = (	ext{Total Binding Energy of Products}) - (	ext{Total Binding Energy of Reactants})$.$Q = 4x_2 - 4x_1 = 4(x_2 - x_1)$.

The binding energies per nucleon for a deuteron $(_{1}H^{2})$ and an $\alpha$-particle $(_{2}He^{4})$ are $x_1$ and $x_2$ respectively. What will be the energy $Q$ released in the reaction $_{1}H^{2} + _{1}H^{2} \to _{2}He^{4} + Q$?

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