_{1}^{2} H + _{1}^{3} H to _{2}^{4} He + _{0} | vedclass

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Question

(B) 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

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$c - a - b$

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(B) 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.Reactants: $_{1}^{2} H$ (binding energy $= a$) and $_{1}^{3} H$ (binding energy $= b$).Total binding energy of reactants $= a + b$.Products: $_{2}^{4} He$ (binding energy $= c$) and $_{0}^{1} n$ (binding energy $= 0$ as it is a single nucleon).Total binding energy of products $= c + 0 = c$.Energy released $Q = (	ext{Total binding energy of products}) - (	ext{Total binding energy of reactants})$$Q = c - (a + b) = c - a - b$.

$_{1}^{2} H + _{1}^{3} H \to _{2}^{4} He + _{0}^{1} n$
If the binding energies of $_{1}^{2} H, _{1}^{3} H$ and $_{2}^{4} He$ are $a, b$ and $c$ (in $MeV$) respectively, then the energy released in the reaction is:

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$_{1}^{2} H + _{1}^{3} H \to _{2}^{4} He + _{0}^{1} n$If the binding energies of $_{1}^{2} H, _{1}^{3} H$ and $_{2}^{4} He$ are $a, b$ and $c$ (in $MeV$) respectively, then the energy released in the reaction is: