A deuterium nucleus ${}_{1}^{2}H$ combines with a tritium nucleus ${}_{1}^{3}H$ to form a heavier helium nucleus ${}_{2}^{4}{He}$ with the release of a neutron ( ${}_{0}^{1}n$ ). The fusion reaction is represented by the equation: ${}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}{He} + {}_{0}^{1}n$
In this reaction, the mass of ${}_{2}^{4}{He} +$ mass of ${}_{0}^{1}n$ is:

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greater than the mass of ${}_{1}^{2}H +$ mass of ${}_{1}^{3}H$ .

the same as the mass of ${}_{1}^{2}H +$ mass of ${}_{1}^{3}H$ .

twice the mass of ${}_{1}^{2}H +$ mass of ${}_{1}^{3}H$ .

less than the mass of ${}_{1}^{2}H +$ mass of ${}_{1}^{3}H$ .

answer is A.

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Detailed Solution

When deuterium nucleus combines with tritium nucleus to form helium nucleus with a release of a neutron, the energy is released. The rest mass energy of products is smaller than rest mass energy of reactant. This difference of energy is due to mass difference of reactants and products in accordance with Einstein's mass-energy equivalence E = mc².

Hence, the sum of masses of products is less than sum of masses of reactants.

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