In the reaction ₁² H + ₁³ H → ₂⁴ He + ₀¹ n, if the binding energies of ₁² H, ₁³ H and ₂⁴ He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is (a) a + b + c (b) a + b - c (c) c - a - b (d) c + a - b

Answer: (c). Energy released = (total BE of products) - (total BE of reactants). The neutron has zero binding energy, so Q = c - (a + b). So the answer is (c) c - a - b.

In the reaction ₁² H + ₁³ H → ₂⁴ He + ₀¹ n, if the binding energies… | NEET 2005 (AIPMT)

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Nuclear Reactions & Q-value

In the reaction ${}_{1}^{2}$ H + ${}_{1}^{3}$ H $\rightarrow {}_{2}^{4}$ He + ${}_{0}^{1}$ n, if the binding

energies of ${}_{1}^{2}$ H, ${}_{1}^{3}$ H and ${}_{2}^{4}$ He are respectively a, b and c (in MeV),

then the energy (in MeV) released in this reaction is

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Answer: (c)

Energy released $=$ (total BE of products) $-$ (total BE of reactants).
The neutron has zero binding energy, so $Q = c - (a + b)$ .
So the answer is (c) $c - a - b$ .

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