Energy released when two deuterons \(({{}_{1}\mathrm{H}}^{2})\) fuse to form a helium nucleus \(({{}_{2}\mathrm{He}}^{4}…
Energy released when two deuterons \(({{}_{1}\mathrm{H}}^{2})\) fuse to form a helium nucleus \(({{}_{2}\mathrm{He}}^{4})\) is :
(Given : Binding energy per nucleon of \({{}_{1}\mathrm{H}}^{2}=1.1\mathrm{MeV}\) and binding energy per nucleon of \({{}_{2}\mathrm{He}}^{4}=7.0\mathrm{MeV}\) )
[JEE Main 2025, 2 Apr (Shift 2)]
\(23.6MeV\)
Binding energy per nucleon of deuteron \( \left(^2_1\text{H}\right) = 1.1\, \text{MeV} \)
Binding energy per nucleon of helium-4 \( \left(^4_2\text{He}\right) = 7.0\, \text{MeV} \)
Two deuterons \( \Rightarrow \) total nucleons = 2 × 2 = 4
\[
\text{BE}_{\text{reactants}} = 2 \times (1.1 \times 2) = 4.4\, \text{MeV}
\]
\[
\text{BE}_{\text{product}} = 4 \times 7.0 = 28.0\, \text{MeV}
\]
\(Q={\text{BE}}_{\text{product }}−{\text{BE}}_{\text{reactants }}=28.0−4.4\\ =23.6\text{MeV}\)
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