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प्रश्न
Calculate the energy released in the following nuclear fusion reaction:
\[\ce{^2_1H + ^2_1H -> ^4_2He + energy}\]
Mass of \[\ce{^2_1H = 2.014102 u}\]
Mass of \[\ce{^4_2He = 4.002604 u}\]
उत्तर
\[\ce{^2_1H + ^2_1H -> ^4_2He + energy}\]
Mass of \[\ce{^2_1H + ^2_1H = 2 × (2.014102)u}\]
= 4.028204u
Mass defect (Δm) = (4.028204 – 4.002604)u
= 0.0256u
∴ Energy released = (Δm) × c2
= 0.0256 × 931 MeV
= 23.8336 MeV (1u × c2 = 931 MeV)
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