Question

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 \[\ce{^4_2He = 4.002604 u}\]

Answer 1

\[\ce{^2_1H + ^2_1H -> ^4_2He + energy}\]

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)