It is proposed to use nuclear fusion reaction;

²H_{1 +} ²H_{1 ------->} ⁴He₂

in a nuclear reactor 200 MW rating. If the energy from the above reaction is used with a 25 % efficiency in the reactor, how many grams of deuterium fuel will be needed per day? ( The masses of ²H₁ and ⁴He₂ are 2.0141 atomic mass units and 4.0026 atomic mass units respectively)

Mass defect in the given nuclear reaction:

                Dm = 2(mass of deuterium) – (mass of helium)

                        = 2(2.0141) – (4.0026)

                        = 0.0256

        Therefore, energy released

                DE = (Dm) (931.48) MeV = 23.85 MeV

                = 23.85 × 1.6 × 10^–13 J

                = 3.82 × 10^–12 J

        Efficiency is only 25%, therefore,

                25% of ?E = (25/100) (3.82 × 10^–12) J = 9.55 × 10^–13 J

i.e., by the fusion of two deuterium nuclei, 9.55 × 10^–13 J energy is available to the nuclear reactor.

Total energy required in one day to run the reactor with a given power of 200 MW:

                E_Total = 200 × 10⁶ × 24 × 3600 = 1.728 × 10¹³ J

        \ Total number of deuterium nuclei required for this purpose:

                n = ^Etotal/?E/2 = 2 x 1.728 x 10¹³ / 9.55 x 10¹³= 0.326 × 10²⁶

        \ Mass of deuterium required

                = (Number of gm-moles of deuterium required) × 2 g

                = (0.362 x 10¹⁶)/(6.02 x 10²³)× 2 = 120.26 g.