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Question

# It is proposed to use the nuclear fusion reaction 1H2+ 1H2→ 2He4 in a nuclear reactor of 200 MW rating. If 25% of energy from the above reaction is used in the reactor. It is found that nearly 12x grams of deuterium will be needed per day. The value of x is (integer only) [Take, mass of 1H2=2.0141 amu,mass of 2He4=4.0026 amu,1amu c2=931.5 MeV]

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Solution

## Energy released in the nuclear fusion reaction is given by, Q=(Δm)c2 Here, Δm=2M(1H2)−M(2He4) ⇒Δm=(2×2.0141 u)−4.00264 u=0.0256 u=0.0256 u×c2 ⇒Q=0.0256×931.5 MeV ⇒Q=23.8464 MeV=23.8464×1.6×10−13 J ⇒Q=38.15424×10−13 J Since, only 25% of energy is used in the reactor, The effective energy used, Q′=25100×Q ⇒Q′=25100×38.15424×10−13 ⇒Q′=9.53856×10−13 J Two deuterium nuclei are involved in a fusion reaction, ∴ The energy released per deuterium E1=9.53856×10−132 J=4.76928×10−13 J For 200 MW power per day, the number of deuterium nuclear required, n=P×tE1=200×106×864004.76928×10−13 [∵1 day=86400 sec] ⇒n=3.62319×1025 We know that, 6×1023 atoms are present in 2 g of deuterium. Then, the mass of deuterium that contains 3.623×1025 atoms is given by, m=2×3.62319×10256×1023 g=120.77 g (or) m≈120 g ∴ x=10

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