Q. A nuclear reactor generated power at 50% efficiency by fission of _{92}U^{235} into equal fragment of _{46}Pd^{116} with the emission of two γ-rays of 5.2 MeV each and three neutrons. The average B.E. per particle of U^{235} and Pd^{116} is 7.2 MeV and 8.2 MeV respectively. The amount of U^{235} consumed per hour to produce 1600

Mw power is

(a) 128 gm

(b) 1.4 kg

(c) 140.5 gm

(d) 281 gm

Ans: (a)

Sol: _{92}U^{235} → 2 _{46}Pd^{116} + 3 _{0}n^{1} + 2 γ

B.E of _{92}U^{235} = 235 × 7.2 = 1692 MeV

B.E of _{46}Pd^{116} = 2(116 × 8.2 ) = 1902 .4 MeV

Energy of 2 γ photon = 2 × 5.2 = 10.4 MeV

Energy released per fission = (1902.4 + 10.4 ) – 1692 = 220 MeV

Number of atoms in m gram of _{92}U^{235} is

$ \displaystyle \frac{m}{235}\times 6.023 \times 10^{23}\times 220 \times 1.6 \times 10^{-13} $

Output power = 1600 MW = 1.6 × 10^{9} J/sec

Input Power = 1.6 × 10^{9}/(50/100) = 3.2 × 10^{9} J/sec

Input energy = 3.2 × 10^{9} × 60 × 60 J

$ \displaystyle \frac{m}{235}\times 6.023 \times 10^{23}\times 220 \times 1.6 \times 10^{-13} = 3.2 \times 10^9 \times 3600 $

m = 128 g