# A nuclear reactor generated power at 50% efficiency by fission of 92U235 into equal fragment of …..

Q. A nuclear reactor generated power at 50% efficiency by fission of 92U235 into equal fragment of 46Pd116 with the emission of two γ-rays of 5.2 MeV each and three neutrons. The average B.E. per particle of U235 and Pd116 is 7.2 MeV and 8.2 MeV respectively. The amount of U235 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: 92U235  → 2 46Pd116 + 3 0n1 + 2 γ

B.E of 92U235 = 235 × 7.2 = 1692 MeV

B.E of 46Pd116 = 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 92U235 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 × 109 J/sec

Input Power = 1.6 × 109/(50/100) = 3.2 × 109 J/sec

Input energy = 3.2 × 109 × 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