Question

If 200 MeV energy is released in the fission of a single $U^{235}$nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given$1\mspace{6mu} eV = 1.6 \times 10^{- 19}J$).

• A. $3.125 \times 10^{13}$
• B. $3.125 \times 10^{14}$
• C. $3.125 \times 10^{15}$
• D. $3.125 \times 10^{16}$

Answer 1

Answer: (A)

$3.125 \times 10^{13}$

Answer 2

$P = n\left( \frac{E}{t} \right) \Rightarrow 1000 = \frac{n \times 200 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$

$\Rightarrow \frac{n}{t} = 3.125 \times 10^{13}.$