Question

200 MeV of energy may be obtained per fission of $U^{235}$. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is.

• A. 1000
• B. $2 \times 10^{8}$
• C. $3.125 \times 10^{16}$
• D. 931

Answer 1

Answer: (C)

$3.125 \times 10^{16}$

Answer 2

Power $\Rightarrow \left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{t/48} \Rightarrow t = 192\mspace{6mu} hour.$

Rate of nuclear fission

$= \frac{10^{6}}{200 \times 1.6 \times 10^{- 13}}$= 3.125 × 10¹⁶.