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 reaction (in per second) is

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

Answer 1

Answer: (C)

Energy $=200\, MeV$
$200 \times 1.6 \times 10^{-19} \times 10^{6}$
$=200 \times 1.6 \times 10^{-13}$
Power $P =1000\, kW$
$=1000 \times 10^{3} W$
Rate of fission $=\frac{\text { Power }}{\text { Energy }}$
$=\frac{10^{6}}{200 \times 1.6 \times 10^{-13}}$
$=3.125 \times 10^{16} / s$