Question

If $200MeV$ 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 1eV=$1.6\times {10}^{-19}$ $J$ )

• 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)

$3.125\times {10}^{13}$

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

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