Question

If $200MeV$ energy is released in the fission of a single nucleus of ${}_{92}{U}^{235}$ . How many fissions must occur per second to produce a power of $1kW$ ?

• A. $3.125\times 10^{13}$
• B. $6.250\times 10^{13}$
• C. $1.525\times 10^{13}$
• D. None of these

Answer 1

Answer: (A)

We know that $1kW=1\times 10^{3}J{s}^{-1}$
Also, $1.6\times 10^{-9}J=1eV$
$\therefore 200MeV=200\times 1.6\times 10^{-19}\times 10^{6}J$
Number of fissions = $\frac{\text{Power}}{\text{Energy released}}$
$=\frac{10^3}{200\times 1.6\times 10^{-13}}=3.125\times 10^{13}$