Question

Assuming that $200MeV$ of energy is released per fission of ${}_{92}{U}^{235}$ atom. Find the number of fission per second ,required to release $1kW$ power.

• 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

The correct option is A $3.125\times {10}^{13}$

$\begin{array}{c}\text{Given,}\\ \text{Energy released per fission}=200\times {10}^{6}eV\\ \text{Power require}=1\times {10}^3\text{watt}\\ \text{Energy in Joule}=200\times {10}^6\times 1.6\times {10}^{-19}J\\ \therefore 1ev=1.6\times {10}^{-19}J\text{oule}\\ =3.2\times {10}^{-11}\text{Joule}\end{array}$

$P=\frac{\text{Energy}}{time}$

$\text{let n : no of fission occur per second}$

$\text{Energy released will be}=n(3.2\times {10}^{-11}J)$

$\begin{array}{cc}{10}^3& =n(3.2\times {10}^{-11})\\ n& =\frac{1}{3.2}\times {10}^{14}\\ & =0.3125\times {10}^{14}\\ n& =3.125\times {10}^{13}\end{array}$