Question

${2\times 10}^8$ eV energy is released due to the fission of one nucleus of ${}_{92}^{235}U$, then claculate the number of fission which must take place per second in order to produce a power of 1 Kilowatt

• A. $1.562{\times 10}^{13}$
• B. $15.62{\times 10}^8$
• C. ${3.125\times 10}^{13}$
• D. ${3.125\times 10}^8$

Answer 1

Answer: (C)

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

$\begin{array}{c}\text{Given,}\\ 2\times {10}^{8}ev{\text{energy released due to fission of one}}_{92}^{235}U\\ \text{Power =1000 watt}\\ \text{let number of fission take place =N}\\ \text{the Energy release will be}\end{array}$

$N\times 2\times {10}^{8}ev$

$or$

$\begin{array}{c}N\times 2\times {10}^8\times 1.6\times {10}^{-19}J\\ \therefore \text{1 ev}=1.6\times {10}^{-19}Joule\end{array}$

$Now,$

$\text{N}\times 2\times {10}^8\times 1.6\times {10}^{-19}=1000$

$\begin{array}{cc}N& =\frac{1000}{3.2\times {10}^{-11}}\\ N& =3.125\times {10}^{13}\end{array}$