Physics - Modern Physics Question with Solution | TestHub

Mass of ${}_{92}^{235}\mathrm{U}$ per second in the reactor is $m=\frac{2\times 10^3}{30\times 24\times 60\times 60}=7.72\times 10^{-4}\mathrm{g}/\sec$ $\therefore$ Number of fissions reaction per second $\begin{array}{rl}& =\frac{6\times 10^{23}}{235}\times m=\frac{6\times 10^{23}\times 7.72\times 10^{-4}}{235}\\ & =1.97\times 10^{18}/\sec \end{array}$ Power of nuclear reactor $=1.97\times 10^{18}\times 185\mathrm{MeV}/\mathrm{s}$ $=1.97\times 10^{18}\times 185\times 10^6\times 1.6\times 10^{-19}\mathrm{J}/\mathrm{s}$. $=58.3\mathrm{MW}$.