Physics - Modern Physics Question with Solution | TestHub

Let assume power $P=1000\mathrm{W}$ Energy per hour $=1000\times 3600\mathrm{J}$ Energy per fission $=200\mathrm{MeV}$ $=200\times 1.6\times 10^{-13}\mathrm{J}$ $\therefore$ Number of fission per hour $n=\frac{1000\times 3600}{200\times 1.6\times 10^{-13}}$ $\begin{array}{rl}& \text{Number of mole per hour }=\frac{n}{N}\\ & \therefore \text{ Mass per hour }=\frac{n}{N}\times 235\end{array}$ $\begin{array}{rl}& =\frac{1000\times 3600\times 235}{200\times 1.6\times 10^{-13}\times 6.02\times 10^{23}}\\ & =43.9\times 10^{-6}\mathrm{g}\end{array}$ This $43.9\times 10^{-6}\mathrm{g}$ is nearest value of 40 micro gram so option (b) is correct.