Physics - Modern Physics Question with Solution | TestHub

Now, number of atoms in 1g of$U^{235}$

$=\frac{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{ }\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}}{\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\frac{6.023 \times {10}^{23}}{235}$

Energy released per fission = 200 MeV

Therefore, the energy released on the fission of 1 g of$U^{235}$

$=\frac{6.023 \times {10}^{23}\times 200}{235}=5.126 \times {10}^{23} \mathrm{MeV}$

$=5.126\mathrm{ }\times {10}^{23}\times 1.6\mathrm{ }\times {10}^{-13} \mathrm{J}=8.2\mathrm{ }\times {10}^{10}\mathrm{ }\mathrm{W}\mathrm{ }\mathrm{s}$$\left(\because 1 \mathrm{J}=1\mathrm{ }\mathrm{W}\mathrm{ }\mathrm{s}\right)$

$=\frac{8.2 \times {10}^7}{3600}=2.278 \times {10}^4 \mathrm{kWh}$