In each fission of U²³⁵, 200 MeV of energy is released. If a reactor produces 100MW power the rate of fission in it will be

$\mathrm{p}=\frac{\mathrm{nE}}{\mathrm{t}}\Rightarrow 100\times {10}^6=\frac{\mathrm{n}\times 32\times {10}^{-12}}{1}$

${\text{(200MeV=200×1.6×10}}^{-13} \mathrm{J}/\mathrm{s})$
(1min=60 sec)

$\mathrm{n}=\frac{{10}^8}{32\times {10}^{-12}}=3.125\times {10}^{18}/\sec$
$=3.125\times {10}^{18}\times 60/\min ;\therefore \mathrm{n}=1.9\times {10}^{20}/\min$