Question

10^14 fissions per second are taking place in a nuclear reactor having efficiency 40%. The energy released per fission is 250 MeV. The power output of the reactant is

• A. 2000W
• B. 4000W
• C. 1600W
• D. 3200W

Answer 1

The correct option is C 1600W

Energy released in ${10}^{14}fissions$ is $E={10}^{14}\times 250Mev=2.5\times {10}^{16}Mev=2.5\times {10}^{16}\times 1.6\times {10}^{-13}Joule$

or $E=4000Joule$

Output energy=$E_0=4000\times \frac{40}{100}Joule=1600Joule$ i.e only $40$ %

Now this, $1600Joule$ is recieved in $1$ second so power$=Enrgy/time=1600Joule/s=1600Watt$

Option C is correct.