Question

If 200 MeV of energy is released in the fission of one nucleus of $U_{92}^{235}$, how many nuclei must fission per second to produce a power of 1 kW?

• A. $3.125\times {10}^{13}$
• B. $3.125\times {10}^{11}$
• C. $3.215\times {10}^{13}$
• D. $3.215\times {10}^{11}$

Answer 1

The correct option is A $3.125\times {10}^{13}$

Energy released per fission $=200MeV=200\times 1.6\times {10}^{-13}=3.2\times {10}^{-11}J$

Total energy released per second $=1kJ={10}^{3}J$

Therefore number of nuclei fissioned per second:

$n=\frac{{10}^3}{3.2\times {10}^{-11}}=3.125\times {10}^{13}$