Question

When a slow neutron is captured by a ${}_{92}^{235}U$ nucleus, a fission energy released is $200MeV$. If power of nuclear reactor is $100W$ then rate of nuclear fission is:

• A. $3.6\times {10}^6{s}^{-1}$
• B. $3.1\times {10}^{12}{s}^{-1}$
• C. $1.8\times {10}^4{s}^{-1}$
• D. $4.1\times {10}^6{s}^{-1}$

Answer 1

The correct option is B $3.1\times {10}^{12}{s}^{-1}$

Energy released per fission is $200MeV=200\times 1.6\times {10}^{-19}\times {10}^{6}J$.

Therefore rate of fission is $\frac{100}{200\times 1.6\times {10}^{-19}\times {10}^6}=3.125\times {10}^{12}$ per second.