Question

200 Mev of energy may be obtained per fission of $U^{235}$. A reactor is generating 1000 KW of power the rate of nuclear fission in the reactor is:

• A. 1000
• B. $2\times {10}^8$
• C. $3.125\times {10}^{16}$
• D. 931

Answer 1

The correct option is C $3.125\times {10}^{16}$

C

The energy released in every fission of $U^{236}$ is given in the question.

We have $P=n\frac{E}{t}$

Hence, substituting the values in the formula, we get

$\frac{n}{t}=\frac{P}{E}=1000\times {10}^3/200[1.6\times {10}^{13}]$

$=3.125{\times }^{16}$