Question

$200\text{MeV}$ of energy may be obtained per atom, in the nuclear fission reaction of $U^{235}$. A nuclear reactor is generating a constant power of $1000\text{kW}$. The number of atoms undergoing fission per second is,

• A. $3.125\times {10}^{16}$
• B. $6.25\times {10}^{16}$
• C. $1.25\times {10}^{16}$
• D. $2.5\times {10}^{16}$

Answer 1

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

Here, we can write,

$\text{Power}=\frac{\text{Energy produced}}{\text{time}}$

$=\frac{\text{No. of fissions}\times \text{Energy released in one fission}}{\text{time}}$

$=\text{No. of fissions per sec.}\times \text{Energy released in one fission}$

$\Rightarrow 1000\times {10}^3=n\times 200\times {10}^6\times 1.6\times {10}^{-19}$

$\therefore n=3.125\times {10}^{16}\text{per sec.}$

Hence, $\left(A\right)$ is the correct answer.