Question

$200\text{MeV}$ energy is released when one nucleus of $U^{235}$ undergoes fission. The number of fissions per second required for producing a power of $16\text{MW}$ is-

• A. $5\times {10}^{17}$
• B. $5\times {10}^{18}$
• C. $5\times {10}^{16}$
• D. $5\times {10}^{19}$

Answer 1

The correct option is A $5\times {10}^{17}$

$\text{Energy released in one fission}=200\text{MeV}$

$=200\times {10}^6\times 1.6\times {10}^{-19}\text{J}$

$=3.2\times {10}^{-11}\text{J}$

Given, $\text{Power}=16\text{MW}=16\times {10}^6\text{W}$

$\text{Number of fissions per second}\left(n\right)=\frac{\text{Total Energy}}{\text{Energy released in one fission}}$

$\frac{N}{t}=\frac{P}{3.2\times {10}^{-11}}$

$=\frac{16\times {10}^6}{3.2\times {10}^{-11}}=\frac{1}{2}\times {10}^{18}$

$=5\times {10}^{17}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.