Question

Energy released in the fission of a single ${}_{92}{\text{U}}^{235}$ nucleus is $200\text{MeV}$. The fission rate of a ${}_{92}{\text{U}}^{235}$ fuel reactor operating at a power level of $5\text{W}$ is $1.56\times {10}^{5.5k}{\text{s}}^{-1}$. The value of $k$ is .(Integer only)

Answer 1

$\text{power}P=\frac{n{E}_1}{t}$

Where,
$n\to \text{Number of fission reaction}$
$E_1\to \text{Energy released in a single fission reaction}$

$\text{Fission rate}\frac{n}{t}=\frac{P}{E_1}$

We know that, $1\text{MeV}=1.6\times {10}^{-13}\text{J}$

$\Rightarrow \frac{n}{t}=\frac{5}{200\times 1.6\times {10}^{-13}}$

$\Rightarrow \frac{n}{t}=1.56\times {10}^{11}{\text{s}}^{-1}$

Given, $\frac{n}{t}=1.56\times {10}^{5.5k}{\text{s}}^{-1}$

$\therefore k=2$