Question

If $40\text{\%}$ energy released during fission would be converted into electrical energy, then the number of fission in a second in a nuclear reactor of $6.4\text{MW}$ output is
$[\text{Energy per fission}=200\text{MeV}]$

• A. $5\times {10}^{15}$
• B. $5\times {10}^{16}$
• C. $5\times {10}^{17}$
• D. $5\times {10}^{18}$

Answer 1

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

Energy released is one fission,

$E_1=200\text{MeV}=200\times 1.6\times {10}^{-13}\text{J}$

Energy utilized to generate electrical energy in one fission is

$E_1=\frac{40}{100}\times 200\times 1.6\times {10}^{-13}=1.28\times {10}^{-11}\text{J}$

Total energy per second converted into electrical energy by the nuclear reactor is,

$E_0=\text{Power output}=6.4\text{MW}=6.4\times {10}^6\text{J}$

Thus, number of fission required per second,

$n=\frac{E_0}{E_1}$

$\Rightarrow n=\frac{6.4\times {10}^6}{1.28\times {10}^{-11}}$

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

Hence, option $\left(C\right)$ is correct.

Why this question? To understand, the concept of the total energy released and the number of fission per second required in fission process.