Question

Calculate the power output of ${}_{92}{\text{U}}^{235}$ reactor, if it takes $30$ days to use up $2\text{kg}$ of fuel, and if each fission gives $185\text{MeV}$ of useable energy.
$(\text{Avogadro number}=6\times {10}^{23})$

• A. $58.3\text{MW}$
• B. $70.3\text{MW}$
• C. $44.3\text{MW}$
• D. $66.3\text{MW}$

Answer 1

The correct option is A $58.3\text{MW}$

Power of the reactor,

$P=\frac{E}{t}$

Here, $t=30\text{days}=30\times 24\times 60\times 60=2.592\times {10}^6\text{s}$

Number of ${\text{U}}^{235}$ atoms in $2\text{kg}$ is,

$n=\frac{m}{M}{N}_A$

Here, $m=2\text{kg}=2000\text{g},M=235\text{g}$

$\Rightarrow n=\frac{6\times {10}^{23}\times 2000}{235}$

$\Rightarrow n=51.06\times {10}^{23}$

Given that, energy released in one fission,

$E_1=185\text{MeV}$

$\Rightarrow E_1=185\times 1.6\times {10}^{-13}\text{J}$

$\therefore$ Total Energy released by $2\text{kg}$ of ${\text{U}}^{235}$ is

$E=n{E}_1$

$\Rightarrow E=51.06\times {10}^{23}\times 185\times 1.6\times {10}^{-13}$

$\therefore E=1.5114\times {10}^{14}\text{J}$

Power of the reactor,

$P=\frac{E}{t}=\frac{1.5114\times {10}^{14}}{2.59\times {10}^6}$

$\Rightarrow P=0.583\times {10}^8=58.3\times {10}^6\text{W}$

$\Rightarrow P=58.3\text{MW}$

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