Question

In a reactor, $2\text{kg}$ of ${}_{92}{\text{U}}^{235}$ fuel is fully used up in $30\text{days}$. The energy released per fission is $200\text{MeV}$. Given that the Avogadro number, $N=6.023\times {10}^{26}$$\text{per kilo mole}$ and $1\text{eV}=1.6\times {10}^{–19}\text{J}$. The power output of the reactor is close to:

• A. $35\text{MW}$
• B. $60\text{MW}$
• C. $125\text{MW}$
• D. $54\text{MW}$

Answer 1

The correct option is B $60\text{MW}$

Power output of the reactor,

$P=\frac{\text{energy}}{\text{time}}=\frac{E}{t}$

Given that, Energy released in one fission reaction is,

$E_1=200\text{MeV}=200\times 1.6\times {10}^{-19}\text{MJ}$

The number of atoms present in $2\text{kg}$ of ${}_{92}{\text{U}}^{235}$ is given by,

$N=\frac{2}{235}\times 6.023\times {10}^{26}$

$\therefore \text{Total energy},E=N{E}_1$

$\text{Here,}t=30\text{days}=30\times 24\times 60\times 60\text{s}$

$\Rightarrow P=\frac{2}{235}\times \frac{6.023\times {10}^{26}\times 200\times 1.6\times {10}^{-19}}{30\times 24\times 60\times 60}$

$\Rightarrow P\approx 60\text{MW}$.

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