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

We are given that the entire mass of $$\,2 \text{ kg}$$ of uranium-235 is consumed in $$30 \text{ days}$$. In a nuclear reactor the energy is obtained because every nucleus that fissions releases energy, so our first task is to count how many nuclei there are in the 2 kg of fuel.

The molar (more precisely, kilomolar) mass of $$U^{235}$$ is numerically $$235 \text{ kg kmol}^{-1}$$, because $$1 \text{ kmol}$$ contains $$1000 \text{ mol}$$ and each mole weighs $$235 \text{ g}$$.

Using the relation

$$\text{number of kilomoles} = \frac{\text{mass (kg)}}{\text{kilomolar mass (kg kmol}^{-1})}$$

we have

$$n = \frac{2 \text{ kg}}{235 \text{ kg kmol}^{-1}} = 0.00851064 \text{ kmol}.$$

Avogadro’s constant is supplied as $$N = 6.023 \times 10^{26} \text{ nuclei kmol}^{-1}$$. Hence the number of nuclei (and therefore the number of fissions) is

$$N_{\text{fission}} = nN = 0.00851064 \times 6.023 \times 10^{26}$$

$$\phantom{N_{\text{fission}}} = 5.126 \times 10^{24} \text{ nuclei}.$$

Each fission of $$U^{235}$$ liberates $$200 \text{ MeV}$$. To convert this to joules we write the relation

$$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}.$$

Therefore

$$E_{\text{per\,fission}} = 200 \times 10^{6} \text{ eV} \times 1.6 \times 10^{-19} \text{ J eV}^{-1} = 3.2 \times 10^{-11} \text{ J}.$$

The total energy released in the 30-day period is consequently

$$E_{\text{total}} = N_{\text{fission}} \times E_{\text{per\,fission}} = 5.126 \times 10^{24} \times 3.2 \times 10^{-11} \text{ J}.$$

Multiplying the mantissas and adding the exponents we obtain

$$E_{\text{total}} = 1.64032 \times 10^{14} \text{ J}.$$

Power is energy divided by time. First we convert the operating time of 30 days to seconds:

$$t = 30 \text{ days} \times 24 \text{ h day}^{-1} \times 3600 \text{ s h}^{-1} = 2\,592\,000 \text{ s} = 2.592 \times 10^{6} \text{ s}.$$

Now, using the formula $$P = \dfrac{E}{t}$$, the average power is

$$P = \frac{1.64032 \times 10^{14} \text{ J}} {2.592 \times 10^{6} \text{ s}} = 0.6329 \times 10^{8} \text{ W} = 6.329 \times 10^{7} \text{ W}.$$

Since $$1 \text{ MW} = 10^{6} \text{ W}$$, we translate this to

$$P \approx 63 \text{ MW}.$$

The closest value among the given choices is $$60 \text{ MW}$$.

Hence, the correct answer is Option B.