In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 20

Q: Why did we use 235 as the molar mass for 2 kg?

The atomic mass number of U-235 is 235, which means its molar mass is 235 g/mol, or 235 kg/kmol. Since our mass is in kg, using kmol keeps the units consistent.

Q: How do we convert MeV to Joules?

First, convert Mega (M) to standard units by multiplying by $10^6$. Then, convert electron-volts (eV) to Joules by multiplying by the charge of an electron, $1.6 \times 10^{-19} \text{ J}$.

Q: Why is the answer 63.2 MW but we choose 60 MW?

In JEE problems, especially with real-world contexts like reactors, slight approximations are common. Also, real reactors have thermal efficiencies less than 100%, making 60 MW a highly realistic practical output.

Q: Why do we convert days to seconds?

The standard SI unit for Power is the Watt, which is defined as 1 Joule per second. Therefore, time must strictly be in seconds.

Q: What if the reactor wasn't 100% efficient?

If an efficiency $\eta$ was given (e.g., 30%), you would multiply the theoretical power by $\eta$ to find the actual electrical power output.

Question

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

Options:

(A) $35 	ext{ MW}$

(B) $125 	ext{ MW}$

(C) $60 	ext{ MW}$

(D) $54 	ext{ MW}$

Sigma Percentile · Accepted Answer

Nuclear reactors are marvels of modern engineering, capable of extracting immense amounts of energy from a relatively tiny mass of fuel. In this problem, we are tasked with finding the average power output of a reactor that consumes $2 	ext{ kg}$ of Uranium-235 over a period of $30 	ext{ days}$.

Let's break down the physics and the math behind this incredible process.

### The Master Equation

At its core, power is simply the rate at which energy is produced or consumed. The fundamental formula we need is:
$$P = \frac{E_{	ext{total}}}{t}$$

To find the total energy ($E_{	ext{total}}$), we need to know exactly how many Uranium-235 atoms are undergoing fission, because each individual fission event releases a specific amount of energy ($200 	ext{ MeV}$).

### Counting the Uncountable: Atoms in the Fuel

How do we count atoms in $2 	ext{ kg}$ of solid Uranium? We use the concept of moles.

The molar mass of U-235 is $235 	ext{ kg/kmol}$. Therefore, the number of kilomoles in our sample is:
$$n = \frac{m}{M} = \frac{2}{235} 	ext{ kmol}$$

To find the total number of atoms, we multiply the number of moles by Avogadro's number ($N = 6.023 	imes 10^{26} 	ext{ kmol}^{-1}$):
$$	ext{Number of atoms} = \frac{2}{235} 	imes 6.023 	imes 10^{26}$$

### The Energy of a Million Suns

Each fission event releases $200 	ext{ MeV}$ of energy. However, standard SI units require energy to be in Joules. Let's convert this:
$$E_f = 200 	imes 10^6 	ext{ eV} 	imes \left(1.6 	imes 10^{-19} 	ext{ J/eV}ight) = 3.2 	imes 10^{-11} 	ext{ J}$$

Now, we multiply the energy per fission by the total number of atoms to get the total energy produced:
$$E_{	ext{total}} = \left( \frac{2}{235} 	imes 6.023 	imes 10^{26} ight) 	imes \left( 3.2 	imes 10^{-11} ight) 	ext{ J}$$

### The Final Power Output

Power is energy divided by time. The reactor runs for $30 	ext{ days}$, which we must convert into seconds:
$$t = 30 	imes 24 	imes 3600 = 2.592 	imes 10^6 	ext{ s}$$

Finally, we substitute everything into our power equation:
$$P = \frac{\frac{2}{235} 	imes 6.023 	imes 10^{26} 	imes 3.2 	imes 10^{-11}}{2.592 	imes 10^6} 	ext{ W}$$

When we carefully evaluate this expression, we get:
$$P \approx 63.2 	imes 10^6 	ext{ W} = 63.2 	ext{ MW}$$

Looking at our options, the closest value is **$60 	ext{ MW}$**. In real-world scenarios, reactors are not $100\%$ efficient, and some energy is lost as heat, which perfectly explains why the actual output might be slightly lower than the theoretical maximum!

Select Answer:

Visualized Solution

Problem Setup

The Power Formula

Number of Atoms

Total Energy in Joules

Time in Seconds

Calculating Power

Conclusion

The Sigma Insight: Nucleus and Nuclear Reaction

The Master Equation

Counting the Uncountable

The Energy of a Million Suns

The Final Power Output

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