In a nuclear reactor `^235U` undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 yr, find the total mass of uranium required.

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Calculate the total energy required for 10 years
The power of the reactor is given as 1000 MW, which is equivalent to \(1000 \times 10^6\) watts.

The time for which the reactor operates is 10 years. We need to convert this time into seconds:
\[ \text{Time} = 10 \text{ years} = 10 \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \]
Calculating this gives:
\[ \text{Time} = 10 \times 365 \times 24 \times 3600 = 3.1536 \times 10^{17} \text{ seconds} \]

Now, we can calculate the total energy supplied by the reactor:
\[ \text{Energy} = \text{Power} \times \text{Time} = (1000 \times 10^6) \times (3.1536 \times 10^{17}) = 3.1536 \times 10^{26} \text{ joules} \]

### Step 2: Calculate the actual energy needed considering efficiency
The efficiency of the reactor is given as 10%, or 0.1. Therefore, the actual energy needed is:
\[ \text{Actual Energy} = \frac{\text{Energy}}{\text{Efficiency}} = \frac{3.1536 \times 10^{26}}{0.1} = 3.1536 \times 10^{27} \text{ joules} \]

### Step 3: Energy released per fission of uranium-235
Each fission of \(^{235}U\) releases 200 MeV of energy. We need to convert this to joules:
\[ 200 \text{ MeV} = 200 \times 1.6 \times 10^{-13} \text{ joules} = 3.2 \times 10^{-11} \text{ joules} \]

### Step 4: Calculate the number of uranium atoms needed
To find the number of uranium atoms needed, we divide the actual energy required by the energy released per fission:
\[ \text{Number of Atoms} = \frac{\text{Actual Energy}}{\text{Energy per fission}} = \frac{3.1536 \times 10^{27}}{3.2 \times 10^{-11}} \approx 9.85 \times 10^{37} \text{ atoms} \]

### Step 5: Calculate the number of moles of uranium
Using Avogadro's number (\(6.022 \times 10^{23} \text{ atoms/mole}\)), we can find the number of moles:
\[ \text{Number of Moles} = \frac{\text{Number of Atoms}}{\text{Avogadro's Number}} = \frac{9.85 \times 10^{37}}{6.022 \times 10^{23}} \approx 1.64 \times 10^{14} \text{ moles} \]

### Step 6: Calculate the total mass of uranium required
The molar mass of \(^{235}U\) is approximately 235 g/mol. To find the total mass in kilograms:
\[ \text{Total Mass} = \text{Number of Moles} \times \text{Molar Mass} = 1.64 \times 10^{14} \text{ moles} \times 235 \text{ g/mol} = 3.86 \times 10^{16} \text{ grams} = 3.86 \times 10^{13} \text{ kg} \]

### Final Answer
The total mass of uranium required is approximately \(3.86 \times 10^{13} \text{ kg}\).

---