The amount of `U^(235)` to be fissioned to operate 10 kW nuclear reactor is

Options

A. `1.2 xx 10^(-5) gm//s`
B. ` 1.2 xx 10^(-7) gm//s`
C. `1.2 xx 10^(-9) gm//s`
D. `1.2 xx 10^( -13) gm//s`

Answer

To determine the amount of Uranium-235 (U-235) required to operate a 10 kW nuclear reactor, we can follow these steps:

### Step 1: Understand the Energy Requirement
The power output of the reactor is given as 10 kW, which is equivalent to 10,000 watts. This means the reactor needs to produce 10,000 joules of energy per second.

### Step 2: Determine the Energy Released per Fission
The energy released per fission of U-235 is approximately 200 MeV (mega-electron volts). To convert this energy into joules, we use the conversion factor:
1 eV = \(1.6 \times 10^{-19}\) joules.
Thus,
\[ 200 \text{ MeV} = 200 \times 10^6 \text{ eV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ joules} = 3.2 \times 10^{-11} \text{ joules}. \]

### Step 3: Calculate the Number of Fissions Required per Second
To find out how many fissions are needed to produce 10 kW, we divide the power requirement by the energy released per fission:
\[ \text{Number of fissions per second} = \frac{\text{Power}}{\text{Energy per fission}} = \frac{10,000 \text{ joules/second}}{3.2 \times 10^{-11} \text{ joules}} \approx 3.125 \times 10^{14} \text{ fissions/second}. \]

### Step 4: Calculate the Number of Atoms Required
To find out how many atoms of U-235 are needed to achieve this number of fissions, we need to consider that each atom of U-235 can undergo one fission. Therefore, the number of U-235 atoms required is also approximately \(3.125 \times 10^{14}\) atoms.

### Step 5: Convert Atoms to Moles
Using Avogadro's number, which is approximately \(6.02 \times 10^{23}\) atoms/mole, we can convert the number of atoms to moles:
\[ \text{Moles of U-235} = \frac{3.125 \times 10^{14} \text{ atoms}}{6.02 \times 10^{23} \text{ atoms/mole}} \approx 5.19 \times 10^{-10} \text{ moles}. \]

### Step 6: Convert Moles to Grams
The molar mass of U-235 is approximately 235 g/mole. Thus, the mass of U-235 required is:
\[ \text{Mass of U-235} = \text{Moles} \times \text{Molar mass} = 5.19 \times 10^{-10} \text{ moles} \times 235 \text{ g/mole} \approx 1.22 \times 10^{-7} \text{ grams}. \]

### Final Answer
The amount of U-235 required to operate a 10 kW nuclear reactor is approximately \(1.22 \times 10^{-7}\) grams per second.

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