What is the energy released by fassion of 1 g of `U^(235)`? (Assume 200 Me V energy is liberated on fission of 1 nucleus)

Options

A. `2.26xx10^(4)kWh`
B. `1.2xx10^(4)kWh`
C. `4.5xx10^(2)kWh`
D. `2.5xx10^(2)` k Wh

Explanation

To find the energy released by the fission of 1 gram of Uranium-235 (U-235), we can follow these steps:

### Step 1: Understand the energy released per fission
We know that the energy released by the fission of one nucleus of U-235 is given as 200 MeV (Mega electron volts).

### Step 2: Convert the mass of U-235 to moles
The molar mass of U-235 is approximately 235 g/mol. Therefore, for 1 gram of U-235, we can calculate the number of moles:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{235 \text{ g/mol}} \approx 0.004255 \text{ mol} \]

### Step 3: Calculate the number of U-235 nuclei
Using Avogadro's number, which is approximately \(6.022 \times 10^{23}\) nuclei/mol, we can find the number of U-235 nuclei in 1 gram:

\[ \text{Number of nuclei} = \text{Number of moles} \times \text{Avogadro's number} = 0.004255 \text{ mol} \times 6.022 \times 10^{23} \text{ nuclei/mol} \approx 2.56 \times 10^{21} \text{ nuclei} \]

### Step 4: Calculate the total energy released
Now, we can calculate the total energy released by the fission of all these nuclei:

\[ \text{Total energy} = \text{Number of nuclei} \times \text{Energy per fission} = 2.56 \times 10^{21} \text{ nuclei} \times 200 \text{ MeV} \]

Converting 200 MeV to joules (1 MeV = \(1.6 \times 10^{-13}\) J):

\[ \text{Energy per fission in joules} = 200 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 3.2 \times 10^{-11} \text{ J} \]

Now, substituting this back into the total energy equation:

\[ \text{Total energy} = 2.56 \times 10^{21} \text{ nuclei} \times 3.2 \times 10^{-11} \text{ J} \approx 8.19 \times 10^{10} \text{ J} \]

### Step 5: Convert energy to kilowatt-hours
To convert joules to kilowatt-hours, we use the conversion factor \(1 \text{ kWh} = 3.6 \times 10^{6} \text{ J}\):

\[ \text{Total energy in kWh} = \frac{8.19 \times 10^{10} \text{ J}}{3.6 \times 10^{6} \text{ J/kWh}} \approx 22700 \text{ kWh} \]

### Final Answer
The energy released by the fission of 1 gram of U-235 is approximately **2.27 \times 10^{4} kWh**.

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