1.00 kg of `.^(235)U` undergoes fission process. If energy released per event is `200 MeV`, then the total energy released is

Options

A. `5.12 xx 10^(24)MeV`
B. `6.02 xx 10^(23)MeV`
C. `5.12 xx 10^(16)MeV`
D. `6.02 xx 10^(6)MeV`

Explanation

To solve the problem of calculating the total energy released when 1.00 kg of Uranium-235 undergoes fission, we can follow these steps:

### Step 1: Calculate the number of atoms in 1 kg of Uranium-235
To find the number of atoms in 1 kg of Uranium-235, we use the formula:

\[ \text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \]

Where:
- Mass = 1 kg = 1000 g
- Molar mass of Uranium-235 = 235 g/mol
- \(N_A\) (Avogadro's number) = \(6.022 \times 10^{23} \text{ atoms/mol}\)

Substituting the values:

\[ \text{Number of atoms} = \frac{1000 \text{ g}}{235 \text{ g/mol}} \times 6.022 \times 10^{23} \text{ atoms/mol} \]

Calculating this gives:

\[ \text{Number of atoms} \approx \frac{1000}{235} \times 6.022 \times 10^{23} \approx 2.56 \times 10^{24} \text{ atoms} \]

### Step 2: Calculate the total energy released
The energy released per fission event is given as 200 MeV. To find the total energy released, we multiply the energy per event by the total number of fission events (which is equal to the number of atoms):

\[ \text{Total energy} = \text{Energy per event} \times \text{Number of atoms} \]

Substituting the values:

\[ \text{Total energy} = 200 \text{ MeV} \times 2.56 \times 10^{24} \text{ atoms} \]

Calculating this gives:

\[ \text{Total energy} \approx 5.12 \times 10^{26} \text{ MeV} \]

### Final Result
The total energy released when 1.00 kg of Uranium-235 undergoes fission is approximately:

\[ \text{Total energy} \approx 5.12 \times 10^{26} \text{ MeV} \]