The fission properties of `._94^239Pu` are very similar to those of `._92^235` U. The average energy released per fission is 180 MeV. If all the atoms in 1 kg of pure `._94^239Pu` undergo fission, then the total energy released in MeV is

Options

A. `4.53xx10^26` MeV
B. `2.21xx10^14` MeV
C. `1xx10^13` MeV
D. `6.33xx10^24` MeV

Explanation

To solve the problem, we need to calculate the total energy released when all the atoms in 1 kg of pure Plutonium-239 undergo fission. Here's a step-by-step breakdown of the solution:

### Step 1: Determine the number of moles of Plutonium-239 in 1 kg
The molar mass of Plutonium-239 (Pu) is approximately 239 g/mol. To find the number of moles in 1 kg (1000 g) of Pu, we use the formula:

\[ \text{Number of moles} = \frac{\text{Total mass (g)}}{\text{Molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles} = \frac{1000 \text{ g}}{239 \text{ g/mol}} \approx 4.18 \text{ moles} \]

### Step 2: Calculate the number of atoms (molecules) of Plutonium-239
To find the total number of atoms, we multiply the number of moles by Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol):

\[ \text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's number} \]

Substituting the values:

\[ \text{Number of atoms} = 4.18 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 2.52 \times 10^{24} \text{ atoms} \]

### Step 3: Calculate the total energy released
Given that the average energy released per fission is 180 MeV, the total energy released when all the atoms undergo fission is:

\[ \text{Total energy released (MeV)} = \text{Number of atoms} \times \text{Energy per fission (MeV)} \]

Substituting the values:

\[ \text{Total energy released (MeV)} = 2.52 \times 10^{24} \text{ atoms} \times 180 \text{ MeV} \]

Calculating this gives:

\[ \text{Total energy released (MeV)} = 4.536 \times 10^{26} \text{ MeV} \]

### Final Answer
The total energy released when all the atoms in 1 kg of pure Plutonium-239 undergo fission is approximately \(4.54 \times 10^{26} \text{ MeV}\).

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