Question

If the energy released per fission of a \(^{235}U\) nucleus is 200 MeV, the energy released in the fission of 0.1 kg of \(^{235}U\) in kilowatt-hour is:

• A. \( 22.8 \times 10^5 \, {kWh} \)
• B. \( 22.8 \times 10^7 \, {kWh} \)
• C. \( 11.4 \times 10^5 \, {kWh} \)
• D. \( 820 \times 10^{10} \, {kWh} \)

Hint:

To calculate energy released in fission reactions, multiply the energy per fission event by the number of atoms and then convert the units accordingly (Joules to kilowatt-hours).

Answer 1

The Correct Option is A

We are given that the energy released per fission of \(^{235}U\) is 200 MeV and the mass of the sample is 0.1 kg. 
Step 1: First, let's convert the given energy released per fission into joules. Since \( 1 \, {MeV} = 1.602 \times 10^{-13} \, {J} \), the energy released per fission is:

\( 200 \, {MeV} = 200 \times 1.602 \times 10^{-13} \, {J} = 3.204 \times 10^{-11} \, {J}. \)

Step 2: Now, calculate the number of atoms in 0.1 kg of \(^{235}U\). The atomic mass of \(^{235}U\) is approximately \( 235 \, {g/mol} \), and Avogadro's number is \( 6.022 \times 10^{23} \, {atoms/mol} \). Thus, the number of moles in 0.1 kg of \(^{235}U\) is:

\( {Number of moles} = \frac{0.1 \, {kg}}{235 \, {g/mol}} = \frac{100 \, {g}}{235 \, {g/mol}} \approx 0.4255 \, {mol}. \)

The number of atoms in 0.1 kg of \(^{235}U\) is:

\( {Number of atoms} = 0.4255 \times 6.022 \times 10^{23} \approx 2.56 \times 10^{23} \, {atoms}. \)

Step 3: Now, calculate the total energy released from the fission of all the atoms:

\( {Total energy} = 2.56 \times 10^{23} \times 3.204 \times 10^{-11} \, {J} = 8.2 \times 10^{13} \, {J}. \)

Step 4: Next, we convert the energy from joules to kilowatt-hours. Since \( 1 \, {kWh} = 3.6 \times 10^6 \, {J} \), the energy in kilowatt-hours is:

\( {Energy in kWh} = \frac{8.2 \times 10^{13} \, {J}}{3.6 \times 10^6 \, {J/kWh}} \approx 22.8 \times 10^5 \, {kWh}. \)

Thus, the energy released in the fission of 0.1 kg of \(^{235}U\) is \( 22.8 \times 10^5 \, {kWh} \).