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

To calculate the energy released in the fission of 0.1 kg of \(^{235}U\), we need to follow these steps:

1. First, determine the number of uranium nuclei in 0.1 kg of \(^{235}U\). The molar mass of \(^{235}U\) is 235 g/mol. Therefore, the number of moles of \(^{235}U\) in 0.1 kg (which is 100 g) is given by: \[\text{Number of moles} = \frac{100 \, \text{g}}{235 \, \text{g/mol}} = 0.4255 \, \text{mol}\]
2. Next, calculate the number of nuclei, using Avogadro's number (\(6.022 \times 10^{23}\) nuclei/mol): \[\text{Number of nuclei} = 0.4255 \, \text{mol} \times 6.022 \times 10^{23} \, \text{nuclei/mol} = 2.562 \times 10^{23} \, \text{nuclei}\]
3. Since each nucleus releases 200 MeV, the total energy released is: \[E = 2.562 \times 10^{23} \times 200 \, \text{MeV} = 5.124 \times 10^{25} \, \text{MeV}\]
4. Convert this energy into joules, knowing that 1 MeV = \(1.602 \times 10^{-13}\) J: \[E = 5.124 \times 10^{25} \times 1.602 \times 10^{-13} \, \text{J} = 8.207 \times 10^{12} \, \text{J}\]
5. Since 1 kWh = \(3.6 \times 10^6\) J, convert joules to kilowatt-hour: \[E = \frac{8.207 \times 10^{12} \, \text{J}}{3.6 \times 10^{6} \, \text{J/kWh}} = 2.279 \times 10^6 \, \text{kWh} \approx 22.8 \times 10^5 \, \text{kWh}\]

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

Answer 2

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 \( 22.8 \times 10^5 \, {kWh} \).