Question

The energy released in the fission of 1 kg of $_{92}U^{235}$ is: (energy per fission = 200 MeV)

• A. 5.1 x $10^{26}$ eV
• B. 5.1 x $10^{26}$ J
• C. 8.2 x $10^{13}$ J
• D. 8.2 x $10^{13}$ MeV

Answer 1

Answer: (C)

8.2 x $10^{13}$ J

Answer 2

1. Calculate the number of U-235 atoms in 1 kg:

   • Mass of U-235 = 1 kg = 1000 g.
   • Molar mass of U-235 = 235 g/mol.
   • Avogadro's number ($N_A$) = $6.023 \times 10^{23}$ atoms/mol.
   • Number of moles = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{1000 \text{ g}}{235 \text{ g/mol}}$.
   • Number of atoms = Number of moles $\times N_A = \frac{1000}{235} \times 6.023 \times 10^{23}$ atoms.

2. Calculate the total energy released in Joules:

   • Energy per fission = 200 MeV.
   • Conversion factor: 1 MeV = $1.602 \times 10^{-13}$ J.
   • Total energy (in Joules) = (Number of atoms) $\times$ (Energy per fission in MeV) $\times$ (Conversion factor to Joules).
   • Total energy (J) = $\left(\frac{1000}{235} \times 6.023 \times 10^{23}\right) \times 200 \times (1.602 \times 10^{-13})$ J.
   • Total energy (J) = $\frac{1000 \times 6.023 \times 200 \times 1.602}{235} \times 10^{(23-13)}$ J.
   • Total energy (J) $\approx \frac{1929220}{235} \times 10^{10}$ J.
   • Total energy (J) $\approx 8209.4 \times 10^{10}$ J.
   • Total energy (J) $\approx 8.2 \times 10^{13}$ J.