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The average energy released by the fission of a single atom of uranium-235 is approximately 205 MeV. How much total energy would be released by the complete fission of 4.55 kg of \(^ {235}U\)?

Express your answer in units of Joules.

Answer :

Final answer:

To find the total energy released from 4.55 kg of U-235, multiply the number of atoms by the energy per atom and convert MeV to Joules. The total energy released is approximately 3.824 × 10^13 J.

Explanation:

To calculate the total energy released by the complete fission of 4.55 kg of uranium-235 (U-235), we need to find the number of atoms in the given mass and multiply it by the energy released per atom.

The atomic mass of U-235 is approximately 235.04 g/mol, so 4.55 kg is equal to 4550 g or 19.35 moles (4550 g / 235.04 g/mol).

Utilizing Avogadro's number (6.02 × 10^23 atoms/mol), we find the number of atoms in 4.55 kg of U-235 to be about 1.165 × 10^25 atoms.

Multiplying this by the energy released per fission results in:

Total energy (MeV) = 1.165 × 10^25 atoms × 205 MeV/atom

To convert the energy to Joules, we use the conversion factor, 1 eV = 1.602 × 10^−16 J, so:

Total energy (J) = Total energy (MeV) × 1.602 × 10^−16 J/MeV

Therefore, the total energy released is approximately 3.824 × 10^13 J.

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