Question 10 of 13intermediate🔧 ApplyNumerical3 marks

The fission properties of 94239Pu^{239}_{94}\text{Pu} are very similar to those of 92235U^{235}_{92}\text{U}. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 94239Pu^{239}_{94}\text{Pu} undergo fission?

Correct Answer

The total energy released is approximately 4.54×104.54 \times 10²⁶ MeV.

Exercise: EXERCISES | Q: 13.7 | (Page 17)
For More Understanding

Explanation

This numerical problem requires calculating the total energy released when all atoms in 1 kg of Pu-239 undergo fission. The key is to first determine the number of atoms present in 1 kg of Pu-239 using the atomic mass unit conversion provided in the context, then multiply by the energy released per fission event.

Solution Steps

  1. Step 1: Find the mass of one atom of Pu-239. The atomic mass of Pu-239 is 239 u. From the context, 1 u = 1.6605×101.6605 \times 10⁻²⁷ kg. Therefore, mass of one Pu-239 atom = 239×1.6605239 \times 1.6605 × 10⁻²⁷ kg = 3.97×103.97 \times 10⁻²⁵ kg.

  2. Step 2: Calculate the number of atoms in 1 kg of Pu-239. Number of atoms = Total mass / Mass of one atom = 1 kg / (3.97×103.97 \times 10⁻²⁵ kg) = 2.52×102.52 \times 10²⁴ atoms.

  3. Step 3: Calculate the total energy released. Total energy = Number of atoms × Energy per fission = 2.52×102.52 \times 10²⁴ × 180 MeV = 4.54×104.54 \times 10²⁶ MeV.

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