The power of a nuclear reactor using 0.2 kg of uranium isotope – 235 per day is 32,000 kW. What part

The power of a nuclear reactor using 0.2 kg of uranium isotope – 235 per day is 32,000 kW. What part of the energy released as a result of nuclear fission is used useful if 200 MeV is released during the fission of one uranium-325 nucleus?

Let’s find the amount of an atom contained in 0.2 kg of uranium:
N = m / M * Na, where m is mass, M is molar mass, Na is Avogadro’s number.
N = 0.2 / 0.235 * 6.67 * 10 ^ 23 = 5.68 * 10 ^ 23
At the same time, energy is released:
E = 5.68 * 10 ^ 23 * 200 = 1.1 * 10 ^ 26 MeV = 1.1 * 10 ^ 26: 1.6 * 10 ^ -13 = 6.9 * 10 ^ 12 J
The useful energy is equal to:
Ep = 3200000 * 24 * 3600 = 2.7 * 10 ^ 12 J
Efficiency = Ep / E * 100% = 39%.