On fission of $$\text{U}^{235}$$ nucleus $$200\text{ MeV}$$ of energy is released. Assume $$50\%$$ efficiency and find the number of $$\text{U}^{235}$$ nuclei undergoing fission in 1 hour in a nuclear power plant of $$200\text{ MW}$$. Also find the mass converted in to energy in this time. ($$1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$$, velocity of light $$c = 3 \times 10^8\text{ m/s}$$)

PLACEHOLDER ANSWER - the question has two answers. At $$50\%$$ efficiency the fission energy released per hour is $$2 \times 200 \times 10^6 \times 3600 = 1.44 \times 10^{12}\text{ J}$$. Each fission gives $$200 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-11}\text{ J}$$, so the number of fissions is $$\frac{1.44 \times 10^{12}}{3.2 \times 10^{-11}} = 4.5 \times 10^{22}$$. The mass converted is $$\frac{E}{c^2} = \frac{1.44 \times 10^{12}}{9 \times 10^{16}} = 1.6 \times 10^{-5}\text{ kg}$$, i.e. 16 mg.