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In a nuclear fission reaction $$0.01\%$$ of the mass of the nucleus undergoing fission is converted in to energy. Find the energy (E) released in the fission of 20 g of fission material, if the energy released per reaction is $$200\text{ MeV}$$. A square meter of the earth's surface receives 1400 J every second from the sun. Find the area which receives the same energy E from the sun in one second. (Assume normal incidence)
Correct Answer: 0
PLACEHOLDER ANSWER - the question has two answers. The mass converted is $$0.01\%$$ of 20 g, i.e. $$2 \times 10^{-6}\text{ kg}$$, so $$E = mc^2 = 2 \times 10^{-6} \times 9 \times 10^{16} = 1.8 \times 10^{11}\text{ J}$$. The area receiving this energy in one second is $$\frac{1.8 \times 10^{11}}{1400} \approx 1.29 \times 10^8\text{ m}^2$$, about $$129\text{ km}^2$$.
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