Calculate the energy released by `1 g` of natural uranium assuming `200 MeV` is released in eaech fission event and that the fissionable isotope `U^235` has an abundance of 0.7% by weight in natural uranium.

Answer

A

Explanation

235 gm uranium contains `6.02 xx 10^(23)`
nuclei
So, `(6.02 xx 10^(23))/235 = 2.56 xx 10^(21) nuclei`
So total energy released
`= 2.56 xx 10^(21) xx 200 MeV`
`= 5.12 xx 10^(23) MeV`
According to energy released
`= 5.12 xx 10^(23) xx 1.6 xx 10^(-9)`
`xx 10^(6) xx 0.7/100 joule`
`=5.7 xx 10^8J`.