Question

Fission of one nucleus uranium-235 yields an average of approximately 200 MeV. What mass of uranium produces the energy of 3.30 x 10^10 J in fission?

Answer 1

If one nucleus yields $200MeV=3.2\cdot10^{-11}J$, then we need $N=\frac{3.3\cdot 10^{10}}{3.2\cdot10^{-11}}\approx 1.03\cdot10^{21}$ of nuclei. The molar mass of U-235 is $235 g/mol$, where $1 mol$ contains $6.02\cdot10^{23}$ atoms/molecules. Therefore the mass of $N$ U-235 nuclei (the mass of a nucleus is approximately the mass of an atom, as the electron mass is neglected) is $235\cdot\frac{1.03\cdot10^{21}}{6.02\cdot10^{23}}\approx 0.4 g$.