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

Energy yield $= \frac{3.30 \times 10^{10} \;J}{1.6 \times 10^{-13}\;J/MeV}$

$= 2.0625 \times 10^{23}\;MeV$

Because A = 235 for uranium, one mole of this isotope has a mass of 235 g.

Let us take 1000 kg of uranium and we will calculate what amount of energy can be produced by this amount of uranium.

The number of nuclei in 1000 kg of uranium

$N = \frac{1.0 \times 10^6 \;g}{235 \;g} \times 6.02 \times 10^{23}\;nuclei \\ = 2.56 \times 10^{27} \; nuclei$

The total energy released when all nuclei undergo fission

$E = NQ \\ = 2.56 \times 10^{27} \;muclei \times 208 \;MeV/nucleus \\ = 5.32 \times 10^{29} \;MeV$

Proportion:

1000 kg of uranium = 5.32 \times 10^{29} \;MeV

x kg of uranium = 2.0625 \times 10^{29} \;MeV

$x = \frac{1000 \;kg \times 2.0625 \times 10^{29} \;MeV}{5.32 \times 10^{29} \;MeV} \\ = 0.387 \times 10^{-3} \;kg \\ = 0.387 \;g$

Answer: 0.387 g