Question

What is the energy released by 1 gram of natural Uranium, assuming 200 MeV is released in each fission event and the reasonable isotope \(^{235}\mathrm{U}\) has an abundance of 0.7% by weight in natural Uranium? Choose the correct answer. (Take Avogadro's number, \(N_A = 6.022 \times 10^{23}\ \text{mole}^{-1}\)):

• A. \(5.7 \times 10^{8}\ \text{J}\)
• B. \(7.5 \times 10^{8}\ \text{J}\)
• C. \(5.7 \times 10^{18}\ \text{J}\)
• D. \(5.7 \times 10^{10}\ \text{J}\)

Hint:

Fissile mass = 0.007 g. Atoms = (0.007/235) times N_A. Multiply by 200 MeV, convert MeV to joules.

Answer 1

The Correct Option is A

Start from the energy released per gram of pure $^{235}\mathrm{U}$, then scale by the 0.7% abundance.

One mole of $^{235}\mathrm{U}$ is 235 g and contains $6.022 \times 10^{23}$ atoms. So 1 g of pure $^{235}\mathrm{U}$ contains $6.022 \times 10^{23}/235 = 2.563 \times 10^{21}$ atoms. Each fission gives 200 MeV $= 200 \times 1.602 \times 10^{-13} = 3.204 \times 10^{-11}$ J. Hence 1 g of pure $^{235}\mathrm{U}$ would yield
\[ (2.563 \times 10^{21})(3.204 \times 10^{-11}) \approx 8.21 \times 10^{10}\ \text{J}. \]

But natural uranium is only 0.7% $^{235}\mathrm{U}$ by weight, so 1 g of natural uranium behaves like 0.007 g of the fissile isotope:
\[ E = 0.007 \times 8.21 \times 10^{10} \approx 5.7 \times 10^{8}\ \text{J}. \]
This matches option (A).
\[\boxed{E \approx 5.7 \times 10^{8}\ \text{J}}\]