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

Step 1: Only the \(^{235}\mathrm{U}\) fraction fissions. In 1 g of natural uranium the fissile mass is
\[ m = 0.007 \times 1\ \text{g} = 7 \times 10^{-3}\ \text{g}. \]

Step 2: Number of \(^{235}\mathrm{U}\) atoms, using molar mass \(235\ \text{g/mol}\):
\[ N = \frac{m}{235} \times N_A = \frac{7 \times 10^{-3}}{235} \times 6.022 \times 10^{23} \approx 1.79 \times 10^{19}. \]

Step 3: Total energy in MeV, with 200 MeV per fission:
\[ E = N \times 200\ \text{MeV} = 1.79 \times 10^{19} \times 200 = 3.59 \times 10^{21}\ \text{MeV}. \]

Step 4: Convert to joules using \(1\ \text{MeV} = 1.602 \times 10^{-13}\ \text{J}\):
\[ E = 3.59 \times 10^{21} \times 1.602 \times 10^{-13} \approx 5.7 \times 10^{8}\ \text{J}. \]
\[\boxed{E \approx 5.7 \times 10^{8}\ \text{J}}\]