Question

Energy released per fission of U-235 is 190 MeV, then total energy released by 47 g of U-235 is \(x \times 10^{23}\) MeV. Find the value of x.

• A. 228
• B. 114
• C. 456
• D. 95

Hint:

This type of problem connects macroscopic quantities (mass) to microscopic events (atomic fission) through the concept of the mole. The calculation is straightforward: Mass \(\rightarrow\) Moles \(\rightarrow\) Number of Atoms \(\rightarrow\) Total Energy.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the energy released from the fission of a single Uranium-235 atom. We need to calculate the total energy released from the fission of a given mass (47 g) of U-235.
Step 2: Key Formula or Approach:
1. Calculate the number of moles of U-235 in the given mass. Number of moles \( = \frac{\text{given mass}}{\text{molar mass}} \).
2. Calculate the total number of atoms using Avogadro's number (\(N_A \approx 6.022 \times 10^{23}\) atoms/mol). Number of atoms = (number of moles) \( \times N_A \).
3. Calculate the total energy released. Total Energy = (Number of atoms) \( \times \) (Energy per atom).
Step 3: Detailed Explanation:
Part A: Moles of U-235
The molar mass of U-235 is approximately 235 g/mol.
\[ \text{Number of moles} = \frac{47 \text{ g}}{235 \text{ g/mol}} = 0.2 \text{ mol} \quad \left( \text{or } \frac{1}{5} \text{ mol} \right) \] Part B: Number of Atoms
For simplicity in calculations common in these exams, we can approximate Avogadro's number to \(6 \times 10^{23}\) atoms/mol.
\[ \text{Number of atoms} = 0.2 \text{ mol} \times (6 \times 10^{23} \text{ atoms/mol}) = 1.2 \times 10^{23} \text{ atoms} \] Part C: Total Energy
The energy released per atom (per fission) is 190 MeV.
\[ \text{Total Energy} = (1.2 \times 10^{23} \text{ atoms}) \times (190 \text{ MeV/atom}) \] \[ \text{Total Energy} = (1.2 \times 190) \times 10^{23} \text{ MeV} \] \[ \text{Total Energy} = 228 \times 10^{23} \text{ MeV} \] Step 4: Final Answer:
The total energy released is \(228 \times 10^{23}\) MeV. Comparing this with the given expression \(x \times 10^{23}\) MeV, we find that \(x = 228\).