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

To solve the problem, we need to calculate the total energy released by 47 g of Uranium-235 (U-235) when it undergoes nuclear fission. The energy released per fission of a single U-235 nucleus is given as 190 MeV. 

1. First, we need to determine the number of U-235 nuclei in 47 g of U-235:
   • The molar mass of U-235 is approximately 235 g/mol.
   • Using Avogadro's number, which is \(6.022 \times 10^{23}\) atoms/mol, we can find the number of atoms in 47 g:
   • Number of moles in 47 g of U-235: \(\frac{47}{235}\) mol.
   • Thus, the number of U-235 atoms (or nuclei) is: \(6.022 \times 10^{23} \times \frac{47}{235}\).
2. Now, we calculate the total energy released using the energy per fission:
   • Total energy released (in MeV) = \(190 \times \left(6.022 \times 10^{23} \times \frac{47}{235}\right)\).
3. Simplify the above expression:
   • Calculate the number of fissions: \(= 6.022 \times 10^{23} \times \frac{47}{235} = 1.203 \times 10^{23}\).
   • Then, the total energy released = \(190 \times 1.203 \times 10^{23} = 228.57 \times 10^{23}\) MeV.
4. According to the problem, the total energy released is \(x \times 10^{23}\) MeV. Thus, we find that \(x \approx 228\).

The value of \(x\) is therefore 228.

Option	Value of x
228	Correct
114	Incorrect
456	Incorrect
95	Incorrect

Thus, the correct answer is 228.