Question:
The energy released by $2.35\,g$ of $^{235}U$ by fission in a nuclear reactor (in $MeV$) is (Average energy released per fission is $200\,MeV$):
The energy released by $2.35\,g$ of $^{235}U$ by fission in a nuclear reactor (in $MeV$) is (Average energy released per fission is $200\,MeV$):
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Always convert mass to number of nuclei using Avogadro number.
Updated On: Apr 24, 2026
- $1.2 \times 10^{24}$
- $0.4 \times 10^{24}$
- $0.6 \times 10^{24}$
- $0.8 \times 10^{24}$
- $2.4 \times 10^{24}$
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The Correct Option is A
Solution and Explanation
Concept:
• Number of atoms: \[ N = \frac{m}{M} N_A \]
• Total energy: \[ E = N \times 200\,MeV \]
Step 1: Number of atoms
\[ N = \frac{2.35}{235} \times 6.022\times10^{23} = 0.01 \times 6.022\times10^{23} \approx 6\times10^{21} \]
Step 2: Total energy
\[ E = 6\times10^{21} \times 200 = 1.2\times10^{24}\,MeV \] Final Conclusion:
Option (A)
• Number of atoms: \[ N = \frac{m}{M} N_A \]
• Total energy: \[ E = N \times 200\,MeV \]
Step 1: Number of atoms
\[ N = \frac{2.35}{235} \times 6.022\times10^{23} = 0.01 \times 6.022\times10^{23} \approx 6\times10^{21} \]
Step 2: Total energy
\[ E = 6\times10^{21} \times 200 = 1.2\times10^{24}\,MeV \] Final Conclusion:
Option (A)
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