JEE Mains · Physics · STD 12 - 13. Nuclei
In a reactor, \(2\, kg\) of \({ }_{92} U ^{235}\) fuel is fully used up in \(30\) days. The energy released per fission is \(200\, MeV.\) Given that the Avogadro number, \(N =6.023 \times 10^{26}\) per kilo mole and \(1\, eV =1.6 \times 10^{-19}\, J .\) The power output of the eactor is close to\(.....MW\)
- A \(125\)
- B \(60\)
- C \(35\)
- D \(54\)
Answer & Solution
Correct Answer
(B) \(60\)
Step-by-step Solution
Detailed explanation
Number of uranium atoms in \(2 kg\) \(=\frac{2 \times 6.023 \times 10^{26}}{235}\) energy from one atom is \(200 \times 10^{6}\) e.v. hence total energy from \(2 kg\) uranium \(=\frac{2 \times 6.023 \times 10^{26}}{235} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} J\)…
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