TS EAMCET · Physics · Nuclear Physics
\(\mathrm{AU}^{235}\) nuclear reactor generates energy at a rate of \(3.70 \times 10^7 \mathrm{~J} / \mathrm{s}\). Each fission liberates \(185 \mathrm{MeV}\) useful energy. If the reactor has to operate for \(144 \times 10^4 \mathrm{~s}\), then, the mass of the fuel needed is (Assume Avogadro's number \(\left.=6 \times 10^{23} \mathrm{~mol}^{-1}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)\)
- A \(70.5 \mathrm{~kg}\)
- B \(0.705 \mathrm{~kg}\)
- C \(13.1 \mathrm{~kg}\)
- D \(1.31 \mathrm{~kg}\)
Answer & Solution
Correct Answer
(B) \(0.705 \mathrm{~kg}\)
Step-by-step Solution
Detailed explanation
In \(1 \mathrm{~s}\), energy generated is \(3.7 \times 10^7 \mathrm{~J}\) In \(144 \times 10^4 \mathrm{~s}\), energy generated is \(=3.7 \times 10^7 \times 144 \times 10^4 \mathrm{~J}\) Also energy released in one fission is…
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