TS EAMCET · Physics · Nuclear Physics
In the nuclear fission of one nucleus of \(U^{255}\) the energy released is \(188 \mathrm{MeV}\). The energy released in the nuclear fission of \(235 \mathrm{~g}\) of \(\mathrm{U}^{235}\) is nearly (Avogadro number \(=6.02 \times 10^{23} \mathrm{~mol}^{-1}\) )
- A \(28.8 \times 10^{12} \mathrm{~J}\)
- B \(23.5 \times 10^{12} \mathrm{~J}\)
- C \(36.2 \times 10^{12} \mathrm{~J}\)
- D \(18.11 \times 10^{12} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(18.11 \times 10^{12} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Energy released, is given by \(\begin{aligned} & \Delta \mathrm{E}=188 \times 1.6 \times 10^{-13} \times \frac{235}{235} \times 6.02 \times 10^{23} \\ & =18.11 \times 10^{12} \mathrm{~J} \end{aligned}\)
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