COMEDK · Physics · 29. Nuclear Physics
A nucleus of uranium-235 absorbs a slow neutron and undergoes nuclear fission according to the reaction: \({}^{235}_{92}U + {}^{1}_{0}n \rightarrow {}^{141}_{56}Ba + {}^{92}_{36}Kr + 3{}^{1}_{0}n + Q\)
If the average energy released per fission is 202 MeV, the energy released when 2.35 g of \(U^{235}\) undergoes complete fission is approximately;
[Given \(1eV = 1.6 \times 10^{-19}\ J\), Avogadro number \(= 6.02 \times 10^{23}\)]
- A \(1.945 \times 10^{12}\ J\)
- B \(19.45 \times 10^{12}\ J\)
- C \(19.45 \times 10^{10}\ J\)
- D \(1.945 \times 10^{10}\ J\)
Answer & Solution
Correct Answer
(C) \(19.45 \times 10^{10}\ J\)
Step-by-step Solution
Detailed explanation
Number of moles of \(U^{235} = \dfrac{2.35}{235} = 0.01\) mol Number of atoms of \(U^{235} = 0.01 \times 6.02 \times 10^{23} = 6.02 \times 10^{21}\) Energy released per fission…
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