AP EAMCET · PHYSICS · Nuclear Physics
If \(200 \mathrm{MeV}\) of energy is released in the fission of one nucleus of \({ }_{92}^{235} \mathrm{U}\), then the number of nuclei that must undergo fission to release an energy of \(1000 \mathrm{~J}\) is
- A \(3.125 \times 10^{13}\)
- B \(6.25 \times 10^{13}\)
- C \(12.5 \times 10^{13}\)
- D \(3.125 \times 10^{14}\)
Answer & Solution
Correct Answer
(A) \(3.125 \times 10^{13}\)
Step-by-step Solution
Detailed explanation
\begin{array}{llll}\text {Number of nuclei } =\frac{E_{\text {total }}}{E_1}=\frac{1000 \mathrm{~J}}{200 \mathrm{MeV}} \\ =\frac{1000 \mathrm{~J}}{200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}} \\ \Rightarrow N =\frac{10^{14}}{3.2} \\ \Rightarrow N =3.125 \times…
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