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JEE Mains · Physics · STD 12 - 13. Nuclei

\({ }^{235} U \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+n\) ન્યુક્લિયર વિખંડન પ્રક્રિયા માટ વિખંડન ઊર્જા \(Q\) _______ \(\mathrm{MeV}\) હશે. પરમાણ્વીય દળોઃ \({ }^{235} U: 235.0439 U ;{ }^{140} \mathrm{Ce}: 139.9054 u, { }^{94} \mathrm{Zr}: 93.9063 U ; n: 1.0086 U\) અને \(C^2=931 \mathrm{MeV} / u\) આપેલ છે.

  1. A \(208\)
  2. B \(209\)
  3. C \(210\)
  4. D \(211\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(208\)

Step-by-step Solution

Detailed explanation

\({ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}\) Disintegration energy \(\mathrm{Q} =\left(\mathrm{m}_{\mathrm{k}}-\mathrm{m}_{\mathrm{r}}\right) \mathrm{c}^2\) \(\mathrm{~m}_{\mathrm{n}} =235.0439 \mathrm{u}\)…
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