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JEE Mains · Physics · STD 12 - 13. Nuclei

नाभिकीय विखंडन \({ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}\) के लिए विघटन ऊर्जा \(Q\) _______ \(\mathrm{MeV}\) है। दिए गए परमाणु द्रव्यमान: \({ }^{235} \mathrm{U}: 235.0439 \mathrm{u},{ }^{140} \mathrm{Ce} ; 139.9054 \mathrm{u},\) \({ }^{94} \mathrm{Zr}: 93.9063 \mathrm{u} ; \mathrm{n}: 1.0086 \mathrm{u},\) \(\mathrm{c}^2=931 \mathrm{MeV} / \mathrm{u}\) का मान दिया गया है।

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

Correct Answer

(A) \(208\)

Step-by-step Solution

Detailed explanation

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