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JEE Mains · Physics · STD 12 - 13. Nuclei

In a hypothetical fission reaction \({ }_{92} \mathrm{X}^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} \mathrm{Z}^{92}+3 \mathrm{R}\) The identity of emitted particles \((R)\) is _______.

  1. A Proton
  2. B  Electron
  3. C Neutron
  4. D \(\gamma\)-radiations
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Answer & Solution

Correct Answer

(C) Neutron

Step-by-step Solution

Detailed explanation

\(\mathrm{Z} \text { in LHS }=92\) \(\mathrm{Z} \text { in } \mathrm{RHS}=56+36=92\) \(\mathrm{~A} \text { in } \mathrm{LHS}=236\) \(\mathrm{~A} \text { in } \mathrm{RHS}=141+92=233\) So \(3\) neutrons are released.
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