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 _______.
- A Proton
- B Electron
- C Neutron
- D \(\gamma\)-radiations
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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