Q75 / 90 · JEE Main 2024 (08 Apr Shift 2) · Physics

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

  • A
    Electron
  • Correct answer
    B
    Neutron
  • C
    $\gamma$-radiations
  • D
    Proton

Solution

$\begin{aligned} & \mathrm{Z} \text { in LHS }=92 \\ & \mathrm{Z} \text { in } \mathrm{RHS}=56+36=92 \\ & \mathrm{~A} \text { in } \mathrm{LHS}=236 \\ & \mathrm{~A} \text { in } \mathrm{RHS}=141+92=233 \end{aligned}$
So 3 neutrons are released.

Hence, the correct answer is option: 'B'.

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