In a hypothetical fission reaction ${ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R$ The…
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 :
Electron
Neutron
$\gamma$-radiations
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.