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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 :
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${ }_{92} X^{236} \rightarrow{ }_{56} Y^{141}+{ }_{36} Z^{92}+3 R$
The identity of emitted particles ( $\mathrm{R})$ is :
Solution:
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Verified Answer
The correct answer is:
Neutron
$\begin{aligned}
& \mathrm{Z} \text { in LHS }=92 \\
& \mathrm{Z} \text { in } \mathrm{RHS}=56+36=92 \\
& \mathrm{~A} \text { in } \mathrm{LHS}=236 \\
& \mathrm{~A} \text { in } \mathrm{RHS}=141+92=233
\end{aligned}$
So 3 neutrons are released.
& \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.
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