Calculate the number of neutrons emitted when $_{92} U^{235}$ undergoes controlled nuclear fission to $_{54} X e^{142}$ and $_{38} S r^{90} .$

(IIT-JEE, 2010)

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Solution

The correct option is C 3
$_{92} U^{235} \to_{54} X e^{142} +_{38} S r^{90} + x_{0} n^{1}$

Equating atomic number on both sides

238 = 142 + 90 + x

∴ x = 3
Hence, 3 neutrons are emitted.

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