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

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Solution

$_{92}^{235} U ⟶_{54}^{142} X e +_{38}^{90} S r + 3_{0}^{1} n$
So, number of neutron $=$ 3

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(IIT-JEE, 2010)

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