The number of neutrons emitted when _{92}^{235} U undergoes controlled nuclear fission to _{54}^{142} Xe \textrm{ }\textrm{ } and \left(\textrm{ }\textrm{ }\right)

Engineering

Chemistry

Application of Nuclear Reaction

Question

The number of neutrons emitted when $_{92}^{235} U$ undergoes controlled nuclear fission to $_{54}^{142} Xe and_{38}^{90} Sr$ is

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$_{92}^{235} U +_{0}^{1} n \to_{54}^{142} Xe +_{38}^{90} Sr + x_{0}^{1} n + y_{1}^{1} p$

235 + 1 = 142 + 90 + x

x = 4