The number of neutrons released when {}_{92}^ | vedclass

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(D) The nuclear fission reaction is represented as: ${}_{92}^{235}U + {}_{0}^{1}n 	o {}_{56}^{144}Ba + {}_{36}^{89}Kr + x({}_{0}^{1}n)$.To find the n

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$3$

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(D) The nuclear fission reaction is represented as: ${}_{92}^{235}U + {}_{0}^{1}n 	o {}_{56}^{144}Ba + {}_{36}^{89}Kr + x({}_{0}^{1}n)$.To find the number of neutrons $x$,we balance the mass number on both sides of the equation.The sum of mass numbers on the left side is $235 + 1 = 236$.The sum of mass numbers on the right side is $144 + 89 + x = 233 + x$.Equating both sides: $236 = 233 + x$.Therefore,$x = 236 - 233 = 3$.Thus,$3$ neutrons are released.

The number of neutrons released when ${}_{92}^{235}U$ undergoes fission by absorbing ${}_{0}^{1}n$ and products ${}_{56}^{144}Ba$ and ${}_{36}^{89}Kr$ are formed,is:

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