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

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(D) The nuclear fission reaction is represented by the equation: $_{92}U^{235} + _0n^1 	o _{92}U^{236} 	o _{56}Ba^{144} + _{36}Kr^{89} + x(_0n^1)$.T

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The nuclear fission reaction is represented by the equation: $_{92}U^{235} + _0n^1 	o _{92}U^{236} 	o _{56}Ba^{144} + _{36}Kr^{89} + x(_0n^1)$.To find the number of neutrons $x$,we balance the mass numbers 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 the two sides: $236 = 233 + x$,which gives $x = 3$.Therefore,$3$ neutrons are released.

The number of neutrons released when $_{92}U^{235}$ undergoes fission by absorbing $_0n^1$ and $(_{56}Ba^{144} + _{36}Kr^{89})$ are formed,is

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