A { }_{92} U^{235} nucleus absorbs a neutron | vedclass

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(A) The nuclear reaction is given by: ${ }_{92} U^{235} + { }_{0} n^{1} 	o { }_{54} X e^{139} + { }_{38} S r^{94} + X$.To find $X$,we balance the mas

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$3$ neutrons

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(A) The nuclear reaction is given by: ${ }_{92} U^{235} + { }_{0} n^{1} 	o { }_{54} X e^{139} + { }_{38} S r^{94} + X$.To find $X$,we balance the mass numbers and atomic numbers on both sides.Sum of mass numbers on the left side: $235 + 1 = 236$.Sum of mass numbers on the right side: $139 + 94 + A = 233 + A$,where $A$ is the mass number of $X$.Equating them: $236 = 233 + A \implies A = 3$.Sum of atomic numbers on the left side: $92 + 0 = 92$.Sum of atomic numbers on the right side: $54 + 38 + Z = 92 + Z$,where $Z$ is the atomic number of $X$.Equating them: $92 = 92 + Z \implies Z = 0$.Since $A=3$ and $Z=0$,the product $X$ consists of $3$ neutrons $(3 { }_{0} n^{1})$.

$A$ ${ }_{92} U^{235}$ nucleus absorbs a neutron and disintegrates into ${ }_{54} X e^{139}$,${ }_{38} S r^{94}$,and $X$. What will be the product $X$?

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