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Q.
When ${ }_{92} U ^{235}$ is bombarded with one neutron, fission occurs and the products are three neutrons, ${ }_{36} Kr ^{94}$, and
NTA AbhyasNTA Abhyas 2022
Solution:
When a slow neutron strikes a $U^{235} \, $ nucleus it is absorbed by the nucleus and the following reaction occurs.
$
{ }_{92} U ^{235}+{ }_{0} n^{1} \rightarrow{ }_{36} Kr ^{94}+{ }_{56} Ba ^{139}+3_{0} n^{1}+\text { energy }
$
Hence, ${ }_{56} B a^{139}$ is another product.