Question

When ${ }_{92} U^{235}$ is bombarded with one neutron, fission occurs and the products are three neutrons, ${ }_{36} \mathrm{Kr}^{94}$ , and

• A. ${ }_{56} \mathrm{Ba}^{141}$
• B. ${ }_{54} \mathrm{Xe}^{139}$
• C. ${ }_{56} \mathrm{Ba}^{139}$
• D. ${ }_{58} \mathrm{I}^{142}$

Answer 1

Answer: (C)

When a slow neutron strikes a $U^{235} \, $ nucleus it is absorbed by the nucleus and the following reaction occurs.
${ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{36} \mathrm{Kr}^{94}+{ }_{56} \mathrm{Ba}^{139}+3{ }_0 n^1+$ energy
Hence, ${ }_{56} B a^{139}$ is another product.