When a uranium isotope $_{92}^{235}\mathrm{U}$ is bombarded with a neutron, it generates $_{36}^{89}\mathrm{Kr

Question

When a uranium isotope $_{92}^{235}\mathrm{U}$ is bombarded with a neutron, it generates $_{36}^{89}\mathrm{Kr}$, three neutrons and:

Accepted Answer

1.  **Conservation Laws:** In any nuclear reaction, both the mass number ($A$) and the atomic number ($Z$) must be conserved.
2.  **Reaction Equation:** The given fission reaction can be written as:
    $$ _{92}^{235}\mathrm{U} + _{0}^{1}\mathrm{n} \rightarrow _{36}^{89}\mathrm{Kr} + 3_{0}^{1}\mathrm{n} + _{Z}^{A}\mathrm{X} $$
3.  **Balancing Mass Number ($A$):**
    $$ 235 + 1 = 89 + 3(1) + A $$
    $$ 236 = 92 + A \implies A = 236 - 92 = 144 $$
4.  **Balancing Atomic Number ($Z$):**
    $$ 92 + 0 = 36 + 3(0) + Z $$
    $$ 92 = 36 + Z \implies Z = 92 - 36 = 56 $$
5.  **Identification:** The element with atomic number $Z=56$ is Barium (Ba). Thus, the missing nucleus is $_{56}^{144}\mathrm{Ba}$.

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