Explain the phenomenon of nuclear fission.

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Nuclear fission is a process in which a heavy nucleus $(A > 230)$ is bombarded with neutrons,causing it to split into two smaller nuclei of comparable mass,accompanied by the release of neutrons and a large amount of energy.
Neutrons are electrically neutral,meaning they do not experience Coulomb repulsion when approaching the positively charged nucleus. This makes them ideal projectiles for inducing fission.
When a slow neutron strikes a $^{235}_{92}U$ nucleus,it forms an unstable $^{236}_{92}U$ nucleus,which then undergoes fission. $A$ typical reaction is:
${ }_{92}^{235} U + { }_{0}^{1} n \rightarrow { }_{92}^{236} U \rightarrow { }_{56}^{144} Ba + { }_{36}^{89} Kr + 3({ }_{0}^{1} n) + Q$
Other possible fission products include:
${ }_{92}^{235} U + { }_{0}^{1} n \rightarrow { }_{92}^{236} U \rightarrow { }_{51}^{133} Sb + { }_{41}^{99} Nb + 4({ }_{0}^{1} n) + Q$
${ }_{92}^{235} U + { }_{0}^{1} n \rightarrow { }_{92}^{236} U \rightarrow { }_{54}^{140} Xe + { }_{38}^{94} Sr + 2({ }_{0}^{1} n) + Q$
The fission fragments are typically radioactive and reach stability through successive $\beta$-particle emissions.
The neutrons produced in these reactions are fast,with energies around $2 \text{ MeV}$.
$Q$ represents the energy released,which is approximately $200 \text{ MeV}$ per fission event. This energy is primarily released as the kinetic energy of the fission fragments and $\gamma$-rays.

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The binding energy of deuteron $_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle $_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction $_1^2H + _1^2H \to _2^4He + Q$,the energy $Q$ released is ........ $MeV$.

Complete the equation for the following fission process:
$_{92}U^{235} + _0n^1 \to _{38}Sr^{90} + ....$

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