What are the differences between $E_1$ and $E_2$ reaction?

Hint: Elimination reactions are organic reactions in which two substituents from a molecule are separated in a one- or two-step mechanism. The reactivity of halogens influences the reaction rate, with iodide and bromide being preferred. Fluoride is a poor leaving group, but fluoride-based eliminations are slower than those involving other halogens.

Complete answer:

$E_1$ is a model that explains one kind of chemical removal reaction in particular. The following are the criteria for $E_1$, which stands for unimolecular removal. Ionization and deprotonation are the two steps in the removal process. Ionization occurs as the carbon-halogen bond breaks, resulting in the formation of a carbocation intermediate. The carbocation is deprotonated.
Bimolecular reduction is abbreviated as $E_2$. The reaction occurs in a single step, with the carbon-hydrogen and carbon-halogen bonds breaking to form a double bond (C=C Pi bond). With just one transition state, $E_2$ is a single-step elimination.

Note:
The $E_2$ reaction is the one-step mechanism, while the $E_1$ reaction is the two-step mechanism. The numbers correspond to the kinetics of the reaction, not the number of steps in the mechanism: $E_2$ is bimolecular (second-order), while $E_1$ is unimolecular (first-order). A third form of reaction, $E_1$CB, occurs where a molecule is able to stabilise an anion but has a weak leaving group. Finally, pyrolysis of xanthate and acetate esters occurs through the Ei mechanism, which is a "internal" removal mechanism.