If in Millikan’s oil drop experiment charges on the drop are found to be $8\;{{\mu C}}$, $12\;{{\mu C}}$, $20\;{{\mu C}}$, then quanta of charge is
A) $8\;{{\mu C}}$
B) $4\;{{\mu C}}$
C) $20\;{{\mu C}}$
D) $12\;{{\mu C}}$

Hint: The experiments conducted by Robert Millikan which was awarded by the Nobel prize was to determine the size of the charges on the electron. He inferred that there is a small unit of charge that exists.

Complete step by step answer:
Millikan’s oil drop experiment study about the electric charge of the electron. He observed that the drops have charges which are the multiples of a single number. Let’s have some idea about the experiment.
Millikan put a charge on a drop. And the electric field is applied to it resisting it from falling. The mass of the drop and the electric field applied was known. The gravitational force and electric forces of the drops will be the same when they are falling free due to gravity at a constant rate.
The equation for calculating the charge is given as,
$q = \dfrac{{mg}}{E}$
Where, $m$ is the mass of drop and $E$ is the applied electric field.
 Through calculations Millikan found the charge of the single electron. He found it as $1.6 \times {10^{ - 19}}\;{{C}}$.
Millikan’s oil drop experiment points out that the charges will be the integral multiple of the base values. Here the charges are found to be $8\;{{\mu C}}$, $12\;{{\mu C}}$, $20\;{{\mu C}}$.
These charges are the multiple of $4$. Therefore the quanta of charge are $4\;{{\mu C}}$.

The correct answer is option B.

Note:
The charges present on the drop were quantized. These quantized charges will be integral multiples of the charge of the electron. The smallest unit of charge is the charge of an electron.