The S.I. unit of electron mobility is
A. ${m^2}{s^2}{V^{ - 1}}$
B. ${m^2}{s^2}{V^{ - 1}}$
C. $m{s^{ - 1}}V$
D. ${m^2}{s^{ - 2}}V$

Hint: Electron mobility is defined as the ratio of velocity to the electric field. $\mu = \dfrac{{{v_E}}}{E}$ SI unit of velocity is $\dfrac{m}{s}$ and SI unit of electric field is $\dfrac{V}{m}$ then substituting these values in the mobility equation we get our required result.

Complete step by step answer:
Now from the question, we have
$\mu = \dfrac{{{v_E}}}{E}$, then $\mu = \dfrac{{\dfrac{m}{s}}}{{\dfrac{V}{m}}}$ = $\dfrac{{{m^2}}}{{Vs}}$
= ${m^2}{s^{ - 1}}{V^{ - 1}}$, which is the required answer.
Thus, option B is the correct answer.

Note:
Drift Velocity
Subatomic particles like electrons move in random directions all the time. When electrons are subjected to an electrical field they are moving randomly, but they slowly drift in one direction, within the direction of the electrical field applied. The net velocity at which these electrons drift is understood as drift velocity.
We can use the subsequent formula so as to calculate drift velocity:
I = nAvQ
Where,
I is that the current flowing through the conductor which is measured in amperes
n is that the number of electrons
A is that the area of the cross-section of the conductor which is measured in ${m^2}$
v is that the drift velocity of the electrons
Q is that the charge of an electron which is measured in Coulombs