Question

Dimension of ionic mobility is:
A.$m{V^{ - 1}}{s^{ - 1}}$
B.${m^2}{V^{ - 2}}{s^{ - 1}}$
C.${m^2}{V^{ - 1}}{s^{ - 1}}$
D.${m^2}V{s^{ - 1}}$

Answer 1

Hint: We can define dimensional formulas of physical quantity as the expression describing the powers to which the fundamental units are raised to derive one unit of derived quantity. For example, if Q is unit of derived quantity, we can represent Q has ${M^a}{L^b}{T^c}$, then we call ${M^a}{L^b}{T^c}$ as dimensional formula and the exponents $a,b,c$ are known as dimensions.

Complete answer:
We have to know that dimensions of a physical quantity are the powers to which the basic units are raised to derive one unit of that quantity.
We can define ionic mobility $\left( \mu \right)$ as the velocity reached by an ion passing through a gas under the influence of an electric field. The SI unit of ionic mobility is ${m^2}{s^{ - 1}}vol{t^{ - 1}}$.
So, we have to know that the ratio of speed of the ion to the potential gradient is ionic mobility. $m{s^{ - 1}}$ is the unit of speed of ion and $V{m^{ - 1}}$ is the unit of potential gradient. So, we can give the dimensions of ionic mobility by dividing the speed of the ions to the potential gradient. Therefore,
$Unit = \dfrac{{{\text{Speed of ion}}}}{{{\text{Potential gradient}}}}$
We have to substitute the units of speed of ion and potential gradient in the above expression.
$Unit = \dfrac{{{\text{Speed of ion}}}}{{{\text{Potential gradient}}}}$
$Unit = \dfrac{{m{s^{ - 1}}}}{{V{m^{ - 1}}}}$
$Unit = {m^2}{V^{ - 1}}{s^{ - 1}}$
The dimension of ionic mobility is ${m^2}{V^{ - 1}}{s^{ - 1}}$.
Therefore, the option (C) is correct.

Note:
It is better not to memorize dimensional formulas as students generally tend to do that. It would be good to understand the physical quantities and relate them. Some of the incorrect options could be eliminated using dimensional analysis. So, understanding and then working out this would be good.