Question

The SI unit of mobility of charges $\left(\mu \right)$ is :
$\begin{array}{rl}& A)Coulomb/s/Kg\\ & B)Coulomb/\left(Kg/s\right)\\ & C)Coulomb/Kg/s^{-1}\\ & D)Coulomb/s^{-1}/Kg\end{array}$

Answer 1

Hint: We must know that mobility of charge is the measure of how quickly a charge could move through a metal or semiconductor. It is directly proportional to the drift velocity of the charge through the conductor and inversely proportional to the applied electric field. We will obtain its unit from its formula by giving and equating the basic units of the factors it depends upon.

Formula used:
$\mu ={\displaystyle \frac{v_d}{E}}$

Complete step by step answer:
We know that charge mobility in a conductor is the measure of speed of the electron moving through a conductor or a semiconductor device under the influence of an applied external electric field.
It is directly proportional to the drift velocity of the charge through the conductor and inversely proportional to the applied electric field. So, it is given as,
$\mu ={\displaystyle \frac{v_d}{E}}$
Where, $v_d$ is the drift velocity of the electron through the conductor and E is the electric field applied.
Actually, the SI unit of mobility of charges is given as $m^2{V}^{-1}{s}^{-1}$. But, this could be further changed into the form of units given in the option. For that, we know,
$\mu ={\displaystyle \frac{v_d}{E}}$
Where, drift velocity has a SI unit m/s.
But, E can be defined as $E={\displaystyle \frac{F}{q}}$.
Where, F is the coulomb’s force and q is the charge.
We know that unit of force is Newton (N) or it can be written as $kg-m/s^2$. Also, the unit of charge is coulomb (C).
Then,
$\begin{array}{rl}& \mu ={\displaystyle \frac{v_d}{E}}={\displaystyle \frac{v_d}{{\displaystyle \frac{F}{q}}}}\\ & \Rightarrow \mu ={\displaystyle \frac{q\times v_d}{E}}\end{array}$
Now, substituting the units, we will get,
$\begin{array}{rl}& \mu ={\displaystyle \frac{q\times v_d}{F}}={\displaystyle \frac{\left(Coulomb\right)\left({\displaystyle \frac{m}{s}}\right)}{kg\left({\displaystyle \frac{m}{s^2}}\right)}}\\ & \therefore \mu ={\displaystyle \frac{Coulomb}{\left({\displaystyle \frac{kg}{s}}\right)}}\end{array}$

So, the unit of mobility of charge is found to be $Coulomb/\left(Kg/s\right)$. Therefore, option B is correct.

Note: In the question, we didn’t find the actual SI unit. The mostly SI unit of mobility of charges is $m^2{V}^{-1}{s}^{-1}$, where V is volt. We must know that drift velocity of an electron is defined as the net velocity at which an electron drifts. Because the electron movement will be slow in the direction within the applied electric field direction. We can calculate current from the equation,
$I=nAvQ$
Here, v is drift velocity.