The dimensions of mobility of charge carriers are

Options

A. `[M^(-2)T^(2)A]`
B. `[M^(-1)T^(2)A]`
C. `M^(-2)T^(3)A]`
D. `[M^(-1)T^(3)A]`

Answer

To find the dimensions of mobility of charge carriers, we start with the formula for mobility:

\[ \mu = \frac{v_d}{E} \]

where:
- \( \mu \) is the mobility,
- \( v_d \) is the drift velocity,
- \( E \) is the electric field.

### Step 1: Determine the dimensions of drift velocity (\( v_d \))

Drift velocity is a type of velocity, so its dimensions are the same as those of velocity:

\[ [v_d] = L T^{-1} \]

### Step 2: Determine the dimensions of electric field (\( E \))

The electric field can be defined as the force per unit charge:

\[ E = \frac{F}{q} \]

Where:
- \( F \) is the force,
- \( q \) is the charge.

The dimensions of force (\( F \)) can be expressed as:

\[ [F] = M L T^{-2} \]

The dimensions of charge (\( q \)) can be expressed in terms of current (\( I \)) and time (\( T \)):

\[ [q] = I T \]

Now substituting these into the electric field formula:

\[ [E] = \frac{[F]}{[q]} = \frac{M L T^{-2}}{I T} = M L I^{-1} T^{-3} \]

### Step 3: Calculate the dimensions of mobility (\( \mu \))

Now we can substitute the dimensions of \( v_d \) and \( E \) into the mobility formula:

\[ [\mu] = \frac{[v_d]}{[E]} = \frac{L T^{-1}}{M L I^{-1} T^{-3}} \]

### Step 4: Simplify the expression

When we simplify the above expression, we have:

\[ [\mu] = \frac{L T^{-1}}{M L I^{-1} T^{-3}} = \frac{T^2 I}{M} \]

Thus, the dimensions of mobility (\( \mu \)) are:

\[ [\mu] = M^{-1} L^2 T^{-2} I \]

### Final Answer

The dimensions of mobility of charge carriers are:

\[ [\mu] = M^{-1} L^2 T^{-2} I \]