The mobility of charge carriers increases with

The correct answer is Option (1) → Increase in average collision time interval

Mobility is the drift velocity per unit electric field. With the drift velocity \( v_d = eE\tau/m \),

    \( \displaystyle \mu = \frac{v_d}{E} = \frac{e\tau}{m} \)

where \( \tau \) is the average collision time (relaxation time), \( e \) the charge and \( m \) the mass of the carrier.

Option (1) Increase in average collision time interval — Correct. \( \mu \propto \tau \), so a longer average time between collisions gives a larger mobility.

Option (2) Increase in the electric field — Incorrect. \( E \) cancels out: the drift velocity grows with the field, but the mobility does not.

Option (3) Increase in the mass of the charge carriers — Incorrect. \( \mu \propto 1/m \), so a heavier carrier has a smaller mobility.

Option (4) Decrease in the charge of the mobile carriers — Incorrect. \( \mu \propto e \), so a smaller charge lowers the mobility.

Note: A similar question was also asked in CUET (UG) 2025 on 28-05-2025 (Slot 1) and on 22-05-2025 (Slot 1).

NCERT (2026-27):
Mobility – “An important quantity is the mobility μ defined as the magnitude of the drift velocity per unit electric field:

    \( \displaystyle \mu = \frac{|\mathbf{v}_d|}{E} \qquad (3.24) \)

The SI unit of mobility is m²/Vs and is 10⁴ of the mobility in practical units (cm²/Vs). Mobility is positive. From Eq. (3.17), we have

    \( \displaystyle v_d = \frac{eE\tau}{m} \)

Hence,

    \( \displaystyle \mu = \frac{v_d}{E} = \frac{e\tau}{m} \qquad (3.25) \)

where τ is the average collision time for electrons.”