The mobility of charge carriers increases with

The correct answer is Option (4) → increase in the average collision time

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

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

So the mobility is directly proportional to the average collision time \( \tau \) (relaxation time), inversely proportional to the mass of the carrier, and does not depend on the electric field.

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

Option (2) increase in the electric field — Incorrect. The drift velocity increases with \( E \), but the mobility \( v_d/E \) does not.

Option (3) decrease in the electric field — Incorrect. For the same reason, decreasing the field does not change the mobility.

Option (4) increase in the average collision time — Correct. \( \mu = e\tau/m \) increases with the average collision time \( \tau \).

Note: A similar question was also asked in CUET (UG) 2025 on 28-05-2025 (Slot 1), and in CUET (UG) 2022 on 06-08-2022 (Slot 2).

NCERT (2026-27):
Mobility – “… An important quantity is the mobility μ defined as the magnitude of the drift velocity per unit electric field: … From Eq. (3.17), we have

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

Hence,

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

where τ is the average collision time for electrons.”