Question:

Match List-I with List-II:

List-I (Electric field) List-II (Mobility)
(A) Low electric field (I) Mobility decreases by \( \frac{1}{E} \)
(B) Medium electric field (II) Mobility decreases by \( \frac{1}{\sqrt{E}} \)
(C) High electric field (III) Mobility remains constant
Choose the correct answer:

List-I (Electric field)	List-II (Mobility)
(A) Low electric field	(I) Mobility decreases by \( \frac{1}{E} \)
(B) Medium electric field	(II) Mobility decreases by \( \frac{1}{\sqrt{E}} \)
(C) High electric field	(III) Mobility remains constant

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At low fields, mobility decreases with increasing electric field, but at high fields, it stabilizes.

Updated On: Sep 19, 2025

(A) – (III), (B) – (II), (C) – (I)
(A) – (I), (B) – (III), (C) – (II)
(A) – (II), (B) – (III), (C) – (I)
(A) – (III), (B) – (I), (C) – (II)

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The Correct Option is D

Solution and Explanation

- Low field: mobility decreases with \(E\).
- Medium field: mobility decreases by \(1/\sqrt{E}\).
- High field: mobility remains constant.

Final Answer: \[ \boxed{(4)} \]

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Top Questions on Solid State Physics, Devices, Electronics

Match List-I with List-II

List-I	List-II
(A) Faraday’s Law	(ii) \( \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \)
(B) Conservation Law	(iii) \( \vec{\nabla} \times \vec{A} = 0 \)
(C) Ohm’s Law	(iv) \( \vec{J} = \sigma \vec{E} \)
(D) Gauss’s Law	(i) \( \frac{\rho}{\epsilon_0} \)