In a p-n junction diode, the depletion region is created primarily due to: MCQ with Answer and Explanation

In a p-n junction diode, the depletion region is created primarily due to:
A. Diffusion of majority charge carriers across the junction.
B. Heating of the semiconductor material.
C. Drift of electrons under a strong electric field.
D. Recombination of minority carriers.
Answer: Option A
Solution (By JKExamLibrary)
When a p-type and an n-type semiconductor are joined, electrons from the n-side (majority carriers) diffuse across the junction to the p-side, and holes from the p-side diffuse to the n-side due to concentration gradients. They recombine near the junction, leaving behind immobile ionized atoms (positive donor ions on n-side, negative acceptor ions on p-side). This uncovered charge region is the depletion layer.

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Practice More Physics Questions

Question #1
A body weighs 100 N on Earth's surface. Weight at a height equal to Earth's radius is
A. 0 N
B. 25 N
C. 100 N
D. 50 N

Correct Answer: Option B


Explanation:
g' = g (R/(R+h))² = g (1/2)² = g/4. Weight = 100/4 = 25 N.

This question belongs to: Science Physics
Question #2
The heating effect of electric current is described by:
A. Ohm's law
B. Joule's law
C. Faraday's law
D. Lenz's law

Correct Answer: Option B


Explanation:
Joule's law states that heat produced H = I²Rt, where I is current, R resistance, t time. This quantifies electrical energy converted to thermal energy in resistors. Ohm's law relates V, I, R; Faraday's and Lenz's laws concern electromagnetic induction. Memory aid: 'Joule heating ∝ I²R'. This definition-based question tests knowledge of fundamental laws in electricity, commonly examined in competitive exams. Applications include electric heaters, fuses, and incandescent bulbs. Always distinguish Joule's law (heating) from power formulas (P=VI=I²R=V²/R).

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Question #3
The critical angle for light passing from glass (n=1.5) to air is approximately:
A. 60°
B. 30°
C. 42°
D. 90°

Correct Answer: Option C


Explanation:
Critical angle θ_c for total internal reflection: sinθ_c = n₂/n₁, where n₁ > n₂. Here n₁=1.5 (glass), n₂=1 (air), so sinθ_c = 1/1.5 = 2/3 ≈ 0.6667 ⇒ θ_c = sin⁻¹(0.6667) ≈ 41.8° ≈ 42°. Memory aid: 'sinθ_c = 1/n for glass to air'. This calculation tests understanding of total internal reflection conditions, crucial for optics in competitive exams. Always ensure light travels from denser to rarer medium for TIR to be possible; otherwise, no critical angle exists.

This question belongs to: Science Physics