Q.22
In a p-n junction, the depletion region has a width of 3 × 10-7 m and the intensity of electric field in the depletion region is 106 V/m. An electron approaches the junction from the n-side with velocity v1 and enters the p-side with velocity v2. If v2 = 4 × 105 m/s, the value of v1 is
Given data: Charge of electron = 1.6 × 10-19 C; Mass of electron = 9.1 × 10-31 kg
The electron slows down crossing the p-n junction depletion region due to the built-in electric field
Problem Statement
In a p-n junction, the depletion region has a width of 3 × 10⁻⁷ m and electric field intensity 10⁶ V/m. An electron approaches from n-side with velocity v₁ and enters p-side with v₂ = 4 × 10⁵ m/s. Find v₁.
Given: q = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg
Physics Concept
The depletion region features a uniform electric field E = 10⁶ V/m directed from n-side (positive) to p-side (negative) [web:6][web:9]. Electron decelerates across width w = 3 × 10⁻⁷ m.
½mv₁² = ½mv₂² + qVWhere
V = E × w = 10⁶ × 3×10⁻⁷ = 0.3 VStep-by-Step Calculation
- Potential difference:
V = E × w = 10⁶ × 3×10⁻⁷ = 0.3 V - Final KE:
½mv₂² = 0.5 × 9.1×10⁻³¹ × (4×10⁵)² = 7.28×10⁻²⁰ J - Energy loss:
qV = 1.6×10⁻¹⁹ × 0.3 = 4.8×10⁻²⁰ J - Initial KE:
7.28×10⁻²⁰ + 4.8×10⁻²⁰ = 1.208×10⁻¹⁹ J - Initial velocity:
v₁ = √(2×KE₁/m) ≈ 5.15×10⁵ m/s ≈ 5.2×10⁵ m/s
Options Analysis
| Option | Velocity (m/s) | Deviation from 5.15×10⁵ | Reason Incorrect |
|---|---|---|---|
| (C) | 5.2 × 10⁵ | ✅ Matches (rounded) | Correct via energy conservation |
| (A) | 3.2 × 10⁵ | Too low | Ignores full qV or wrong V |
| (B) | 4.2 × 10⁵ | Low | Partial energy accounting |
| (D) | 6.2 × 10⁵ | Too high | Assumes acceleration (wrong field direction) |
Key Formula
v₁ = √(v₂² + 2qEw/m)Where E = electric field, w = depletion width
Exam Tips
- ✅ Field direction: n→p (decelerates electrons)
- ✅ Use V = Ew for uniform field
- ❌ Avoid parabolic field assumption
- Practice similar problems with different barrier heights
