---
title: "Electron transmission through potential barrier"
subject: "Engineering Physics"
chapter: 10
key: "semester_1/engineering_physics:electron-barrier-transmission-percent"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/electron-barrier-transmission-percent
site: BCT Engineering
---

# Electron transmission through potential barrier

Engineering Physics · Chapter 10: Photon and Matter Waves

- Asked in: 2072 Chaitra, 5 marks; 2069 Chaitra, 5 marks (asked 2 times)
- Video: [Electron tunneling through a barrier](https://www.youtube.com/watch?v=8-pBXRQIq6M) — Winfernity
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/electron-barrier-transmission-percent
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-10

## Question

A beam of electrons having energy of 3eV is incident on a potential barrier of height 4 eV. If the width of the barrier is 20 Å, calculate the percentage transmission (transmission coefficient) of the beam through the barrier.

## Short answer

Using T ≈ 16(E/V₀)(1−E/V₀)e^{-2κa} with κ = √(2m(V₀−E))/ħ, κ = 5.12×10⁹ m⁻¹, 2κa = 20.48, prefactor 3, gives T ≈ 3.8×10⁻⁹, i.e. about 3.8×10⁻⁷% transmission.

## Answer

A particle with E < V₀ can still pass through a barrier of width a (quantum tunnelling). For a wide barrier (κa ≳ 1), the transmission coefficient is

**T ≈ 16 (E/V₀)(1 − E/V₀) e^{−2κa}**, where κ = √(2m(V₀ − E))/ħ.

**Given:** E = 3 eV, V₀ = 4 eV, a = 20 Å = 2×10⁻⁹ m, m = 9.109×10⁻³¹ kg, ħ = 1.055×10⁻³⁴ J·s.

```text
V₀ − E = 1 eV = 1.602×10⁻¹⁹ J
κ  = √(2 × 9.109×10⁻³¹ × 1.602×10⁻¹⁹) / 1.055×10⁻³⁴
   = √(2.919×10⁻⁴⁹) / 1.055×10⁻³⁴
   = 5.403×10⁻²⁵ / 1.055×10⁻³⁴
   = 5.12×10⁹ m⁻¹

2κa = 2 × 5.12×10⁹ × 2×10⁻⁹ = 20.48
e^(−20.48) = 1.27×10⁻⁹

Prefactor: 16 × (3/4) × (1/4) = 3

T = 3 × 1.27×10⁻⁹ = 3.8×10⁻⁹
```

Percentage transmission = 3.8×10⁻⁹ × 100 ≈ 3.8×10⁻⁷ %.

If a textbook uses the simpler form T ≈ e^(−2κa), the result is 1.27×10⁻⁹, that is, about 1.3×10⁻⁷ %.

**Answer: T ≈ 3.8×10⁻⁹, i.e. about 3.8×10⁻⁷ % of the electrons are transmitted.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How can an electron with E < V₀ cross the barrier?
   A. It gains energy from walls
   B. It goes over the barrier classically
   C. It cannot cross
   D. By quantum tunnelling, as ψ decays exponentially inside the barrier
   Answer: D. The wave function is non-zero beyond a barrier of finite width. (Key point: Tunnelling)

2. The decay constant κ is given by:
   A. κ = √(2m(V₀ − E))/ħ
   B. κ = √(2mE)/ħ
   C. κ = √(2m(V₀ + E))/ħ
   D. κ = 2m(V₀ − E)/ħ
   Answer: A. κ depends on how far E is below the barrier top. (Key point: Decay constant)

3. With V₀ − E = 1 eV and a = 20 Å, what is 2κa?
   A. About 5.1
   B. About 20.5
   C. About 1.0
   D. About 2.0×10⁻⁹
   Answer: B. κ ≈ 5.12×10⁹ m⁻¹, so 2κa = 2 × 5.12×10⁹ × 2×10⁻⁹ ≈ 20.5. (Key point: Exponent value)

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