---
title: "Transmission through CuO layer (7 eV electrons)"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:tunneling-cuo-layer-7ev"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-cuo-layer-7ev
site: BCT Engineering
---

# Transmission through CuO layer (7 eV electrons)

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2071 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-cuo-layer-7ev
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

Consider two copper wires separated only by their surface oxide layer (CuO) of thickness 3 nm. The surface oxide layer offer potential barrier of height 10eV to the conduction electrons in copper. What is the transmission probability for conduction electrons in copper, which have kinetic energy of about 7eV?

## Short answer

Using tunneling probability T ≈ T₀ exp(−2αa) with α = √[2m(V₀−E)]/ħ and T₀ = 16(E/V₀)(1−E/V₀). For V₀=10eV, E=7eV, a=3nm, α=8.86×10⁹ m⁻¹, T₀=3.36, so T ≈ 2.8×10⁻²³ (dimensionless).

## Answer

The thin CuO layer acts as a rectangular potential barrier. An electron with energy $E < V_0$ can cross it only by **tunneling**, with probability

$$
T \approx T_0\,e^{-2\alpha a}, \qquad \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}, \qquad T_0 = 16\frac{E}{V_0}\left(1 - \frac{E}{V_0}\right)
$$

**Given:** $V_0 = 10$ eV, $E = 7$ eV, $a = 3$ nm $= 3\times10^{-9}$ m, $m = 9.1\times10^{-31}$ kg, $\hbar = 1.055\times10^{-34}$ J s.

#### Decay constant

$$
\begin{aligned}
V_0 - E &= 3\ \text{eV} = 3\times1.6\times10^{-19} = 4.8\times10^{-19}\ \text{J} \\
\alpha &= \frac{\sqrt{2\times9.1\times10^{-31}\times4.8\times10^{-19}}}{1.055\times10^{-34}} \\
&= \frac{9.347\times10^{-25}}{1.055\times10^{-34}} = 8.86\times10^{9}\ \text{m}^{-1}
\end{aligned}
$$

#### Exponent

$$
2\alpha a = 2\times8.86\times10^{9}\times3\times10^{-9} = 53.16
$$

#### Transmission probability

$$
\begin{aligned}
T_0 &= 16\times\frac{7}{10}\left(1 - \frac{7}{10}\right) = 16\times0.7\times0.3 = 3.36 \\
T &\approx 3.36\times e^{-53.16} = 3.36\times8.21\times10^{-24} \\
&= 2.76\times10^{-23}
\end{aligned}
$$

**Answer:** $T \approx 2.8\times10^{-23}$ (if the prefactor $T_0$ is ignored, $T \approx e^{-2\alpha a} = 8.2\times10^{-24}$).

The probability is extremely small, so a 3 nm oxide layer is practically an insulator for these electrons. Because $T$ depends exponentially on thickness, a thinner oxide (about 1 nm) would let a significant tunneling current through, which is why lightly oxidised copper contacts still conduct.

## Check yourself

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

1. An electron with E < V0 crosses the barrier by:
   A. Classical motion over it
   B. Tunneling with a probability that decays exponentially with thickness
   C. Reflection only
   D. Heating
   Answer: B. T ~ exp(-2 alpha a). (Key point: Tunneling)

2. alpha is given by:
   A. sqrt(2mE)/hbar
   B. sqrt(2m(V0 - E))/hbar
   C. sqrt(2m(V0 + E))/hbar
   D. hbar/sqrt(2m(V0 - E))
   Answer: B. Here V0 - E = 3 eV gives alpha about 8.86 x 10^9 m^-1. (Key point: Decay constant)

3. 2 alpha a for a = 3 nm is about:
   A. 5.3
   B. 53.2
   C. 8.9
   D. 26.6
   Answer: B. 2 x 8.86e9 x 3e-9 = 53.2. (Key point: Exponent)

4. The transmission probability is about:
   A. 0.34
   B. 10^-5
   C. 2.7 x 10^-23
   D. 1
   Answer: C. T = 3.36 x e^-53.18 = 2.7 x 10^-23. (Key point: Result)

## Related in Electrical Engineering Material

- [Transmission through CuO layer (5 eV, 1 nm)](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Atunneling-cuo-layer-5ev) — Electrical Engineering Material, chapter 1
- [Derivation of tunneling probability through barrier](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Atunneling-probability-derivation) — Electrical Engineering Material, chapter 1

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