---
title: "Transmission through CuO layer (5 eV, 1 nm)"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:tunneling-cuo-layer-5ev"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/tunneling-cuo-layer-5ev
site: BCT Engineering
---

# Transmission through CuO layer (5 eV, 1 nm)

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2072 Kartik, 6 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-5ev
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

A 3nm thick oxide layer of CuO separates two copper conductors providing a barrier height of 10eV for the conduction of electrons in copper. Determine the transmission coefficient if the energy of electron is 5eV. What will be the new transmission coefficient if the thickness of CuO was reduced to 1nm.

## Short answer

Applying T ≈ 16(E/V₀)(1−E/V₀) exp(−2√[2m(V₀−E)] a/ħ) for E=5eV, V₀=10eV: α≈1.14×10¹⁰ m⁻¹. For a=3nm, T ≈ 6.1×10⁻³⁰; for a=1nm, T ≈ 4.6×10⁻¹⁰.

## Answer

The CuO layer is a rectangular barrier. The transmission (tunneling) coefficient for $E < V_0$ and a wide barrier is

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

**Given:** $V_0 = 10$ eV, $E = 5$ eV, $a_1 = 3$ nm, $a_2 = 1$ nm, $m = 9.1\times10^{-31}$ kg, $\hbar = 1.055\times10^{-34}$ J s.

#### Decay constant

$$
\begin{aligned}
V_0 - E &= 5\ \text{eV} = 8\times10^{-19}\ \text{J} \\
\alpha &= \frac{\sqrt{2\times9.1\times10^{-31}\times8\times10^{-19}}}{1.055\times10^{-34}} \\
&= \frac{1.2066\times10^{-24}}{1.055\times10^{-34}} = 1.144\times10^{10}\ \text{m}^{-1}
\end{aligned}
$$

$$
T_0 = 16\times\frac{5}{10}\times\left(1 - \frac{5}{10}\right) = 4
$$

#### (a) Thickness 3 nm

$$
\begin{aligned}
2\alpha a_1 &= 2\times1.144\times10^{10}\times3\times10^{-9} = 68.65 \\
T_1 &\approx 4\times e^{-68.65} = 4\times1.53\times10^{-30} = 6.1\times10^{-30}
\end{aligned}
$$

#### (b) Thickness 1 nm

$$
\begin{aligned}
2\alpha a_2 &= 2\times1.144\times10^{10}\times1\times10^{-9} = 22.88 \\
T_2 &\approx 4\times e^{-22.88} = 4\times1.15\times10^{-10} = 4.6\times10^{-10}
\end{aligned}
$$

| Oxide thickness | $2\alpha a$ | $T$ |
|---|---|---|
| 3 nm | 68.65 | $6.1\times10^{-30}$ |
| 1 nm | 22.88 | $4.6\times10^{-10}$ |

**Answer:** $T \approx 6.1\times10^{-30}$ for 3 nm and $T \approx 4.6\times10^{-10}$ for 1 nm. (Without the prefactor: $1.5\times10^{-30}$ and $1.2\times10^{-10}$.)

Reducing the thickness to one third increases the transmission by a factor of about $e^{45.8} \approx 7.5\times10^{19}$, showing the very strong (exponential) dependence of tunneling on barrier width.

## Check yourself

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

1. Which expression is used for a wide rectangular barrier with E < V0?
   A. T = T0 exp(+2 alpha a)
   B. T = T0 exp(-2 alpha a)
   C. T = exp(-alpha a / 2)
   D. T = 1 - exp(-2 alpha a)
   Answer: B. T falls exponentially with width: T = T0 e^(-2 alpha a). (Key point: Transmission formula)

2. Which quantity sets alpha?
   A. sqrt(2mE)/hbar
   B. sqrt(2m(V0 - E))/hbar
   C. sqrt(2m(V0 + E))/hbar
   D. (V0 - E)/hbar
   Answer: B. alpha = sqrt(2m(V0 - E))/hbar, here 1.144e10 per m for 5 eV. (Key point: Decay constant)

3. What is T0 = 16 (E/V0)(1 - E/V0) for E = 5 eV, V0 = 10 eV?
   A. 1
   B. 2
   C. 4
   D. 16
   Answer: C. 16 x 0.5 x 0.5 = 4. (Key point: Prefactor)

4. What happens to T when the thickness is reduced from 3 nm to 1 nm?
   A. It decreases
   B. It stays same
   C. It rises from about 6e-30 to about 5e-10
   D. It becomes 3 times larger
   Answer: C. 2 alpha a falls from 68.65 to 22.88, so T rises by about e^45.8, not simply threefold. (Key point: Effect of thinning)

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