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Transmission through CuO layer (7 eV electrons)

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).

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?

The thin CuO layer acts as a rectangular potential barrier. An electron with energy E<V0E < V_0 can cross it only by tunneling, with probability

T≈T0 e−2αa,α=2m(V0−E)ℏ,T0=16EV0(1−EV0)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: V0=10V_0 = 10 eV, E=7E = 7 eV, a=3a = 3 nm =3×10−9= 3\times10^{-9} m, m=9.1×10−31m = 9.1\times10^{-31} kg, ℏ=1.055×10−34\hbar = 1.055\times10^{-34} J s.

Decay constant

V0−E=3 eV=3×1.6×10−19=4.8×10−19 Jα=2×9.1×10−31×4.8×10−191.055×10−34=9.347×10−251.055×10−34=8.86×109 m−1\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

2αa=2×8.86×109×3×10−9=53.162\alpha a = 2\times8.86\times10^{9}\times3\times10^{-9} = 53.16

Transmission probability

T0=16×710(1−710)=16×0.7×0.3=3.36T≈3.36×e−53.16=3.36×8.21×10−24=2.76×10−23\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≈2.8×10−23T \approx 2.8\times10^{-23} (if the prefactor T0T_0 is ignored, T≈e−2αa=8.2×10−24T \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 TT 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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