Skip to main content

Transmission through CuO layer (5 eV, 1 nm)

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.

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.

The CuO layer is a rectangular barrier. The transmission (tunneling) coefficient for E<V0E < V_0 and a wide barrier is

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=5E = 5 eV, a1=3a_1 = 3 nm, a2=1a_2 = 1 nm, 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=5 eV=8×10−19 Jα=2×9.1×10−31×8×10−191.055×10−34=1.2066×10−241.055×10−34=1.144×1010 m−1\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} T0=16×510×(1−510)=4T_0 = 16\times\frac{5}{10}\times\left(1 - \frac{5}{10}\right) = 4

(a) Thickness 3 nm

2αa1=2×1.144×1010×3×10−9=68.65T1≈4×e−68.65=4×1.53×10−30=6.1×10−30\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

2αa2=2×1.144×1010×1×10−9=22.88T2≈4×e−22.88=4×1.15×10−10=4.6×10−10\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 thickness2αa2\alpha aTT
3 nm68.656.1×10−306.1\times10^{-30}
1 nm22.884.6×10−104.6\times10^{-10}

Answer: T≈6.1×10−30T \approx 6.1\times10^{-30} for 3 nm and T≈4.6×10−10T \approx 4.6\times10^{-10} for 1 nm. (Without the prefactor: 1.5×10−301.5\times10^{-30} and 1.2×10−101.2\times10^{-10}.)

Reducing the thickness to one third increases the transmission by a factor of about e45.8≈7.5×1019e^{45.8} \approx 7.5\times10^{19}, showing the very strong (exponential) dependence of tunneling on barrier width.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Engineering Material chapter 1 questions →