Worked example: An electron (E = 1 eV) hits a barrier of height V₀ = 5 eV and width a = 0.5 nm. Estimate the tunnelling probability. Step 1: V₀ − E = 4 eV = 4 × 1.602×10⁻¹⁹ = 6.41×10⁻¹⁹ J. Step 2: κ = √(2m(V₀−E))/ℏ = √(2 × 9.11×10⁻³¹ × 6.41×10⁻¹⁹) / 1.055×10⁻³⁴. Step 3: Numerator = √(1.168×10⁻⁴⁸) = 1.081×10⁻²⁴; κ = 1.081×10⁻²⁴/1.055×10⁻³⁴ ≈ 1.02×10¹⁰ m⁻¹. Step 4: 2κa = 2 × 1.02×10¹⁰ × 0.5×10⁻⁹ = 10.2; T ≈ e^{−10.2} ≈ 3.7×10⁻⁵. Conclusion: about 1 electron in 27,000 tunnels through — widen the barrier to 1 nm and T plunges to ~10⁻⁹. Exponential sensitivity in one calculation.