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- 2072 Chaitra · 5 marks
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Electron transmission through potential barrier
Short answer
Using T ≈ 16(E/V₀)(1−E/V₀)e^{-2κa} with κ = √(2m(V₀−E))/ħ, κ = 5.12×10⁹ m⁻¹, 2κa = 20.48, prefactor 3, gives T ≈ 3.8×10⁻⁹, i.e. about 3.8×10⁻⁷% transmission.
A beam of electrons having energy of 3eV is incident on a potential barrier of height 4 eV. If the width of the barrier is 20 Å, calculate the percentage transmission (transmission coefficient) of the beam through the barrier.
A particle with E < V₀ can still pass through a barrier of width a (quantum tunnelling). For a wide barrier (κa ≳ 1), the transmission coefficient is
T ≈ 16 (E/V₀)(1 − E/V₀) e^{−2κa}, where κ = √(2m(V₀ − E))/ħ.
Given: E = 3 eV, V₀ = 4 eV, a = 20 Å = 2×10⁻⁹ m, m = 9.109×10⁻³¹ kg, ħ = 1.055×10⁻³⁴ J·s.
V₀ − E = 1 eV = 1.602×10⁻¹⁹ J
κ = √(2 × 9.109×10⁻³¹ × 1.602×10⁻¹⁹) / 1.055×10⁻³⁴
= √(2.919×10⁻⁴⁹) / 1.055×10⁻³⁴
= 5.403×10⁻²⁵ / 1.055×10⁻³⁴
= 5.12×10⁹ m⁻¹
2κa = 2 × 5.12×10⁹ × 2×10⁻⁹ = 20.48
e^(−20.48) = 1.27×10⁻⁹
Prefactor: 16 × (3/4) × (1/4) = 3
T = 3 × 1.27×10⁻⁹ = 3.8×10⁻⁹
Percentage transmission = 3.8×10⁻⁹ × 100 ≈ 3.8×10⁻⁷ %.
If a textbook uses the simpler form T ≈ e^(−2κa), the result is 1.27×10⁻⁹, that is, about 1.3×10⁻⁷ %.
Answer: T ≈ 3.8×10⁻⁹, i.e. about 3.8×10⁻⁷ % of the electrons are transmitted.
✓ Answer checked by Claude Opus 5.5 (high effort)
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