Quantum tunneling — concept, transmission coefficient: Worked examples

This is the worked-example workshop for the parent note. There we derived the transmission coefficient. Here we use it on every kind of input the formula can be handed — thin barriers, thick barriers, heavy particles, the moment $E$ creeps up toward $V_{0}$ , the degenerate width-zero case, and a couple of exam-style twists.

Before we start, let us re-anchor the three symbols so you never have to scroll back. Picture a particle (a little wave) coming in from the left, hitting a wall of height $V_{0}$ and width $L$ .

The two formulas we will lean on the entire page:

The wall picture we keep referring to:

Figure — Quantum tunneling — concept, transmission coefficient

The scenario matrix

Every problem this topic throws at you lands in exactly one of these cells. The last column names the example that covers it.

#	Case class	What is special about it	Covered by
A	Thin barrier, $κ L$ modest	Approximation risky — check against exact	Ex 1
B	Thick barrier, $κ L ≫ 1$	Approximation excellent; exponential rules	Ex 2
C	Width scaling ( $L \to 2 L$ , etc.)	Exponential sensitivity to $L$	Ex 3
D	$E \to V_{0}^{-}$ (energy near top)	$κ \to 0$ : degenerate/limiting; $T \to$ max	Ex 4
E	$L \to 0$ (no barrier)	Degenerate: must recover $T = 1$	Ex 4
F	Heavy particle ( $m$ large)	$κ \propto m$ : tunneling collapses	Ex 5
G	Height scaling ( $V_{0}$ up)	Isolate the $V_{0}$ dependence	Ex 6
H	Real-world word problem (STM)	Translate physics → tiny $Δ L$ , big $Δ$ current	Ex 7
I	Exam twist: $E > V_{0}$	Formula breaks; $κ$ imaginary → resonance	Ex 8

Prerequisites if any symbol feels new: Schrödinger Equation, Wavefunction and Boundary Conditions, de Broglie Wavelength. Applications live at Alpha Decay and Scanning Tunneling Microscope.

Worked examples

Example 1 — Cell A: thin barrier, is the approximation safe?

Worked example Electron, thin wall

Electron ( $m = 9.11 \times 1 0^{- 31}$ kg), $E = 1$ eV, $V_{0} = 2$ eV, $L = 0.5$ nm. Find $T$ both ways and compare.

Forecast: $κ L$ will be a few — not huge. Guess: does the simple $G e^{- 2 κ L}$ overshoot or undershoot the exact answer? Write your guess before reading on.

Step 1 — Convert energies to joules. $V_{0} - E = 1 eV = 1.602 \times 1 0^{- 19}$ J. Why this step? $κ$ 's formula uses SI units; eV would give nonsense metres.

Step 2 — Compute $κ$ . $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 1.602 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}} \approx 5.12 \times 1 0^{9} m^{- 1}$ Why this step? $κ$ is the master quantity — the decay rate that sets everything.

Step 3 — Exponent $2 κ L$ . $2 (5.12 \times 1 0^{9}) (0.5 \times 1 0^{- 9}) = 5.12$ . Why this step? Both formulas need $κ L = 2.56$ ; the probability rides on $e^{- 2 κ L}$ .

Step 4 — Approximate $T$ . $G = \frac{16 ( 1 ) ( 1 )}{2 ^{2}} = 4$ , so $T \approx 4 e^{- 5.12} \approx 4 (0.00598) \approx 0.0239$ , i.e. 2.4%. Why this step? This is the number a fast estimate gives.

Step 5 — Exact $T$ . $sinh (2.56) = \frac{1}{2} (e^{2.56} - e^{- 2.56}) \approx 6.44$ , so $T = [1 + \frac{2 ^{2} ( 6.44 ) ^{2}}{4 ( 1 ) ( 1 )}]^{- 1} = [1 + 41.5]^{- 1} \approx 0.0235.$ Why this step? $κ L = 2.56$ is "medium", so we must check whether dropping the " $+ 1$ " was legal.

Verify: approximate $0.0239$ vs exact $0.0235$ — agree to within $\sim 2%$ . So even here the approximation is fine, and it slightly overshoots (because it threw away the " $+ 1$ " which only ever reduces $T$ ). Units: $κ$ has $m^{- 1}$ , $L$ has m, product dimensionless ✓. $T$ dimensionless and $< 1$ ✓.

Example 2 — Cell B: thick barrier, exponential dominates

Worked example Same wall, tripled width

Electron as above but $L = 1.5$ nm. Find $T$ .

Forecast: $L$ tripled from Ex 1. Does $T$ drop by a factor of 3? Guess an order of magnitude.

Step 1 — Reuse $κ = 5.12 \times 1 0^{9} m^{- 1}$ . Why this step? $κ$ depends only on $m, V_{0}, E$ — none changed, so no recompute.

Step 2 — Exponent. $2 κ L = 2 (5.12 \times 1 0^{9}) (1.5 \times 1 0^{- 9}) = 15.36$ . Why this step? Now $κ L = 7.68 ≫ 1$ → deep in approximation territory.

Step 3 — Approximate $T$ . $G = 4$ still, so $T \approx 4 e^{- 15.36} \approx 4 (2.14 \times 1 0^{- 7}) \approx 8.6 \times 1 0^{- 7}$ . Why this step? With $κ L$ this large the " $+ 1$ " is utterly negligible, so the approximation is the exact answer to many digits.

Verify: Not "3× smaller" but ~28000× smaller than Ex 1's $0.024$ . That is the whole lesson: $T$ falls exponentially, not linearly, with $L$ . Check: $0.024/8.6 \times 1 0^{- 7} \approx 2.8 \times 1 0^{4}$ , and $e^{15.36 - 5.12} = e^{10.24} \approx 2.8 \times 1 0^{4}$ ✓ (the ratio is purely the exponential difference).

Example 3 — Cell C: doubling the width

Worked example The STM sensitivity number

Take Ex 1's setup ( $2 κ L = 5.12$ at $L = 0.5$ nm). By what factor does $T$ change if $L$ doubles to $1.0$ nm?

Forecast: Guess the factor. Most people say "2×". Watch.

Step 1 — New exponent. $L \to 1.0$ nm gives $2 κ L = 10.24$ . Why this step? Only the exponent changes; $G$ is unchanged since $E, V_{0}$ fixed.

Step 2 — Ratio of the two $T$ values. $\frac{T ( 1.0 )}{T ( 0.5 )} = \frac{G e ^{- 10.24}}{G e ^{- 5.12}} = e^{- 5.12} \approx 0.00598.$ Why this step? Dividing kills $G$ entirely — the sensitivity is pure exponential, independent of the prefactor.

Step 3 — Invert. $1/0.00598 \approx 167$ . Why this step? We want "how many times smaller", so we invert.

Verify: Doubling the width suppresses $T$ by ~167×, not 2×. This is exactly why a Scanning Tunneling Microscope resolves single atoms: a height bump of $0.1$ nm changes the tunneling current by an order of magnitude. Units cancel in a ratio ✓.

Example 4 — Cells D & E: the two degenerate limits

Now the delicate cases the naive user forgets: what happens as the barrier disappears? There are two ways it can disappear — the wall gets thin ( $L \to 0$ ) or the particle's energy climbs to the wall top ( $E \to V_{0}$ ). The approximation $G e^{- 2 κ L}$ is illegal in both (it needs $κ L ≫ 1$ ), so we must use the exact $sinh$ form. This is where $sinh$ earns its place.

Worked example Cell D — energy creeps to the top,

$E \to V_{0}^{-}$ Fix $L$ and let $E$ approach $V_{0}$ from below. What does $T$ approach? (Use $V_{0} = 2$ eV, $L = 0.5$ nm as before.)

Forecast: The wall is barely too tall now. Guess: does $T \to 1$ , or to some number below 1?

Step 1 — Watch $κ$ . $κ = 2 m (V_{0} - E) /ℏ \to 0$ as $E \to V_{0}$ . Why this step? The height the wave must fight, $V_{0} - E$ , shrinks to zero — the decay rate vanishes.

Step 2 — Small- $κ L$ form of the correction. With $κ L$ tiny, $sinh (κ L) \approx κ L$ , so $sinh^{2} (κ L) \approx κ^{2} L^{2} = \frac{2 m ( V _{0} - E )}{ℏ ^{2}} L^{2}$ . Why this step? We must resolve a " $0/0$ " — the numerator has $(V_{0} - E)$ from $sinh^{2}$ and the denominator has $(V_{0} - E)$ too, so they cancel cleanly.

Step 3 — Cancel. $\frac{V _{0}^{2} s i n h ^{2} ( κ L )}{4 E ( V _{0} - E )} \approx \frac{V _{0}^{2} \cdot \frac{2 m ( V _{0} - E )}{ℏ ^{2}} L ^{2}}{4 E ( V _{0} - E )} = \frac{V _{0}^{2} m L ^{2}}{2 E ℏ ^{2}} .$ As $E \to V_{0}$ : $\to \frac{V _{0} m L ^{2}}{2 ℏ ^{2}}$ . Why this step? This is the finite leftover after the $(V_{0} - E)$ 's cancel — $T$ does not blow up or vanish.

Step 4 — Number. $\frac{V _{0} m L ^{2}}{2 ℏ ^{2}}$ with $V_{0} = 2 \times 1.602 \times 1 0^{- 19}$ J, $L = 0.5 \times 1 0^{- 9}$ m: $= \frac{( 3.204 \times 1 0 ^{- 19} ) ( 9.11 \times 1 0 ^{- 31} ) ( 0.5 \times 1 0 ^{- 9} ) ^{2}}{2 ( 1.055 \times 1 0 ^{- 34} ) ^{2}} \approx 3.28.$ So $T \to [1 + 3.28]^{- 1} \approx 0.234$ , i.e. 23%. Why this step? Even with the wall barely taller than the particle, transmission is only ~23% for this width — a striking, non-obvious limiting value.

Verify: $T$ stays finite and $< 1$ as $E \to V_{0}$ , as it must (a wall of finite width still reflects some wave even when the particle nearly clears it). No division-by-zero blowup — the cancellation in Step 2 was the whole point.

Example 5 — Cell F: heavy particle (mass dependence)

Worked example Proton instead of electron

Repeat Ex 1's numbers ( $E = 1$ eV, $V_{0} = 2$ eV, $L = 0.5$ nm) but with a proton, $m_{p} = 1836 m_{e}$ .

Forecast: Heavier particle. Guess: does $T$ drop by a factor of 1836, or far more?

Step 1 — How $κ$ scales. $κ \propto m$ , so $κ_{p} = 1836 κ_{e} \approx 42.85 κ_{e}$ . Why this step? Mass enters only through $m$ ; scaling beats recomputing from scratch.

Step 2 — New exponent. From Ex 1, electron's $2 κ_{e} L = 5.12$ . So $2 κ_{p} L = 42.85 \times 5.12 \approx 219$ . Why this step? The exponent, not the prefactor, is where mass detonates.

Step 3 — Estimate $T$ . $T \approx 4 e^{- 219} \approx 4 \times (2 \times 1 0^{- 96}) \approx 1 0^{- 95}$ . Why this step? This is effectively zero — a proton simply does not tunnel this wall.

Verify: $e^{- 219}$ is unimaginably small. This is why macroscopic objects never tunnel: mass sits under a square root inside an exponential, so even 1836× the mass (still tiny!) sends $T$ from 2% to $1 0^{- 95}$ . Units: $m$ scaling is dimensionless in the ratio ✓.

Example 6 — Cell G: raising the barrier height

Worked example Taller wall

Electron, $E = 1$ eV, $L = 0.5$ nm, but raise $V_{0}$ from $2$ eV to $5$ eV. Find $T$ and compare to Ex 1.

Forecast: Wall more than doubled in height. Guess the order of magnitude of $T$ .

Step 1 — New $κ$ . $V_{0} - E = 4$ eV $= 6.408 \times 1 0^{- 19}$ J. $κ = \frac{2 ( 9.11 \times 1 0 ^{- 31} ) ( 6.408 \times 1 0 ^{- 19} )}{1.055 \times 1 0 ^{- 34}} \approx 1.024 \times 1 0^{10} m^{- 1} .$ Why this step? Both $G$ and the exponent depend on $V_{0}$ ; $κ$ changes because $V_{0} - E$ grew.

Step 2 — Exponent. $2 κ L = 2 (1.024 \times 1 0^{10}) (0.5 \times 1 0^{- 9}) = 10.24$ . Why this step? Note it doubled from Ex 1's $5.12$ because $V_{0} - E$ quadrupled and $κ \propto V_{0} - E$ .

Step 3 — Prefactor. $G = \frac{16 ( 1 ) ( 4 )}{5 ^{2}} = \frac{64}{25} = 2.56$ . Why this step? $G$ shrank (from 4) because the $V_{0}^{2}$ in the denominator grew faster than the numerator.

Step 4 — Combine. $T \approx 2.56 e^{- 10.24} \approx 2.56 (3.58 \times 1 0^{- 5}) \approx 9.2 \times 1 0^{- 5}$ , i.e. 0.009%. Why this step? Delivers the final number.

Verify: From $2.4%$ (Ex 1) down to $0.009%$ — a ~260× drop — from raising the wall alone. Confirms the parent-note warning that " $T$ depends only on width" is wrong: height matters just as strongly through $κ$ .

Example 7 — Cell H: real-world STM word problem

Worked example Reading an atom with current

In a Scanning Tunneling Microscope the tunneling current $I \propto T \approx G e^{- 2 κ L}$ , where $L$ is the tip–surface gap. For an electron with $V_{0} - E \approx 4$ eV (typical work-function gap), by what factor does the current change when the tip moves $0.1$ nm closer?

Forecast: $0.1$ nm is a fraction of an atom's size. Guess: does current change by a few percent, or a factor of several?

Step 1 — Compute $κ$ . With $V_{0} - E = 4$ eV, from Ex 6 $κ \approx 1.024 \times 1 0^{10} m^{- 1}$ . Why this step? $κ$ sets how sharply current responds to gap changes.

Step 2 — Current ratio for a gap change $Δ L$ . $\frac{I ( L - Δ L )}{I ( L )} = e^{- 2 κ (L - Δ L)} / e^{- 2 κ L} = e^{+ 2 κ Δ L} .$ Why this step? Moving closer by $Δ L$ shortens the tunnel path; the prefactor $G$ (same $E, V_{0}$ ) cancels.

Step 3 — Number. $2 κ Δ L = 2 (1.024 \times 1 0^{10}) (0.1 \times 1 0^{- 9}) = 2.048$ , so factor $= e^{2.048} \approx 7.75$ . Why this step? Turns the physics into a measurable multiplier.

Verify: A mere $0.1$ nm approach makes the current jump ~7.8×. That is why an STM feels single-atom bumps: the exponential turns sub-atomic height differences into large, easily-measured current swings. Units: $κ Δ L$ dimensionless ✓, ratio dimensionless ✓.

Example 8 — Cell I: exam twist, $E > V_{0}$ (formula breaks)

Worked example "Just plug in

$E > V_{0}$ "? A student sets $E = 3$ eV, $V_{0} = 2$ eV (so $E > V_{0}$ ) into the tunneling formula. What goes wrong, and what really happens?

Forecast: The particle has more energy than the wall is tall. Classically it always passes. Guess: is $T$ exactly 1?

Step 1 — Look at $κ$ . $κ = 2 m (V_{0} - E) /ℏ$ with $V_{0} - E = - 1$ eV $< 0$ . Why this step? A negative inside the square root means $κ$ becomes imaginary — the $E < V_{0}$ formula is out of its domain.

Step 2 — Convert $κ$ to a real wavenumber. Write $κ = i k^{'}$ with $k^{'} = 2 m (E - V_{0}) /ℏ$ real. Why this step? Imaginary decay = real oscillation. Inside the barrier the wave now oscillates instead of decaying.

Step 3 — $sinh$ becomes $sin$ . Using $sinh (i u) = i sin (u)$ , so $sinh^{2} (κ L) = - sin^{2} (k^{'} L)$ : $T = [1 + \frac{V _{0}^{2} s i n ^{2} ( k ^{'} L )}{4 E ( E - V _{0} )}]^{- 1} .$ Why this step? The correct over-barrier formula — and it is oscillatory in $L$ .

Step 4 — The resonance. Whenever $k^{'} L = nπ$ ( $n = 1, 2, \dots$ ), $sin = 0$ and $T = 1$ exactly (perfect transmission). At other widths $T < 1$ even though $E > V_{0}$ ! Why this step? This is the Ramsauer–Townsend effect: over a barrier, quantum reflection still happens except at resonant widths.

Verify: So $T$ is not always 1 for $E > V_{0}$ — quantum mechanics reflects some wave even over a short wall, with perfect-transmission windows at $k^{'} L = nπ$ . Sanity: at $sin = 0$ , $T = [1 + 0]^{- 1} = 1$ ✓; at $sin^{2} = 1$ with $E = 3, V_{0} = 2$ , $T = [1 + \frac{4}{4 ( 3 ) ( 1 )}]^{- 1} = [1 + \frac{1}{3}]^{- 1} = 0.75$ , genuinely below 1 ✓. Never blindly plug $E > V_{0}$ into the $sinh$ form.

Recall Scenario-matrix self-test

Cover the "Covered by" column and name which example handles each cell. Thin barrier check ::: Ex 1 (approx vs exact agree ~2%) Thick barrier ::: Ex 2 (approximation exact to many digits) Doubling width ::: Ex 3 (~167× drop) $E \to V_{0}$ limit ::: Ex 4 (finite $T \approx 0.23$ , no blowup) $L \to 0$ limit ::: Ex 4 ( $T \to 1$ ) Heavy particle ::: Ex 5 ( $T \sim 1 0^{- 95}$ ) Raising $V_{0}$ ::: Ex 6 (~260× drop) STM word problem ::: Ex 7 (~7.8× current per 0.1 nm) $E > V_{0}$ twist ::: Ex 8 ( $sinh \to sin$ , resonances)