Quantum Mechanics 101 · Unit 3: The Mathematical Machinery

Quantum Academy · Chapter 12: Quantum Tunneling

Walking through walls you don't have the energy to climb

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Instructions. Answer every question with the lesson closed, then check your work. Use ħ = 1.055×10⁻³⁴ J·s, electron mass 9.109×10⁻³¹ kg, 1 eV = 1.602×10⁻¹⁹ J and Boltzmann's constant k = 8.617×10⁻⁵ eV per kelvin. Once you have derived it, you may use the shortcut that an electron's κ is 5.12 per nanometre times the square root of the energy deficit in eV. Show every step in the calculation questions. One point for each vocabulary match and each question; the challenge is a bonus.

Part A Vocabulary matching

Write the letter of the matching definition in the blank beside each term.

1.Potential barrier
2.Classically forbidden region
3.Turning point
4.Decay constant
5.Transmission probability
6.Quantum tunnelling
7.WKB approximation
8.Coulomb barrier
9.Gamow peak
10.Work function
11.Josephson junction
12.Macroscopic quantum tunnelling

A.κ, with ħκ = √(2m(V − E)); the wavefunction in a barrier fades as e^(−κx), and 1/κ is the decay length.
B.A method, introduced in 1926 by Wentzel, Kramers and Brillouin, that estimates tunnelling through smooth barriers by adding up κ(x) across the forbidden region.
C.The energy hill created by electric repulsion between positively charged nuclear particles, which an alpha particle tunnels out through and fusing protons tunnel in through.
D.Tunnelling of a collective variable describing enormous numbers of particles, such as the state of a whole superconducting circuit; recognized by the 2025 Nobel Prize in Physics.
E.The position where E = V, at which a classical particle stops and reverses.
F.The minimum energy needed to remove an electron from a metal surface; it sets the barrier height in a scanning tunnelling microscope.
G.T, the chance that a particle sent at a barrier is found beyond it; with the reflection probability R, R + T = 1.
H.A region where the potential energy rises above a particle's total energy, over a finite width.
I.A region where V exceeds E, which a classical particle can never enter because its kinetic energy would be negative.
J.Two superconductors separated by a thin insulating barrier through which Cooper pairs tunnel; the heart of superconducting qubits and voltage standards.
K.The narrow energy window in which fusion reactions in a star actually occur, where the falling number of fast particles balances their rising tunnelling probability; near 6 keV in the Sun's core.
L.Passage of a particle through a classically forbidden barrier, possible because its wavefunction fades inside the barrier rather than vanishing.

Part B Questions

1.
A thick barrier transmits a tiny fraction of the particles that hit it. Which single change raises the transmission probability the most?Circle one
1. AHalving the particle's mass
2. BHalving the energy deficit V − E
3. CHalving the barrier width
4. DDoubling the particle's mass
2.
What does the wavefunction of a particle with E below V do inside a wide barrier?Circle one
1. AIt decays exponentially with distance
2. BIt drops to exactly zero at the edge
3. CIt oscillates with a shorter wavelength
4. DIt grows without limit
3.
An STM operates with κ = 10 per nanometre, and its tunnelling current suddenly doubles. What happened to the tip-surface gap?Circle one
1. AIt grew by about 0.035 nm
2. BIt shrank by about 0.035 nm
3. CIt shrank by about 0.35 nm
4. DIt was cut in half
4.
A particle that tunnels through a barrier that does not change in time emerges with less energy than it had before.Circle one
TrueFalse
5.
Without quantum tunnelling, hydrogen fusion in the Sun's core would essentially stop.Circle one
TrueFalse
6.
The 2020 Toronto experiment found that rubidium atoms cross a tunnelling barrier in zero time.Circle one
TrueFalse
7.
The distance 1/κ over which ψ shrinks by a factor of e inside a barrier is called the length.Fill in the blank
8.
In 1928 George Gamow explained decay as tunnelling through the Coulomb barrier of the nucleus.Fill in the blank
9.
Alpha energies of common emitters differ by only about a factor of two, yet their half-lives differ by more than twenty orders of magnitude. Explain why in two or three sentences.Short answer
10.
A popular explanation says a tunnelling particle briefly borrows the energy it needs from the uncertainty principle. In two or three sentences, explain what is wrong with this and what actually makes tunnelling possible.Short answer
11.
An electron with E = 2.0 eV meets a rectangular barrier of height 5.0 eV and width 0.40 nm. Find κ, the bare exponential e^(−2κL), and the transmission probability including the thick-barrier prefactor.Show your working
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Answer:
12.
Barrier design: an electron with E = 1.0 eV meets a 2.0 eV barrier. Using the thick-barrier formula, what width gives a transmission probability of exactly 1 percent?Show your working
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Answer:
13.
The Sun's core is at 1.571×10⁷ K. Find the thermal energy scale kT in keV, the ratio of a 1 MeV Coulomb barrier to kT, and the Boltzmann factor e^(−1 MeV/kT) as a power of ten.Show your working
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Answer:
14.
Treat a 70 kg person as a single particle facing a wall 0.10 m thick with an energy deficit of just 1.0 J. Estimate 2κL and express T as a power of ten.Show your working
Working
Answer:

Part C Mad scientist challenge bonus

Reproduce Gamow's 1928 triumph. For an alpha particle of energy Q escaping a daughter nucleus of charge number Z, the barrier runs from the nuclear radius R out to the turning point b = 2Z × 1.440 MeV·fm ÷ Q, and the WKB exponent is G = (2b/ħ)√(2μQ) × [arccos√(R/b) − √((R/b)(1 − R/b))], with T = e^(−G). Use an alpha mass of 6.645×10⁻²⁷ kg, reduced mass μ = (alpha mass) × A ÷ (A + 4) for a daughter of mass number A, 1 MeV = 1.602×10⁻¹³ J, and a strike rate of 10²¹ per second. (a) Uranium-238: Q = 4.27 MeV, Z = 90, A = 234, R = 9.3 fm. Find b, G, T and the predicted half-life. (b) Polonium-212: Q = 8.95 MeV, Z = 82, A = 208, R = 9.0 fm. (c) Compare both with the measured half-lives, 4.47×10⁹ years and 294 ns.

Working

Memory aid

Mnemonic Tunnelling loves the THIN, the LIGHT and the NEARLY THERE. Thin: a narrow barrier, small L. Light: a small mass m. Nearly there: an energy just below the top, so the deficit V − E is small. All three sit together in the exponent, 2κL = 2L√(2m(V − E)) divided by ħ, so getting any one of them wrong is punished exponentially. For the applications, remember FAST JS: Flash memory, Alpha decay, the Sun, Tunnel diodes, Josephson junctions, Scanning tunnelling microscope. And remember the joke hidden in the name: tunnelling is never faster than light. To use it: before any tunnelling problem, ask whether the barrier is thin, the particle light and the energy nearly there. A loud no to any of them means a vanishing transmission.

In one line Where a particle lacks the energy to climb, its wave fades but never stops; thin, light, nearly-there particles leak through with odds of about e^(−2κL).