01Quantum Physics
Explore the photoelectric effect, photon energy, wave-particle duality, the de Broglie wavelength, and the Heisenberg uncertainty principle.
Quantum physics explains phenomena that classical physics cannot — from the photoelectric effect to the behavior of electrons in atoms. It is the foundation of modern technology.
Why can't classical physics explain the behavior of light and matter at the atomic scale — and what does quantum mechanics reveal instead?
At the turn of the 20th century, physicists encountered phenomena that defied every classical prediction: hot objects emitted light in patterns that classical wave theory could not reproduce, and shining light on metal surfaces ejected electrons in ways that depended on frequency rather than intensity. Max Planck resolved the first puzzle in 1900 by proposing that energy is quantized — emitted and absorbed only in discrete packets called quanta. Albert Einstein extended this idea in 1905 to explain the photoelectric effect, establishing that light itself consists of particle-like packets of energy called photons. Louis de Broglie then turned the argument around: if waves can behave like particles, perhaps particles can behave like waves. Werner Heisenberg formalized the limits of simultaneous measurement with his uncertainty principle, and Erwin Schrödinger provided the mathematical framework — the wave function — that describes quantum systems probabilistically. Together these ideas form the foundation of modern physics, underpinning semiconductors, lasers, MRI machines, and our entire understanding of atomic structure.
Classical vs. Quantum Physics
Classical Physics — Assumptions
- •Energy is continuous — any value is allowed
- •Light is purely a wave (electromagnetic)
- •Particles have definite position and momentum simultaneously
- •Measurement does not disturb the system
- •Brighter light → more energy per photon
- •Objects follow deterministic trajectories
Quantum Physics — Revelations
- •Energy is quantized: E = nhf (n = 1, 2, 3…)
- •Light has particle-like properties (photons)
- •ΔxΔp ≥ ℏ/2 — uncertainty is fundamental
- •Measurement collapses the wave function
- •Higher frequency → more energy per photon
- •Outcomes are probabilistic, not deterministic
Key Equations
E = hfKEₘₐₓ = hf − φf₀ = φ / hp = h / λ = E / cλᵈᴹ = h / mvΔλ = (h / mₑc)(1 − cosθ)ΔxΔp ≥ ℏ / 2ΔEΔt ≥ ℏ / 2Eₙ = n²h² / 8mL²eVₛ = KEₘₐₓh = 6.626 × 10⁻³⁴ J·s | ℏ = h/2π = 1.055 × 10⁻³⁴ J·s | m_e = 9.11 × 10⁻³¹ kg | c = 3 × 10⁸ m/s
Worked Examples
Calculate the energy of a photon with frequency f = 6 × 10¹⁴ Hz. Express your answer in both joules and electron-volts.
Identify the equation: E = hf
Substitute values: E = (6.626 × 10⁻³⁴ J·s)(6 × 10¹⁴ Hz)
Multiply: E = 6.626 × 6 × 10⁻³⁴⁺¹⁴ = 39.756 × 10⁻²⁰ J
E = 3.976 × 10⁻¹⁹ J
Convert to eV: E = 3.976 × 10⁻¹⁹ J ÷ 1.602 × 10⁻¹⁹ J/eV
E ≈ 2.485 eV
Light of frequency f = 8 × 10¹⁴ Hz strikes a metal with work function φ = 2.3 eV. Find (a) the photon energy, (b) the maximum kinetic energy of ejected electrons, and (c) the stopping potential.
Find photon energy: E = hf = (6.626 × 10⁻³⁴)(8 × 10¹⁴) = 5.301 × 10⁻¹⁹ J
Convert to eV: 5.301 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 3.31 eV
Apply photoelectric equation: KE_max = hf − φ = 3.31 eV − 2.3 eV = 1.01 eV
Stopping potential: eV_s = KE_max → V_s = 1.01 eV / e = 1.01 V
A metal has a work function φ = 4.5 eV. Find the threshold frequency f₀ below which no electrons are ejected.
Convert work function to joules: φ = 4.5 eV × 1.602 × 10⁻¹⁹ J/eV = 7.209 × 10⁻¹⁹ J
Use threshold frequency formula: f₀ = φ / h
Substitute: f₀ = 7.209 × 10⁻¹⁹ J ÷ 6.626 × 10⁻³⁴ J·s
f₀ = 1.088 × 10¹⁵ Hz
This lies in the ultraviolet region — visible light cannot eject electrons from this metal.
An electron (m = 9.11 × 10⁻³¹ kg) moves at v = 2 × 10⁶ m/s. Calculate its de Broglie wavelength.
Use de Broglie relation: λ = h / mv
Calculate momentum: p = mv = (9.11 × 10⁻³¹ kg)(2 × 10⁶ m/s) = 1.822 × 10⁻²⁴ kg·m/s
Calculate wavelength: λ = h / p = 6.626 × 10⁻³⁴ J·s ÷ 1.822 × 10⁻²⁴ kg·m/s
λ = 3.637 × 10⁻¹⁰ m
Convert: λ = 0.364 nm — comparable to atomic spacing, confirming electron diffraction is observable.
An electron is confined to a one-dimensional box of length L = 1 nm. Calculate the ground-state (n = 1) energy.
Use particle-in-a-box formula: E_n = n²h² / (8mL²)
For ground state n = 1: E₁ = (1)²(6.626 × 10⁻³⁴)² / (8 × 9.11 × 10⁻³¹ × (1 × 10⁻⁹)²)
Numerator: (6.626 × 10⁻³⁴)² = 4.390 × 10⁻⁶⁷ J²·s²
Denominator: 8 × 9.11 × 10⁻³¹ × 10⁻¹⁸ = 7.288 × 10⁻⁴⁸ kg·m²
E₁ = 4.390 × 10⁻⁶⁷ ÷ 7.288 × 10⁻⁴⁸ = 6.024 × 10⁻²⁰ J
Convert to eV: 6.024 × 10⁻²⁰ ÷ 1.602 × 10⁻¹⁹ = 0.376 eV
Guided Practice
X-ray photons have a wavelength of 0.10 nm. Calculate (a) the frequency of the X-rays and (b) the energy of each photon in joules and eV.
Hint: Use c = fλ to find frequency first, then E = hf. Remember 1 eV = 1.602 × 10⁻¹⁹ J.
Ultraviolet light of wavelength 200 nm strikes a sodium surface with work function φ = 2.28 eV. Find the maximum speed of the ejected electrons.
Hint: Find E_photon = hc/λ, then KE_max = hf − φ. Finally use KE_max = ½mv² to solve for v, with m_e = 9.11 × 10⁻³¹ kg.
In a Compton scattering experiment, an X-ray photon is scattered at θ = 90°. Calculate the Compton wavelength shift Δλ.
Hint: Use Δλ = (h/m_ec)(1 − cosθ). At θ = 90°, cos 90° = 0, so Δλ = h/m_ec. The Compton wavelength h/m_ec = 2.426 × 10⁻¹² m.
A proton (m = 1.673 × 10⁻²⁷ kg) has a de Broglie wavelength of 1.0 × 10⁻¹⁰ m. What is its kinetic energy in eV?
Hint: Find momentum from λ = h/p → p = h/λ. Then KE = p²/2m. Convert to eV at the end.
An electron's position is measured with an uncertainty of Δx = 0.50 nm. What is the minimum uncertainty in its momentum? What does this imply about its speed?
Hint: Use ΔxΔp ≥ ℏ/2, so Δp_min = ℏ/(2Δx). Then Δv = Δp/m_e. ℏ = 1.055 × 10⁻³⁴ J·s.
Key Vocabulary
Photon
A discrete quantum (packet) of electromagnetic energy. Photons are massless particles that travel at the speed of light and carry energy E = hf and momentum p = h/λ.
Example: A green photon (f ≈ 5.5 × 10¹⁴ Hz) carries about 2.3 eV of energy — just enough to eject electrons from some metals.
Photoelectric effect
The emission of electrons from a metal surface when light of sufficient frequency strikes it. The effect cannot be explained by classical wave theory — only the photon model accounts for the frequency threshold.
Example: Zinc emits electrons when illuminated by UV light but not by visible light, regardless of intensity.
Work function (φ)
The minimum energy required to liberate an electron from the surface of a metal. It is a material-specific constant measured in electron-volts.
Example: Cesium has φ ≈ 2.0 eV (ejects electrons with visible light); platinum has φ ≈ 5.7 eV (requires UV).
Threshold frequency (f₀)
The minimum photon frequency needed to eject an electron from a metal surface. Below f₀, no electrons are emitted regardless of light intensity. f₀ = φ/h.
Example: For sodium (φ = 2.28 eV), f₀ = 5.51 × 10¹⁴ Hz — just at the edge of the visible spectrum.
Wave-particle duality
The principle that all matter and radiation exhibit both wave-like and particle-like properties. Which behavior is observed depends on the experimental setup.
Example: Electrons produce interference patterns (wave behavior) in double-slit experiments but land at discrete points (particle behavior) on a detector.
de Broglie wavelength
The wavelength associated with a moving particle, given by λ = h/p = h/mv. Larger momentum means shorter wavelength, making quantum effects negligible for macroscopic objects.
Example: A baseball (0.145 kg at 40 m/s) has λ ≈ 10⁻³⁴ m — far too small to observe. An electron at 10⁶ m/s has λ ≈ 0.7 nm — comparable to atomic spacings.
Heisenberg uncertainty principle
A fundamental limit on the precision with which complementary quantities (position–momentum, energy–time) can be simultaneously known: ΔxΔp ≥ ℏ/2. This is not a measurement limitation — it is an intrinsic property of quantum systems.
Example: Confining an electron to a smaller box (smaller Δx) necessarily increases the spread in its momentum (larger Δp).
Quantum tunneling
The quantum mechanical phenomenon in which a particle passes through a potential energy barrier that it classically could not surmount. It arises because the wave function does not drop to zero at the barrier boundary.
Example: Alpha decay, tunnel diodes, and scanning tunneling microscopes all rely on quantum tunneling.
Workbook Check
Interactive Practice — 5 Questions
Which of the following correctly describes the photoelectric effect?
A photon has energy E = 5.0 eV. What is its frequency?
The de Broglie wavelength of a particle is inversely proportional to its:
Which statement about the Heisenberg uncertainty principle is correct?
In the particle-in-a-box model, if the box length L is halved, the ground-state energy E₁:
Independent Practice
Calculate the energy of a photon with frequency 6.0 × 10¹⁴ Hz. Express your answer in both joules and eV. (h = 6.626 × 10⁻³⁴ J·s, 1 eV = 1.6 × 10⁻¹⁹ J)
Light of wavelength 400 nm strikes a metal with work function 2.0 eV. Find (a) the photon energy in eV and (b) the maximum kinetic energy of the emitted electrons.
Calculate the de Broglie wavelength of an electron moving at 2.0 × 10⁶ m/s. (m_e = 9.11 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ J·s)
An electron is confined to a region of width Δx = 1.0 × 10⁻¹⁰ m. Use the Heisenberg uncertainty principle to find the minimum uncertainty in its momentum and speed.
★ In a photoelectric experiment, light of wavelength 300 nm is shone on a sodium surface (work function = 2.28 eV). (a) Calculate the photon energy in eV. (b) Find the maximum kinetic energy of emitted electrons. (c) Find the stopping potential. (d) Calculate the maximum speed of the emitted electrons. (e) What is the threshold wavelength for sodium? (h = 6.626 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s, m_e = 9.11 × 10⁻³¹ kg)
ChallengeCommon Mistakes
Thinking brighter light (higher intensity) ejects faster electrons in the photoelectric effect
Electron KE depends only on FREQUENCY (E = hf − φ), not intensity. Higher intensity means MORE electrons ejected, not faster ones
Confusing the work function φ (energy to remove an electron) with the threshold frequency f₀
They are related: φ = hf₀. The work function is in joules or eV; threshold frequency is in Hz. Both describe the minimum energy needed to eject an electron
Using the de Broglie formula λ = h/mv with the wrong mass (e.g., using atomic mass in kg without converting from amu)
Always use mass in kg. 1 amu = 1.66×10⁻²⁷ kg. Electron mass = 9.11×10⁻³¹ kg. Proton mass = 1.67×10⁻²⁷ kg
Thinking the uncertainty principle is just about measurement error or instrument limitations
The uncertainty principle ΔxΔp ≥ ℏ/2 is a fundamental property of quantum systems — not a limitation of our tools. A particle genuinely does not have a precise position AND momentum simultaneously
Forgetting to convert photon energy from joules to eV (or vice versa) when comparing to work functions
1 eV = 1.6×10⁻¹⁹ J. Work functions are typically given in eV; photon energies from E = hf are in joules. Convert before comparing
Math Tips
Photon energy shortcut in eV: E(eV) = 1240/λ(nm). Visible light at 500 nm → E = 1240/500 = 2.48 eV. Memorize this formula for quick calculations
Stopping potential: eV_s = KE_max = hf − φ. The stopping potential in volts numerically equals KE_max in eV. No unit conversion needed when both are in eV
de Broglie wavelength for an electron accelerated through voltage V: λ = h/√(2meV). At V = 100V: λ ≈ 0.123 nm — comparable to atomic spacings, explaining electron diffraction
Particle-in-a-box energy levels: E_n = n²h²/(8mL²) = n²E₁. The energy levels scale as n². The gap between levels n and n+1 is (2n+1)E₁
Heisenberg uncertainty: ΔxΔp ≥ ℏ/2 ≈ 5.3×10⁻³⁵ J·s. If Δx = 1 nm (10⁻⁹ m), then Δp ≥ 5.3×10⁻²⁶ kg·m/s — significant for electrons, negligible for macroscopic objects