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University32 min43 / 44

Foundations of quantum physics

Learn about photons, the photoelectric effect, de Broglie waves, the uncertainty principle, the idea of the Schrödinger equation, energy levels in a box and atomic spectra.

Check yourself
In this lesson you will learn
  • Calculate photon energy and momentum and apply Einstein's photoelectric equation
  • Find de Broglie wavelengths and the limits set by the uncertainty principle
  • Derive energy levels in a box and calculate the hydrogen spectrum

At the end of the 19th century classical physics could not explain a few simple experiments: the spectrum of light from hot bodies, light knocking electrons out of metals, and atoms emitting light only in certain colours. This “crisis” led to a new theory — quantum physics. Today lasers, LEDs, solar panels, transistors and MRI scanners all work on quantum principles.

  1. 1900
    Max Planck proposes that radiation is emitted in portions — quanta
  2. 1905
    Einstein explains the photoelectric effect with light quanta — photons
  3. 1913
    Niels Bohr creates his model of the hydrogen atom
  4. 1924
    Louis de Broglie proposes that particles have wave properties
  5. 1926
    Erwin Schrödinger writes down his wave equation
  6. 1927
    Heisenberg's uncertainty principle; electron diffraction is observed

Photons and the photoelectric effect

E = h · f = h · c / λ, p = h / λE = h · f = h · c / λ, p = h / λ
where:
  • hPlanck constant, 6.63 · 10⁻³⁴ J · s
  • ffrequency of the light, in Hz
  • λwavelength, in m

In atomic physics energy is measured in electronvolts: 1 eV = 1.6 · 10⁻¹⁹ J.

Experiments on the photoelectric effect show three facts: the maximum kinetic energy of the electrons depends on the frequency of the light, not its intensity; each metal has a threshold frequency below which no electrons come out at all; and electrons are emitted without delay. Einstein's explanation: one photon gives all its energy to one electron; part is spent escaping from the metal (the work function A), the rest becomes kinetic energy.

h · f = A + Ek,max, Ek,max = e · U₀, f₀ = A / hh · f = A + Ek,max, Ek,max = e · U₀, f₀ = A / h
where:
  • Awork function of the metal, in eV or J
  • U₀stopping voltage — the voltage that stops the photocurrent, in V
  • f₀threshold frequency, in Hz
Example 1: violet light on a metal

A metal with a work function of 2.3 eV is lit with 400 nm light. Find the photon energy, the maximum kinetic energy of the electrons, the stopping voltage and the threshold wavelength.

Show solution
E = hc/λ = 6.63 · 10⁻³⁴ · 3 · 10⁸ / (4 · 10⁻⁷) ≈ 4.97 · 10⁻¹⁹ J ≈ 3.1 eV.
Ek,max = 3.1 − 2.3 = 0.8 eV, so U₀ ≈ 0.8 V.
λ₀ = hc / A ≈ 1243 eV · nm / 2.3 eV ≈ 540 nm — green light. Longer wavelengths (yellow, red) eject no electrons, however bright they are.
Interactive
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Ek,max (eV) as a function of frequency; x is the frequency in units of 10¹⁴ Hz, a is the work function (eV). The slope is h — the same for every metal — and the x-intercept is the threshold frequency; the negative part means “no electrons”.

The wave nature of particles

λ = h / p = h / (m · v), Δx · Δp ≥ ħ / 2, ħ = h / (2π)λ = h / p = h / (m · v), Δx · Δp ≥ ħ / 2, ħ = h / (2π)
where:
  • λde Broglie wavelength, in m
  • Δx, Δpuncertainties in position and momentum
  • ħreduced Planck constant, ≈ 1.055 · 10⁻³⁴ J · s
Example 2: an electron and a ball

a) Find the de Broglie wavelength of an electron accelerated through 100 V. b) What is the wavelength of a 0.1 kg ball moving at 10 m/s?

Show solution
a) Ek = eU = p²/(2m) → p = √(2meU) = √(2 · 9.11 · 10⁻³¹ · 1.6 · 10⁻¹⁹ · 100) ≈ 5.4 · 10⁻²⁴ kg · m/s.
λ = h/p ≈ 6.63 · 10⁻³⁴ / 5.4 · 10⁻²⁴ ≈ 1.2 · 10⁻¹⁰ m = 0.12 nm — about the spacing of atoms, which is why electrons diffract in crystals.
b) λ = 6.63 · 10⁻³⁴ / (0.1 · 10) = 6.6 · 10⁻³⁴ m — far too small for any instrument, so everyday objects show no wave behaviour.
Example 3: can an electron in an atom be at rest?

An electron is confined inside an atom about 10⁻¹⁰ m across. Estimate the minimum uncertainty in its velocity.

Show solution
Δp ≥ ħ / (2Δx) = 1.055 · 10⁻³⁴ / (2 · 10⁻¹⁰) ≈ 5.3 · 10⁻²⁵ kg · m/s.
Δv = Δp / m ≈ 5.3 · 10⁻²⁵ / 9.11 · 10⁻³¹ ≈ 5.8 · 10⁵ m/s.
An electron cannot “sit still” in an atom — this is behind the stability of atoms and zero-point energy.

The Schrödinger equation and a particle in a box

In quantum mechanics a particle's state is described by a wave function ψ(x): |ψ(x)|² dx is the probability of finding the particle between x and x + dx. For stationary states ψ satisfies the Schrödinger equation. In an infinitely deep well of width L, U = 0 inside and ψ(0) = ψ(L) = 0 at the walls. The equation becomes d²ψ/dx² = −k²ψ with k² = 2mE/ħ²; its solution is ψ = C sin kx, and the boundary condition gives kL = nπ. The energy can take only discrete values — it is quantised.

−(ħ² / 2m) · d²ψ/dx² + U(x) · ψ = E · ψ, Eₙ = n² · h² / (8mL²), ψₙ = √(2/L) · sin(nπx / L)−(ħ² / 2m) · d²ψ/dx² + U(x) · ψ = E · ψ, Eₙ = n² · h² / (8mL²), ψₙ = √(2/L) · sin(nπx / L)
where:
  • ψwave function; |ψ|² is the probability density
  • U(x)potential energy, in J
  • nquantum number, n = 1, 2, 3, …
  • Lwidth of the well, in m
Example 4: an electron in a 0.5 nm box

An electron is in an infinitely deep well of width 0.5 nm. Find E₁ and E₂ and the wavelength of the photon emitted in the 2 → 1 transition.

Show solution
E₁ = h² / (8mL²) = (6.63 · 10⁻³⁴)² / (8 · 9.11 · 10⁻³¹ · (5 · 10⁻¹⁰)²) ≈ 2.4 · 10⁻¹⁹ J ≈ 1.5 eV.
E₂ = 4E₁ ≈ 6.0 eV.
ΔE = 3E₁ ≈ 4.5 eV → λ = 1243 / 4.5 ≈ 275 nm (ultraviolet).
The narrower the well, the wider the level spacing — this is how the colour of quantum dots is “tuned”.
Interactive
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Probability density |ψₙ|² = 2 sin²(nπx) in a well of width L = 1. Increase n: there are n peaks and n − 1 nodes. For n = 1 the particle is most likely in the middle and never right at the walls.

Atomic spectra

Eₙ = −13.6 eV / n², h · f = Em − EnEₙ = −13.6 eV / n², h · f = Em − En
where:
  • Eₙenergy of level n in hydrogen (negative — the electron is bound)
  • Em, Enenergies of the upper and lower levels in the transition

Each element emits only photons matching differences between its own levels — its spectral “fingerprint”.

Example 5: the red line of hydrogen

In a hydrogen atom an electron drops from level n = 3 to n = 2. Find the energy and wavelength of the photon.

Show solution
ΔE = 13.6 · (1/4 − 1/9) = 13.6 · 5/36 ≈ 1.89 eV.
λ = 1243 / 1.89 ≈ 658 nm — red light. This is the famous Hα line of the Balmer series, measured at 656 nm (the difference comes from rounding the constants).

Key points

  • Photon: E = hf = hc/λ, p = h/λ; E (eV) ≈ 1240 / λ (nm).
  • Photoelectric effect: hf = A + Ek,max; electron energy depends on frequency, their number on intensity.
  • de Broglie: λ = h/p; uncertainty: Δx · Δp ≥ ħ/2.
  • In an infinite well Eₙ = n²h²/(8mL²); in hydrogen Eₙ = −13.6 eV/n², photon hf = Em − En.

Check yourself

10 questions. Every correct answer earns XP.

1 / 10
What is the approximate energy of a photon with a wavelength of 620 nm?