Failures of Classical Physics

Introduction

By 1900 physics seemed nearly complete. Newton's mechanics and Maxwell's electromagnetism explained motion, heat, light and electricity. Yet three stubborn experimental results refused to fit: the colour distribution of light from hot objects, the ejection of electrons from metals by light, and the sharp lines in atomic spectra. Each failure pointed to the same conclusion — energy is exchanged in discrete packets — and together they launched quantum theory.

Core explanation

Black-body radiation. A hot object glows, and the distribution of energy over wavelengths depends only on its temperature. The peak shifts to shorter wavelength as temperature rises, following Wien's law, λₘₐₓT ≈ 2.90 × 10⁻³ m K. The Sun's surface, at about 5800 K, peaks near 500 nm in the visible; a filament at 3000 K peaks in the infrared near 970 nm.

Classical physics treated the radiation as a collection of electromagnetic oscillators, each carrying an average energy kT regardless of frequency. Because the number of possible oscillators rises steeply with frequency, the Rayleigh–Jeans law predicts that energy density grows without limit at short wavelength. This is the ultraviolet catastrophe : every warm object should blaze with ultraviolet and X-rays.

Planck's solution. In 1900 Max Planck proposed that an oscillator of frequency ν can hold energy only in whole-number multiples of hν, where h = 6.626 × 10⁻³⁴ J s. High-frequency oscillators need a large quantum to be excited at all, so at a given temperature they are rarely excited. The high-frequency emission is suppressed and the predicted curve matches experiment exactly.

The photoelectric effect. When light strikes a clean metal surface, electrons can be ejected. The observations are striking: (i) no electrons appear below a threshold frequency, however intense the light; (ii) above threshold, electrons appear immediately, even at very low intensity; (iii) the maximum kinetic energy of the electrons rises linearly with frequency but does not depend on intensity; (iv) intensity controls only the number of electrons. A classical wave, spreading energy continuously, predicts the opposite on almost every point.

Einstein explained this in 1905 by treating light itself as a stream of photons, each of energy E = hν. One photon ejects one electron, and energy conservation gives:

Eₖ,ₘₐₓ = hν − Φ

where Φ is the work function of the metal. The threshold frequency is ν₀ = Φ/h.

Line spectra. Excited atoms emit light only at particular wavelengths. For hydrogen, all lines fit the Rydberg formula, 1/λ = R H(1/n₁² − 1/n₂²), with R H = 1.097 × 10⁷ m⁻¹. Such sharp lines mean that the atom has discrete energy levels, and each photon carries exactly the energy difference: ΔE = hν. Classical physics has no mechanism that produces discrete levels.

Formulae

Wien's law: λₘₐₓT = 2.90 × 10⁻³ m K. Planck: E = nhν (n = 0, 1, 2, …). Photoelectric: Eₖ,ₘₐₓ = hν − Φ, ν₀ = Φ/h. Rydberg: 1/λ = R H(1/n₁² − 1/n₂²). Bohr condition: ΔE = hν = hc/λ.

Step-by-step reasoning

To analyse a photoelectric problem:

1. Convert the wavelength to photon energy using E = hc/λ. 2. Convert the work function to joules if it is given in eV (1 eV = 1.602 × 10⁻¹⁹ J). 3. If E < Φ, no electrons are ejected. 4. Otherwise subtract: Eₖ,ₘₐₓ = E − Φ. 5. If required, find the speed from Eₖ = ½mv².

Visual explanation

Sketch the black-body curves for 3000 K, 4000 K and 5000 K: each rises from zero, peaks and falls, with higher temperatures giving taller curves peaked further left. Draw the Rayleigh–Jeans curve as a line that climbs without end towards short wavelengths. The gap between them is the ultraviolet catastrophe.

Real-world analogy

A vending machine that accepts only whole coins behaves like a quantised oscillator. If the cheapest item costs a large coin and you hold only small change, you get nothing, however many small coins you have. Likewise, many low-energy photons cannot eject an electron that one high-energy photon can.

Real-world example

Infrared thermometers and thermal cameras measure the radiation emitted by skin or machinery and apply black-body relationships to deduce temperature. Astronomers use the same physics to find the surface temperatures of stars from their colours.

Why?

Why does increasing light intensity not increase the electrons' kinetic energy? Intensity measures the number of photons arriving each second, not the energy of each one. Each electron absorbs one photon, so its energy depends only on that photon's frequency.

Common misconception

"Bright light of any colour will eventually eject electrons." Below the threshold frequency no single photon has enough energy, and electrons do not accumulate energy from several photons under normal conditions, so no emission occurs.

Worked example

Question: Light of wavelength 250 nm falls on a metal with work function 4.30 eV. Find the maximum kinetic energy of the ejected electrons in eV.

Reasoning: E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (250 × 10⁻⁹) = 7.95 × 10⁻¹⁹ J. In eV: 7.95 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 4.96 eV. Eₖ,ₘₐₓ = 4.96 − 4.30 = 0.66 eV.

Answer: About 0.66 eV (1.06 × 10⁻¹⁹ J).

Quick check

1. What did Planck assume about the energy of an oscillator of frequency ν? Answer: That its energy can only take whole-number multiples of hν, so energy is exchanged in discrete quanta.

Exam focus

Examiners expect you to state the four photoelectric observations and explain each with photons. Convert units carefully: nm to m, and eV to J. Remember that a line spectrum is evidence for discrete energy levels, not merely for photons.

Advanced insight

Planck's full distribution law is ρ(λ,T) = (8πhc/λ⁵) × 1/(e^(hc/λkT) − 1). At long wavelengths, where hc/λkT is small, it reduces to the classical Rayleigh–Jeans law — an early example of the correspondence principle, in which quantum results merge smoothly into classical ones in the appropriate limit.

Summary

Three experiments broke classical physics. Black-body radiation required Planck's quantised oscillators with E = nhν to avoid the ultraviolet catastrophe. The photoelectric effect required Einstein's photons, with Eₖ,ₘₐₓ = hν − Φ. Atomic line spectra required discrete energy levels, with ΔE = hν for each line. All three share one lesson: energy at the atomic scale comes in packets.

Practice questions

1. The work function of caesium is 2.1 eV. Calculate its threshold wavelength. Answer: Φ = 2.1 × 1.602 × 10⁻¹⁹ = 3.36 × 10⁻¹⁹ J; λ₀ = hc/Φ = 1.99 × 10⁻²⁵ ÷ 3.36 × 10⁻¹⁹ ≈ 5.9 × 10⁻⁷ m, about 590 nm. 2. Use Wien's law to estimate the peak wavelength emitted by a human body at 310 K. Answer: λₘₐₓ = 2.90 × 10⁻³ ÷ 310 ≈ 9.4 × 10⁻⁶ m, about 9.4 μm, in the infrared. 3. State how doubling the intensity of light above threshold affects (a) the number of photoelectrons and (b) their maximum kinetic energy. Answer: (a) The number roughly doubles; (b) the maximum kinetic energy is unchanged. 4. Calculate the wavelength of the hydrogen line for the transition from n = 3 to n = 2. Answer: 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.524 × 10⁶ m⁻¹, so λ ≈ 656 nm (the red Balmer line).