Classical physics under stress

Blackbody radiation and the ultraviolet catastrophe. Classical (Rayleigh–Jeans) theory gave every electromagnetic mode inside a hot cavity an equal share $k_BT$ of thermal energy. Since a cavity supports unlimited high-frequency modes, this predicted infinite radiated energy at short wavelengths — an absurdity nicknamed the ultraviolet catastrophe. Planck resolved it by postulating that a cavity oscillator can only exchange energy in discrete packets,

where: $E$ is the oscillator's energy, $f$ is its frequency, $n$ is a non-negative integer (the quantum number), and $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ is Planck's constant. When $hf \gg k_BT$, exciting even one quantum costs far more energy than is thermally available, so high-frequency modes are exponentially suppressed rather than each carrying $k_BT$ — the catastrophe simply disappears.

The photoelectric effect. Shining light on a metal surface can eject electrons, but classical wave theory predicts two things that experiment flatly denies: a time delay before weak light accumulates enough energy to eject an electron, and a maximum kinetic energy that grows with the light's intensity. Instead, ejection is instantaneous, there is a sharp threshold frequency $f_0$ below which nothing is ejected no matter how bright the light, and the maximum kinetic energy grows only with frequency. Einstein explained this by treating light itself as a stream of quanta, each of energy $hf$, absorbed one at a time by one electron:

where: $V_0$ is the stopping potential needed to halt the fastest ejected electrons, $\phi$ is the metal's work function (the minimum energy binding an electron to the surface), and the threshold condition is $hf_0 = \phi$.

The Compton effect. X-rays scattered off free (or loosely bound) electrons return with a longer wavelength than they arrived with, and the shift depends only on the scattering angle — never on the incident wavelength or the target material. That is inexplicable for a classical wave, which should simply re-radiate at the same frequency it absorbs. It is natural, however, if the photon behaves as a particle carrying momentum

and the photon-electron collision conserves momentum and energy exactly as a two-body particle collision would. Carrying this through (relativistically, and not required here) gives the Compton shift formula

where: $\Delta\lambda$ is the wavelength shift, $\theta$ is the scattering angle, $m_e = 9.11\times10^{-31}\ \text{kg}$ is the electron mass, and $c = 3\times10^8\ \text{m/s}$. The quantity $h/m_ec$ is a fixed length (about $2.43\ \text{pm}$) called the Compton wavelength of the electron.

The Bohr model, offered as historical motivation, quantized the hydrogen atom's orbital angular momentum, $L = n\hbar$, and from that single postulate correctly predicted the discrete spectral lines of hydrogen — another result classical physics, which allows any orbit and any energy, could not deliver.

A metal has work function $\phi = 2.3\ \text{eV}$. Light of wavelength $\lambda = 400\ \text{nm}$ is incident on it. Find the photon energy in eV, whether photoelectrons are emitted, and their maximum kinetic energy in eV.

Step 1. Find the photon's frequency and energy. $f = c/\lambda = (3\times10^8)/(400\times10^{-9}) = 7.5\times10^{14}\ \text{Hz}$, so $E_{photon} = hf = (6.63\times10^{-34})(7.5\times10^{14}) = 4.97\times10^{-19}\ \text{J}$. Converting with $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ gives $E_{photon} = 3.11\ \text{eV}$.

Step 2. Compare to the work function. Since $E_{photon} = 3.11\ \text{eV}$ exceeds $\phi = 2.3\ \text{eV}$, the photon carries enough energy to free an electron, so photoelectrons ARE emitted.

Step 3. Apply the photoelectric equation to find the maximum kinetic energy: $K_{max} = E_{photon} - \phi = 3.11 - 2.3 = 0.81\ \text{eV}$. The ejected electrons leave with at most 0.81 eV of kinetic energy, a value fixed entirely by frequency and unaffected by how bright the light source is.

Competitive stretch

The same beam is now replaced by an X-ray photon of wavelength $\lambda = 0.100\ \text{nm}$ that Compton-scatters off a free electron through $\theta = 90°$. Using $\Delta\lambda = (h/m_ec)(1-\cos\theta)$ with $\cos 90° = 0$, the Compton wavelength $h/m_ec = (6.63\times10^{-34})/[(9.11\times10^{-31})(3\times10^8)] = 2.43\times10^{-12}\ \text{m}$, so $\Delta\lambda = 2.43\ \text{pm}$, a fractional shift of $\Delta\lambda/\lambda = 2.43\times10^{-12}/1.00\times10^{-10} = 2.43\%$. This shift depends only on $\theta$, never on $\lambda$ or the target — the fingerprint of photons colliding like particles.

• Thinking a brighter light alone can eject photoelectrons below the threshold frequency — intensity sets the number of photons arriving per second, not each individual photon's energy $hf$; only frequency decides whether ejection is possible at all.
• Confusing the stopping potential equation $eV_0 = hf - \phi$ with the photon's total energy — $eV_0$ measures only the maximum kinetic energy left over after freeing the electron, not the photon energy itself.
• Assuming the Compton shift $\Delta\lambda$ depends on the incident wavelength $\lambda$ — it depends only on the scattering angle $\theta$, a hallmark of a genuine particle collision rather than classical scattering.
• Treating Planck's quantization $E = nhf$ as a mathematical trick rather than a physical postulate — it is the quantization of energy exchange itself, not any property of classical waves, that removes the ultraviolet catastrophe.

Ask the tutor: want a walk-through of why intensity never changes a single photon's energy, or another Compton-shift calculation at a different angle?