In Physics 2 you used two rules for glowing objects: hotter objects radiate much more ($P = \sigma AT^4$) and peak at shorter wavelengths ($\lambda_{\max}T = 2.898 \times 10^{-3}$ m·K). Around 1900 these were measured facts that classical physics could not explain. Max Planck's fix, energy that comes only in lumps of $hf$, was the first step into quantum physics, and it also explains the photon.
🔑 Key idea
Planck assumed that the energy of radiation at frequency $f$ can change only in steps of $hf$. Then each mode of radiation in a hot cavity holds, on average, $$\langle E\rangle = \frac{hf}{e^{hf/k_BT} - 1}$$ instead of the classical $k_BT$. This gives the blackbody spectrum $$I(\lambda, T) = \frac{2\pi hc^2}{\lambda^5}\,\frac{1}{e^{hc/\lambda k_BT} - 1},$$ which peaks at $\lambda_{\max} = \frac{2.898 \times 10^{-3}\ \text{m·K}}{T}$ (Wien) and has total power $\sigma T^4$ per square meter (Stefan–Boltzmann).
💡 Why it works
Classical physics gives every mode of the radiation in a hot box the same average energy, $k_BT$. But the number of modes per wavelength interval grows like $1/\lambda^4$ as the wavelength shrinks, so the predicted energy grows without limit toward the ultraviolet: the Rayleigh–Jeans law, which works at long wavelengths and fails badly at short ones. Planck's steps change the average. A mode can hold $0, hf, 2hf, \ldots$, each with Boltzmann weight $e^{-nhf/k_BT}$. Averaging, $$\langle E\rangle = \frac{\sum_n nhf\,e^{-nhf/k_BT}}{\sum_n e^{-nhf/k_BT}} = \frac{hf}{e^{hf/k_BT} - 1}.$$ When $hf \ll k_BT$ this is close to $k_BT$, the classical value. When $hf \gg k_BT$ even one step costs too much, the weight $e^{-hf/k_BT}$ is tiny, and the mode is nearly empty. High frequencies are frozen out, and the spectrum turns down.
Planck curves for the Sun's surface (5772 K) and a cooler star (3500 K), with the classical Rayleigh–Jeans prediction for the first. The classical curve matches at long wavelengths and shoots upward at short ones.Average energy per mode, in units of $k_BT$. Classical physics says 1 for every mode. Planck's steps agree for small $hf/k_BT$ and fall off fast once a single step $hf$ is several times $k_BT$.
Worked examples
✏️ Worked example: The Sun's peak
The Sun's surface is at about 5772 K. Where does its spectrum peak?
Answer: About 502 nm, blue-green light (the Sun still looks white because it emits strongly at all visible wavelengths)
✏️ Worked example: You glow too
Skin is at about 306 K. At what wavelength does it radiate most strongly?
$\lambda_{\max} = \frac{2.898 \times 10^{-3}}{306} = 9.47 \times 10^{-6}$ m.
Answer: About 9.5 μm, in the infrared, which is what thermal cameras see
✏️ Worked example: A frozen-out mode
At $T = 5000$ K, compare the average energy of a mode of wavelength 500 nm with the classical $k_BT$. ($hc = 1240$ eV·nm, $k_B = 8.617 \times 10^{-5}$ eV/K.)
$hf = \frac{1240}{500} = 2.48$ eV and $k_BT = (8.617 \times 10^{-5})(5000) = 0.431$ eV, so $\frac{hf}{k_BT} = 5.76$.
Answer: Under 2% of the classical value: the steps are too big for the heat to fill
⚠️ Common mistake
Thinking the “ultraviolet catastrophe” was something measured. It was a prediction of classical physics: infinite energy in the short-wavelength radiation of any warm object. Experiments showed the opposite, a spectrum that turns down and goes to zero at short wavelengths. The failed prediction (named later, in 1911) is what pointed to quantized energy.
⚠️ Common mistake
Using $\lambda_{\max}T$ with the temperature in degrees Celsius. Wien's law needs kelvins: a 25 °C room is 298 K.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.