Planck’s Quantum Hypothesis

By the late nineteenth century, classical physics had a serious problem: it could not explain the spectrum of light emitted by a blackbody — an idealized object that absorbs and re-emits all radiation falling on it. The classical Rayleigh–Jeans formula predicted that a hot object would radiate infinite power at short wavelengths, a disaster known as the ultraviolet catastrophe.

Planck's radical fix

In 1900, Max Planck found that he could match experimental blackbody spectra exactly by making one deeply non-classical assumption: the electromagnetic oscillators inside the body could only exchange energy with radiation in discrete chunks. An oscillator of frequency $\nu$ can only have energies

$E_n = n h \nu, \quad n = 0, 1, 2, \ldots,$

where $h$ is a new constant of nature, now called Planck's constant:

$h = 6.626 \times 10^{-34}\ \text{J·s}.$

The smallest non-zero energy a single oscillator can have is therefore $h\nu$ , and it can only gain or lose energy in integer multiples of that amount. This is the first statement of energy quantization in physics.

The energy–frequency relation

The key formula that follows from Planck's hypothesis is

$E = h\nu,$

where $E$ is the energy of one quantum (one "packet") of radiation at frequency $\nu$ . It tells us that higher-frequency (shorter-wavelength) light carries more energy per quantum. Ultraviolet light, for example, carries more energy per packet than infrared.

Why quantization solves the ultraviolet catastrophe

Classically, every oscillation mode of a cavity is expected to hold thermal energy $k_B T$ regardless of its frequency. At very high frequencies that prediction diverges — there are infinitely many such modes. Planck's quantization changes the weighting: a high-frequency mode of energy $h\nu \gg k_B T$ is very rarely excited (the probability of populating it goes as $e^{-h\nu / k_B T}$ ), so it contributes little to the total power. The spectrum therefore peaks and then falls off at short wavelengths, exactly as measured.

Working in electron-volts

Joules are convenient for calculations, but photon energies in the visible and near-ultraviolet range work out to numbers between about $1.5$ and $6\ \text{eV}$ , where $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$ is the energy gained by one electron accelerated through a one-volt potential difference. Converting:

$E_{\text{eV}} = \frac{h\nu}{1.602 \times 10^{-19}}.$

For example, a photon at $\nu = 6.0 \times 10^{14}\ \text{Hz}$ (green light, $\lambda \approx 500\ \text{nm}$ ) has energy

$E = 6.626 \times 10^{-34} \times 6.0 \times 10^{14} = 3.976 \times 10^{-19}\ \text{J} \approx 2.48\ \text{eV}.$

This value is derived directly from $E = h\nu$ .

Try it

This is a numerical exercise — your code should return a number. Apply $E = h\nu$ to find the energy (in eV) of a photon whose frequency is $\nu = 6.0 \times 10^{14}\ \text{Hz}$ .

Run your code to see the quantum state.