Black-body radiation and Planck's quantum hypothesis E = hν

WHAT is a black body?

WHY do we care? Because its emission spectrum is universal — every hot body (a star, a filament, a furnace) follows the same curve. If we can explain this curve, we understand thermal light itself.

WHAT the experiment shows

When you plot intensity (energy emitted) vs wavelength at fixed temperature you get a humped curve:

• The curve rises, peaks at some $\lambda_{max}$, then falls.
• As temperature increases, the peak shifts to shorter wavelength (redhot → whitehot → bluehot). This is Wien's displacement law: $\lambda_{max} T = \text{constant}$.
• Total area under the curve (total energy) grows as $T^4$ (Stefan–Boltzmann).

WHY classical physics FAILED — the Ultraviolet Catastrophe

Classical physics (Rayleigh–Jeans) treated the cavity as containing standing electromagnetic waves, each an oscillator sharing energy equally (equipartition: $kT$ per mode).

The disaster: as $\nu \to \infty$ (short wavelength / UV), $u(\nu) \to \infty$. The theory predicts infinite energy radiated at high frequency — the ultraviolet catastrophe. Reality says the curve falls to zero there.

HOW Planck fixed it — the quantum hypothesis

Planck's radical assumption: an oscillator of frequency $\nu$ cannot have any energy. It can only have energy in whole-number multiples of a basic packet:

Derivation of the average energy per mode (from first principles):

Instead of $\langle E\rangle = kT$, compute the average using Boltzmann weights over the discrete levels $E_n = nh\nu$:

$\langle E\rangle = \frac{\sum_{n=0}^{\infty} nh\nu\, e^{-nh\nu/kT}}{\sum_{n=0}^{\infty} e^{-nh\nu/kT}}$

Let $x = e^{-h\nu/kT}$. Denominator is a geometric series: $\sum x^n = \frac{1}{1-x}$ Numerator: $h\nu \sum n x^n = h\nu \dfrac{x}{(1-x)^2}$.

Divide: $\langle E\rangle = h\nu\,\frac{x}{1-x} = \frac{h\nu}{\tfrac{1}{x}-1} = \frac{h\nu}{e^{h\nu/kT}-1}$

WHY this cures the catastrophe: at high $\nu$, the factor $e^{h\nu/kT}$ becomes huge, so $\langle E\rangle \to 0$. High-frequency modes are frozen out — you need a whole packet $h\nu$ to excite them, and $kT$ isn't enough. The curve falls, matching experiment.

Check the classical limit: for small $\nu$ (or large $T$), $e^{h\nu/kT} \approx 1 + \frac{h\nu}{kT}$, so $\langle E\rangle \approx \frac{h\nu}{h\nu/kT} = kT$ recovering Rayleigh–Jeans. Planck's law contains the classical result as a special case.

Worked Examples

Common Mistakes (Steel-man + fix)

Flashcards

Connections

• Photoelectric Effect — Einstein extended $E=h\nu$ to light itself (photons).
• Bohr Model of the Atom — quantized energy levels build on Planck's quanta.
• de Broglie Wavelength — wave–particle duality follows from quantized energy.
• Wien's Displacement Law & Stefan-Boltzmann Law — empirical facts Planck's law reproduces.
• Boltzmann Distribution — supplies the $e^{-E/kT}$ weighting used in the derivation.

Concept Map

Hinglish (regional understanding)