Questionnaire 1.1.3 — Light

Three questions on the quantum nature of light: what a black body actually is, what the ultraviolet catastrophe actually claimed, and which knob — intensity or frequency — sets a photon's energy.

Key Ideas

A black body is a perfect absorber: it soaks up every wavelength that hits it, which is why it looks black when cold. By Kirchhoff's law, a perfect absorber is also the best possible thermal emitter — no object at the same temperature can outglow it at any wavelength. Its emission spectrum depends on temperature alone, not on what the body is made of.

Wien's displacement law puts the peak of the blackbody spectrum at a wavelength inversely proportional to temperature — hotter bodies glow bluer, and the Planck curves for different temperatures never cross (a hotter body emits more at every wavelength):

\lambda_{\text{peak}} = \frac{b}{T}, \qquad b \approx 2.898\times 10^{-3}\ \text{m·K}

The ultraviolet catastrophe is the failure mode of classical physics applied to thermal radiation: the Rayleigh–Jeans law $B_\lambda \propto kT/\lambda^4$ grows without bound as $\lambda \to 0$ , predicting that any warm body radiates infinite power in the ultraviolet. The catastrophe is too much predicted UV, not too little. Planck cured it by quantizing the energy exchange in units of $E = h f$ , which starves the high-frequency modes.

A photon carries energy set by frequency alone, $E = hf$ — the intensity of a beam sets how many photons arrive per second, never how energetic each one is. In the photoelectric effect this splits cleanly: intensity controls the ejected-electron current, frequency controls the maximum kinetic energy per electron,

K_{\max} = hf - \phi

Exercises

E1 (medium). Which one of the following statements about a black body is true?

a) A black body never emits any light b) A black body is the best emitter of light possible from simply heating up an object c) The color of a black body doesn't depend on its temperature d) A black body at the temperature of the Sun emits less red light than the same black body at the temperature of a light bulb filament

Solution

Answer: b) a black body is the best emitter of light possible from simply heating up an object.

Why, step by step:

"Black" describes how the body absorbs: absorptivity 1 at every wavelength.
Kirchhoff's law ties emission to absorption at thermal equilibrium: emissivity equals absorptivity. A perfect absorber is therefore a perfect emitter — emissivity 1, the theoretical maximum.
So at any given temperature, nothing outglows a black body at any wavelength — it is the benchmark every real glowing object (filament, star, ember) is measured against.

Why the tempting options fail:

a) confuses absorbing light with not emitting it. A cold black body looks black, but a heated one glows — the Sun is, to excellent approximation, a black body.
c) contradicts Wien's law: $\lambda_{\text{peak}} = b/T$ . Red-hot → white-hot → blue-hot is the color–temperature dependence.
d) inverts the never-crossing property of Planck curves: the hotter body emits more at every wavelength — the ~5800 K Sun out-emits a ~3000 K filament even in the red.

See it: drag the temperature and watch the whole curve rise above the fixed 3000 K reference at every wavelength while the peak dot slides toward the blue.

E2 (medium). True or false: the ultraviolet catastrophe is that, according to classical physics, a warm body would not emit much ultraviolet light.

Solution

Answer: False.

Why, step by step:

Classical physics (equipartition applied to the radiation field) gives the Rayleigh–Jeans law $B_\lambda \propto kT/\lambda^4$ .
As $\lambda \to 0$ this diverges: classical physics predicts any warm body should blast out unbounded ultraviolet power — infinitely much, not "not much".
That absurd prediction — not the modest UV output of real bodies — is the "catastrophe". Planck's $E = hf$ quantization removes it: high-frequency modes cost too much energy to excite, so the real spectrum turns over and dies off in the UV.

Why the tempting reading fails:

Real warm bodies indeed emit little UV, so the statement sounds right — but it attributes the observation to the classical prediction. The catastrophe is precisely that classical physics predicted the opposite of what is observed.

See it: tick "show the classical prediction" — the dashed Rayleigh–Jeans curve hugs the Planck curve at long wavelengths, then explodes off the top of the plot as λ → 0.

E3 (easy). True or false: increasing the intensity of the light in a photoelectric effect experiment increases the photon energy.

Solution

Answer: False.

Why, step by step:

A photon's energy is fixed by its frequency alone: $E = hf$ .
Turning up the intensity at fixed frequency sends more photons per second — each one still carrying exactly $hf$ .
Experimentally: brighter light ejects more electrons (larger current), but the maximum kinetic energy per electron, $K_{\max} = hf - \phi$ , does not budge — only changing the color moves it. This is the observation Einstein's photon picture explained and classical wave theory could not.

Why the tempting reading fails:

Classically, a more intense wave carries more energy, so "brighter → more energetic electrons" feels natural. But the beam's energy is divided into discrete photons; intensity changes the count, frequency changes the denomination.

See it: the two knobs do different jobs — intensity feeds the electron count, frequency feeds the energy per electron.

flowchart TD
    int["turn up INTENSITY<br/>(same color)"] --> np["more photons per second<br/>each still E = hf"]
    np --> cur["more electrons ejected<br/>(bigger current)"]
    np --> same["K per electron<br/>UNCHANGED"]
    freq["turn up FREQUENCY<br/>(bluer light)"] --> ep["each photon carries more:<br/>E = hf"]
    ep --> ke["higher K_max = hf − φ<br/>per electron"]