# 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:** 1. "Black" describes how the body _absorbs_: absorptivity 1 at every wavelength. 2. 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. 3. 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. ```qm-demo blackbody-spectrum ```

**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:** 1. Classical physics (equipartition applied to the radiation field) gives the Rayleigh–Jeans law $B_\lambda \propto kT/\lambda^4$. 2. As $\lambda \to 0$ this _diverges_: classical physics predicts any warm body should blast out unbounded ultraviolet power — infinitely much, not "not much". 3. 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. ```qm-demo blackbody-spectrum ```

**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:** 1. A photon's energy is fixed by its frequency alone: $E = hf$. 2. Turning up the intensity at fixed frequency sends _more photons per second_ — each one still carrying exactly $hf$. 3. 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. ```mermaid flowchart TD int["turn up INTENSITY
(same color)"] --> np["more photons per second
each still E = hf"] np --> cur["more electrons ejected
(bigger current)"] np --> same["K per electron
UNCHANGED"] freq["turn up FREQUENCY
(bluer light)"] --> ep["each photon carries more:
E = hf"] ep --> ke["higher K_max = hf − φ
per electron"] ```