Question bank — Blackbody radiation — Planck's quantum hypothesis

False. "Black" means it absorbs everything at room temperature; when heated it glows brightly — the Sun and a red-hot poker are near-blackbodies that are anything but black to the eye.

False. The whole point is universality: at a given temperature $T$ the spectrum is a fixed function of $T$ alone, identical for iron, carbon, or a cavity hole.

True. The spectral shape is set purely by $T$; size and setting change total power and geometry, not the per-frequency curve shape $u(\nu ,T)$.

False. It is exactly correct in the low-frequency limit $h\nu \ll k_{B}T$, where Planck's formula reduces to it — it only fails (catastrophically) at high $\nu$.

True at fixed $\nu$: high $T$ makes $h\nu /k_{B}T$ small, so $⟨E⟩\to k_{B}T$. The classical world is the "quantum lumps are tiny compared to thermal energy" limit.

False. Wien's law ${\lambda }_{\text{peak}}T=b$ means higher $T$ gives shorter peak wavelength — hotter objects glow bluer, not redder, at the peak.

False. Stefan–Boltzmann says power $\propto T^4$, so doubling $T$ multiplies total power by $2^4=16$.

True. When $h\nu \gg k_{B}T$, the "$-1$" is negligible against $e^{h\nu /k_{B}T}$, so $⟨E⟩\to h\nu e^{-h\nu /k_{B}T}$, exactly Wien's exponential tail.

False. It is a failure of the classical theory, not of nature. Real ovens obey Planck's law and radiate a finite amount; the infinity signals the theory was wrong.

The error is assuming every mode gets $k_{B}T$. High-$\nu$ modes need a whole quantum $h\nu \gg k_{B}T$ to get started, so they are frozen out (exponentially suppressed) — hence no X-ray flood.

Wrong object. He quantized the energy each oscillator can hold ($E_n=nh\nu$), not the allowed frequencies. All frequencies still exist; only their energies come in steps.

The $⟨E⟩$ factor decays like $e^{-h\nu /k_{B}T}$, which crushes the ${\nu }^2$ growth. Exponential decay beats any power of $\nu$, so the product peaks and falls to zero.

They do not correspond to the same frequency. Because $d\nu /d\lambda$ warps the density during the change of variable, the $\lambda$-peak and $\nu$-peak land at different physical points; converting one peak's location naively is a mistake.

Equipartition is only valid when energy is continuous and thermally accessible. For quantized oscillators with $h\nu \gtrsim k_{B}T$ its assumption breaks, so the theorem simply does not apply there.

Anachronism. Planck quantized the energy exchange of material oscillators in the walls; the photon (light itself being quantized) came later, from Einstein's Photoelectric effect work.

You integrate over all frequencies (total energy is an area under the curve), not differentiate. Differentiating (setting slope to zero) is how you get Wien's peak, a different law.

A ray entering the hole bounces internally many times, absorbed with near-certainty before it can escape, so absorptivity ≈ 1; when hot, the hole re-emits the equilibrium radiation trapped inside.

Dropping it gives Wien's law and, at low $\nu$, would miss the classical limit; keeping it makes $⟨E⟩\to k_{B}T$ as $h\nu /k_{B}T\to 0$, correctly recovering the classical regime.

The substitution $x=h\nu /k_{B}T$ pulls out four factors of $T$ — three from the ${\nu }^3$ in the integrand and one from $d\nu$ — leaving a pure dimensionless number (${\pi }^4/15$) times $T^4$.

The geometry of standing waves (how many fit in a cavity) is a real, correct classical result; only the energy per mode was wrong, so Planck kept the count and replaced $k_{B}T$ with $⟨E⟩$.

In thermal equilibrium the probability of a state falls exponentially with its energy; higher rungs $E_n=nh\nu$ are increasingly unlikely, which is precisely what freezes out costly high-$\nu$ quanta.

A quantum costs $h\nu$; when that is small compared to $k_{B}T$ the steps look continuous (classical), but when $h\nu \gg k_{B}T$ even one quantum is unaffordable, so the mode stays dark.

Because its spectrum depends only on $T$ and nothing else, it is a universal function — any theory of light and heat must reproduce it exactly, leaving no wiggle room to hide errors.

$h\nu /k_{B}T\to \infty$, so $e^{h\nu /k_{B}T}\to \infty$ and $⟨E⟩\to 0$ — a cold cavity holds no thermal quanta, every mode is in its ground state.

The quantum $h\nu \to 0$, so $e^{h\nu /k_{B}T}-1\to h\nu /k_{B}T$ and $⟨E⟩\to k_{B}T$ — the classical equipartition value, showing zero-frequency modes are fully classical.

No. It approaches zero only in the limits $\nu \to 0$ and $\nu \to \infty$; for every finite positive $\nu$ the value is strictly positive, so the curve never touches the axis in between.

No. The frequency peak obeys its own relation ${\nu }_{\text{peak}}/T=\text{const}$ with a different numerical constant, because the change of variables reshapes the curve — the two peaks are genuinely different physical points.

Infinite — ${\int }_0^{\infty }{\nu }^{2}d\nu$ diverges. This divergence is the ultraviolet catastrophe and is the clearest sign the classical energy assumption must be abandoned.

The classical Rayleigh–Jeans law with its catastrophe: $h\to 0$ makes quanta infinitesimal so $⟨E⟩\to k_{B}T$ everywhere, restoring the (wrong) classical result — quantization is exactly the $h\ne 0$ effect.

Right at the crossover — $⟨E⟩=k_{B}T/(e-1)\approx 0.58k_{B}T$, already noticeably below the classical $k_{B}T$, so quantization is starting to bite but has not yet frozen the mode out.