Questions: Planck Distribution and Blackbody Radiation

Question 1

Classical physics (Rayleigh-Jeans law) treats each electromagnetic mode as having average energy kT. At what part of the spectrum does this prediction break down most catastrophically?

• A. Low frequencies (radio waves), where classical physics predicts too little energy
• B. High frequencies (ultraviolet and beyond), where classical physics predicts diverging energy
• C. The visible spectrum only, where quantum corrections are largest
• D. Classical physics is accurate at all frequencies — only the total integrated energy is wrong

Answer: (B)

The Rayleigh-Jeans law assigns energy kT to every mode regardless of frequency, so the total radiated power diverges as frequency increases — the 'ultraviolet catastrophe.' At low frequencies (kT ≫ hν), Planck's law reduces to the classical result, so classical physics is actually correct there. It's at high frequencies where the quantum suppression exp(−hν/kT) → 0 cuts off the classical prediction, resolving the catastrophe. A common misconception is that the failure is broad or applies equally at all frequencies.

Question 2

What happens to the mean energy per mode ⟨E⟩ = hν / [exp(hν/kT) − 1] in the high-temperature limit (kT ≫ hν)?

• A. It approaches zero, because high temperatures excite all modes equally
• B. It approaches kT, recovering the classical equipartition result
• C. It approaches hν, because the mode is always singly occupied at high temperature
• D. It diverges, reproducing the ultraviolet catastrophe

Answer: (B)

When kT ≫ hν, we have exp(hν/kT) ≈ 1 + hν/kT, so exp(hν/kT) − 1 ≈ hν/kT. Therefore ⟨E⟩ ≈ hν / (hν/kT) = kT — exactly the classical equipartition result. This is the correct classical limit: quantum mechanics reduces to classical physics at low frequencies (or high temperatures). The Planck distribution is not a replacement that always gives different results from classical physics; it agrees with classical physics where classical physics is valid.

Question 3

Photons in a blackbody cavity have chemical potential μ = 0, unlike most particles in statistical mechanics.

• T. True
• F. False

Answer: True

Photons can be freely created and absorbed by the cavity walls — there is no conservation law fixing their total number. This means there is no free energy cost to changing the photon number, which is precisely what μ = 0 means. Setting μ = 0 in the Bose-Einstein distribution gives ⟨n⟩ = 1/(exp(hν/kT) − 1) — the Planck occupation number. This is a crucial distinction from, say, electrons or atoms, where the chemical potential is determined by particle number conservation.

Question 4

Hotter blackbodies emit radiation that peaks at longer wavelengths (lower frequencies) than cooler blackbodies.

• T. True
• F. False

Answer: False

This is backwards. Wien's displacement law states that the peak frequency of the Planck spectrum is proportional to temperature: ν_max ∝ T. Hotter objects peak at higher frequency (shorter wavelength). This is why a piece of iron glows red at lower temperatures (peak in the infrared, with a visible red tail), then orange, then white (all visible wavelengths), then blue-white as it gets hotter. The confusion likely arises from conflating 'higher temperature = more energy at all frequencies' with 'the peak shifts to lower frequency.'

Question 5

Why does the Planck distribution exponentially suppress high-frequency modes, and how does this resolve the ultraviolet catastrophe?

Model answer: At high frequencies, the quantum of energy hν is much larger than kT. To excite such a mode even once requires a thermal fluctuation of magnitude hν ≫ kT, which is exponentially unlikely. The mean occupation number ⟨n⟩ = 1/(exp(hν/kT) − 1) → exp(−hν/kT) → 0 exponentially as ν → ∞. Classical physics assigned kT to every mode regardless of frequency, ignoring the quantization that makes high-frequency excitation costly. Quantization provides a natural cutoff: modes with hν ≫ kT are effectively frozen out, making the total integrated energy density finite.

The key is understanding what quantization does mechanically. In classical physics, a harmonic oscillator can have any energy and always averages kT. In quantum mechanics, the same oscillator can only have energies 0, hν, 2hν, … and the probability of excited states falls exponentially with ν. This exponential damping — not a gentle correction — is what makes the integral over all frequencies finite and equal to σT⁴.