Blackbody radiation from statistical mechanics — Planck distribution: Worked examples

This page is a drill floor. The parent note built the machine: $u (ν, T) d ν = \frac{8 π h ν ^{3}}{c ^{3}} \frac{1}{e ^{h ν / k_{B} T} - 1} d ν .$ Here we test it against every kind of input a problem can hand you — small $ν$ , large $ν$ , the exact peak, the total integral, real starlight, and a couple of exam traps. Before each worked example you must forecast the answer yourself.

The scenario matrix

Every Planck-distribution problem lives in one of these cells. The examples below are labelled with the cell(s) they hit, and together they cover all of them.

Cell	Input regime	What controls the answer	Example
A low- $ν$ limit	$x ≪ 1$	$⟨ ϵ ⟩ \to k_{B} T$ (Rayleigh–Jeans)	Ex 1
B high- $ν$ limit	$x ≫ 1$	exponential cutoff $e^{- x}$ (Wien)	Ex 2
C mid-range exact	$x \sim 1$	full formula, no approximation	Ex 3
D degenerate $ν \to 0$	$ν = 0$	mode has no energy; check $0/0$	Ex 3b (inside Ex 3)
E the peak (extremum)	$d u / d ν = 0$	transcendental $3 (1 - e^{- x}) = x$	Ex 4
F $ν$ -peak vs $λ$ -peak	Jacobian $c / λ^{2}$	different transcendental, $x = 4.965$	Ex 5
G total (integrate all $ν$ )	$\int_{0}^{\infty}$	Stefan–Boltzmann $T^{4}$	Ex 6
H real-world word problem	numbers with units	pick the right sub-law	Ex 7 (the Sun)
I exam twist / ratio	cancel constants cleverly	dimensionless ratios	Ex 8

We rely on the Density of states, the Partition function, the Rayleigh-Jeans law, Wien's law and the Stefan-Boltzmann law throughout.

Worked examples

Worked example Ex 1 — Cell A (low-frequency limit)

An FM radio wave has $ν = 1.0 \times 1 0^{8} Hz$ (100 MHz). The cavity is at room temperature $T = 300 K$ . What is the average energy per mode $⟨ ϵ ⟩$ , and how close is it to the classical $k_{B} T$ ?

Forecast: radio is very low frequency, so $x$ is minuscule and we should land almost exactly on the classical $k_{B} T$ . Guess: essentially $k_{B} T$ to many digits.

Compute $x = h ν / k_{B} T$ . Why this step? $x$ decides which regime we're in — always first. $x = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 1.0 \times 1 0 ^{8} )}{( 1.381 \times 1 0 ^{- 23} ) ( 300 )} = 1.60 \times 1 0^{- 5} .$ Indeed $x ≪ 1$ : Cell A confirmed.
Use the exact formula $⟨ ϵ ⟩ = \frac{h ν}{e ^{x} - 1}$ (defined above). Why? We want to see how close to $k_{B} T$ we get. Numerically $h ν = 6.626 \times 1 0^{- 26} J$ and $e^{x} - 1 \approx x = 1.60 \times 1 0^{- 5}$ , so $⟨ ϵ ⟩ \approx \frac{6.626 \times 1 0 ^{- 26}}{1.60 \times 1 0 ^{- 5}} = 4.14 \times 1 0^{- 21} J .$
Compare to $k_{B} T$ . Why? That's the classical prediction. $k_{B} T = (1.381 \times 1 0^{- 23}) (300) = 4.14 \times 1 0^{- 21} J .$ They match to ~5 significant figures.

Verify: the fractional gap is about $x /2 \approx 8 \times 1 0^{- 6}$ (from $e^{x} - 1 \approx x + x^{2} /2$ ). Units: J. Forecast confirmed — at radio frequencies equipartition is essentially exact, which is exactly why nobody noticed the catastrophe using long-wavelength measurements.

Worked example Ex 2 — Cell B (high-frequency limit)

A cavity at $T = 3000 K$ (a glowing filament). Compare the energy density $u (ν)$ at a visible-blue frequency $ν = 7.0 \times 1 0^{14} Hz$ against what Rayleigh–Jeans would (wrongly) predict. By what factor does the true Planck value fall short of Rayleigh–Jeans?

Forecast: blue light at 3000 K should be well up the "expensive" side. Rayleigh–Jeans has no cutoff, so it will overshoot. Guess: the true value is smaller by a large factor $\sim e^{- x} \cdot x$ .

Before we start, name the comparison object. Let $u_{RJ}$ denote the energy density predicted by the Rayleigh-Jeans law, i.e. the classical spectrum $u_{RJ} (ν) = \frac{8 π ν ^{2}}{c ^{3}} k_{B} T$ that gives every mode the full $k_{B} T$ . And let $u_{Planck}$ denote the true Planck spectrum. We compare these two.

Compute $x$ . Why? Sets the regime. $x = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 7.0 \times 1 0 ^{14} )}{( 1.381 \times 1 0 ^{- 23} ) ( 3000 )} = 11.2.$ $x ≫ 1$ : Cell B.
Write both energies-per-mode. Why? The density of states $g (ν) = 8 π ν^{2} / c^{3}$ (derived in the parent note — the count of modes per unit volume and frequency) is the same in both the Planck and Rayleigh–Jeans spectra, so when we take a ratio it cancels and only the ratio of $⟨ ϵ ⟩$ survives. $\frac{u _{Planck}}{u _{RJ}} = \frac{⟨ ϵ ⟩ _{Planck}}{k _{B} T} = \frac{h ν / ( e ^{x} - 1 )}{k _{B} T} = \frac{x}{e ^{x} - 1} .$ Why does $h ν / k_{B} T$ become $x$ ? Because that ratio is $x$ by definition.
Plug in $x = 11.2$ . Why? Get the number. $\frac{x}{e ^{x} - 1} = \frac{11.2}{e ^{11.2} - 1} = \frac{11.2}{7.32 \times 1 0 ^{4}} = 1.53 \times 1 0^{- 4} .$

Verify: Rayleigh–Jeans over-predicts by a factor $\sim 6500 \times$ . The exponential $e^{- x}$ crushing the $x$ prefactor is the ultraviolet catastrophe being defeated — precisely the mechanism the parent note described. Units: dimensionless ratio, good.

Worked example Ex 3 — Cell C & D (mid-range exact, plus the

$ν \to 0$ degenerate check) (C) Evaluate $⟨ ϵ ⟩$ exactly when $x = 1$ (i.e. $h ν = k_{B} T$ ). (D) Then check what happens to $⟨ ϵ ⟩$ as $ν \to 0$ — the formula is naively $0/0$ ; resolve it.

Forecast (C): $x = 1$ is the crossover. The lump costs about one thermal kick, so we expect somewhat less than $k_{B} T$ — the mode is partly frozen. Guess: around $0.6 k_{B} T$ .

(C) Exact value. Why the exact formula? At $x = 1$ neither limit applies. $⟨ ϵ ⟩ = \frac{h ν}{e ^{1} - 1} = \frac{k _{B} T \cdot 1}{e - 1} = \frac{k _{B} T}{1.71828} = 0.5820 k_{B} T .$ So a mode at the crossover holds only ~58% of the classical share.
(D) The $ν \to 0$ limit. Why care? Radio, microwave, DC — all live near $ν = 0$ ; the formula must behave. Write $⟨ ϵ ⟩ = \frac{h ν}{e ^{h ν / k_{B} T} - 1}$ . As $ν \to 0$ both numerator and denominator $\to 0$ : a genuine $0/0$ .
Resolve with the Taylor series $e^{x} - 1 = x + \frac{x ^{2}}{2} + \dots$ . Why Taylor and not L'Hôpital? Same thing here, but the series also shows the next correction: $⟨ ϵ ⟩ = \frac{k _{B} T x}{x + \frac{x ^{2}}{2} + \dots} = k_{B} T (1 - \frac{x}{2} + \dots) x \to 0 k_{B} T .$

Verify: the degenerate input $ν = 0$ gives a finite, sensible $⟨ ϵ ⟩ \to k_{B} T$ — no blow-up, no divide-by-zero disaster. The number $1/ (e - 1) = 0.5820$ can be checked directly (see VERIFY). This crossover behaviour is graphed below.

Figure (s01) — Average energy per mode $⟨ ϵ ⟩ / k_{B} T$ versus $x = h ν / k_{B} T$ . The chalk-blue curve is the exact factor

x / (e^{x} - 1)

; the dashed pale-yellow line is the classical ceiling

k_{B} T

. On the left (

x \to 0

, Cell A) the curve hugs

k_{B} T

— every low-frequency mode is fully "on". At the pink dot (

x = 1

, this example's Cell C) it has already dropped to

0.582 k_{B} T

: the mode is partly frozen. On the right (

x ≫ 1

, Cell B) it collapses toward zero — high-frequency modes are frozen out. The single picture is the whole story of why the ultraviolet catastrophe never happens: read it left-to-right as "modes switching off as they get expensive".

Look at that chalk-blue curve: it starts flat at $k_{B} T$ (Cell A), passes through $0.582 k_{B} T$ at the marked $x = 1$ (Cell C), and dives to zero on the right (Cell B). One picture, three cells.

Worked example Ex 4 — Cell E (the frequency peak, an extremum problem)

Find the frequency $ν_{m a x}$ at which $u (ν, T)$ is largest, at $T = 5800 K$ (roughly the Sun's surface).

Forecast: setting a derivative to zero of $ν^{3} / (e^{x} - 1)$ should give a fixed dimensionless $x$ ; the parent note quoted $x \approx 2.82$ . So $ν_{m a x} \propto T$ . Guess: a few $\times 1 0^{14}$ Hz (visible).

Set up the extremum. Why differentiate? A maximum is where the slope is zero. Write $u \propto \frac{ν ^{3}}{e ^{x} - 1}$ with $x = h ν / k_{B} T$ , so $ν \propto x$ at fixed $T$ and we maximize $f (x) = \frac{x ^{3}}{e ^{x} - 1}$ .
Differentiate $f$ and set to zero. Why in $x$ ? It removes all constants — the peak's location in $x$ is universal. $f^{'} (x) = \frac{3 x ^{2} ( e ^{x} - 1 ) - x ^{3} e ^{x}}{( e ^{x} - 1 ) ^{2}} = 0 \Rightarrow 3 (e^{x} - 1) = x e^{x} \Rightarrow 3 (1 - e^{- x}) = x .$
Solve the transcendental equation numerically. Why numerically? It mixes a polynomial and an exponential — no closed form. Iterating gives $x = 2.8214$ .
Convert $x$ to $ν_{m a x}$ . Why? We want a physical frequency. $ν_{m a x} = \frac{x k _{B} T}{h} = \frac{2.8214 ( 1.381 \times 1 0 ^{- 23} ) ( 5800 )}{6.626 \times 1 0 ^{- 34}} = 3.41 \times 1 0^{14} Hz .$

Verify: $ν_{m a x} \propto T$ (Wien) ✓. That frequency corresponds to $λ = c / ν \approx 880 nm$ (near-infrared). Note this is not the Sun's usual quoted 500 nm peak — because that quote is the $λ$ -peak. That mismatch is exactly Cell F, next.

Worked example Ex 5 — Cell F (the

$λ$ -peak vs $ν$ -peak trap) For the same Sun ( $T = 5800 K$ ), find the wavelength $λ_{m a x}$ where $u (λ)$ peaks, and confirm it disagrees with $c / ν_{m a x}$ .

Forecast: the parent note warned the two peaks differ because of a Jacobian. Expect a different transcendental root, $x \approx 4.97$ , giving $λ_{m a x} \approx 500 nm$ (green — the famous number).

Change variables with the Jacobian. Why? A density transforms so total energy is unchanged: $u (ν) d ν = u (λ) d λ$ , hence $u (λ) = u (ν) \frac{d ν}{d λ} = u (ν) \frac{c}{λ ^{2}}$ . That extra $λ^{- 2}$ (i.e. $ν^{2}$ ) shifts the peak — this is the whole trap.
Write $u (λ) \propto \frac{λ ^{- 5}}{e ^{y} - 1}$ , with $y = \frac{h c}{λ k _{B} T}$ . Why $λ^{- 5}$ ? Because $ν^{3} \cdot c / λ^{2} = λ^{- 3} λ^{- 2}$ up to constants. (Here $y$ is just $x$ rewritten for wavelength: since $ν = c / λ$ , we get $y = h ν / k_{B} T = h c / (λ k_{B} T)$ — note the whole product $λ k_{B} T$ sits in the denominator.)
Maximize: same algebra as Ex 4 but with power 5 instead of 3: $5 (1 - e^{- y}) = y \Rightarrow y = 4.9651.$ Why 5 not 3? The Jacobian added two powers of $λ^{- 1}$ .
Solve for $λ_{m a x}$ . Why? We want the physical wavelength that a spectrometer reports, to compare with $c / ν_{m a x}$ . $λ_{m a x} = \frac{h c}{y k _{B} T} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{4.9651 ( 1.381 \times 1 0 ^{- 23} ) ( 5800 )} = 499 nm .$

Verify: $c / ν_{m a x} = (3.0 \times 1 0^{8}) / (3.41 \times 1 0^{14}) = 880 nm \neq = 499 nm$ . The two peaks genuinely differ — same spectrum, different variable. This is Wien's displacement law in wavelength form: $λ_{m a x} T = (h c) / (4.9651 k_{B}) = 2.90 \times 1 0^{- 3} m\cdotK$ ✓ (the textbook Wien constant).

Worked example Ex 6 — Cell G (integrate over all

$ν$ : Stefan–Boltzmann) Derive the total energy density $U (T)$ and evaluate the Stefan–Boltzmann-style prefactor numerically for $T = 5800 K$ .

Forecast: integrating $ν^{3} / (e^{x} - 1)$ over all frequencies must give $\propto T^{4}$ (dimensional argument: only $k_{B} T / h$ can set the scale). Guess: $U \sim$ a few $J/m^{3}$ .

Set up the integral. Why? "Total" = sum over every mode. $U = \int_{0}^{\infty} \frac{8 π h ν ^{3}}{c ^{3}} \frac{d ν}{e ^{h ν / k_{B} T} - 1} .$
Substitute $x = h ν / k_{B} T$ so $ν = (k_{B} T / h) x$ , $d ν = (k_{B} T / h) d x$ . Why? It pulls all the $T$ -dependence out front and makes the remaining integral a pure number. $U = \frac{8 π h}{c ^{3}} (\frac{k _{B} T}{h})^{4} \int_{0}^{\infty} \frac{x ^{3} d x}{e ^{x} - 1} .$
Use the standard integral $\int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = \frac{π ^{4}}{15}$ . Why this value? It is $Γ (4) ζ (4) = 6 \cdot \frac{π ^{4}}{90} = \frac{π ^{4}}{15}$ . $U = \frac{8 π ^{5} k _{B}^{4}}{15 c ^{3} h ^{3}} T^{4} .$
Plug numbers at $T = 5800$ K. Why? Get a physical density. Coefficient $a = \frac{8 π ^{5} k _{B}^{4}}{15 c ^{3} h ^{3}} = 7.566 \times 1 0^{- 16} J\cdotm^{- 3} K^{- 4}$ , so $U = 7.566 \times 1 0^{- 16} (5800)^{4} = 8.57 \times 1 0^{- 1} J/m^{3} .$

Verify: $\int_{0}^{\infty} x^{3} / (e^{x} - 1) d x = π^{4} /15 = 6.4939$ ✓. Units: $[J\cdotm^{- 3} K^{- 4}] \cdot K^{4} = J/m^{3}$ ✓. The related surface flux is $σ T^{4}$ with $σ = a c /4 = 5.67 \times 1 0^{- 8}$ — the Stefan constant, confirming Cell G.

Worked example Ex 7 — Cell H (real-world word problem: the Sun's color)

A telescope measures a distant star's spectrum peaking (in wavelength) at $λ_{m a x} = 290 nm$ (ultraviolet). What is its surface temperature, and how does it compare to the Sun (5800 K)?

Forecast: shorter peak wavelength ⇒ hotter (Wien). At half the Sun's peak wavelength, expect roughly double the temperature, ~10000 K, a blue-white star.

Choose the right law. Why Wien-displacement? We're given a peak, not a total flux, so use $λ_{m a x} T = b$ from Ex 5 with $b = 2.898 \times 1 0^{- 3} m\cdotK$ .
Solve for $T$ . Why rearrange? $T$ is the unknown. $T = \frac{b}{λ _{m a x}} = \frac{2.898 \times 1 0 ^{- 3}}{290 \times 1 0 ^{- 9}} = 9993 K \approx 1.0 \times 1 0^{4} K .$
Ratio to the Sun. Why? The question asks a comparison. $\frac{T _{star}}{T _{⊙}} = \frac{9993}{5800} = 1.72.$
Interpret brightness. Why? Total output scales as $T^{4}$ (Stefan–Boltzmann), so a hotter surface radiates disproportionately more per unit area. $(\frac{T _{star}}{T _{⊙}})^{4} = 1.7 2^{4} = 8.7,$ so per unit area it radiates ~8.7× the Sun's power.

Verify: UV peak ⇒ ~10000 K, hotter than the Sun ✓ (matches the "blue = hot" intuition). Units: $m\cdotK / m = K$ ✓. The Wien's law link supplied the tool; the $T^{4}$ jump is Stefan-Boltzmann law. Cell H fully worked.

Worked example Ex 8 — Cell I (exam twist: a clean ratio that kills the constants)

At temperature $T$ , what is the ratio of energy density at $ν_{2}$ where $x_{2} = 6$ to that at $ν_{1}$ where $x_{1} = 2$ ? (Both at the same $T$ .) Express purely in terms of the $x$ 's — no $h, k_{B}, c$ should survive.

Forecast: higher frequency ( $x = 6$ ) is deep in the expensive region; the ratio should be much less than 1 even though $ν^{3}$ favors it. Guess: order $1 0^{- 2}$ .

Write the full ratio. Why? At fixed $T$ the constants $h, k_{B}, c$ multiply both spectra equally, so a ratio wipes them out — this is the whole trick of the problem. $\frac{u ( ν _{2} )}{u ( ν _{1} )} = \frac{ν _{2}^{3} / ( e ^{x_{2}} - 1 )}{ν _{1}^{3} / ( e ^{x_{1}} - 1 )} .$
Turn $ν$ into $x$ . Why? At fixed $T$ , $x = h ν / k_{B} T$ means $ν \propto x$ , so $ν_{2} / ν_{1} = x_{2} / x_{1}$ and the leftover is pure- $x$ , exactly as demanded. $\frac{u ( ν _{2} )}{u ( ν _{1} )} = (\frac{x _{2}}{x _{1}})^{3} \frac{e ^{x_{1}} - 1}{e ^{x_{2}} - 1} .$
Plug $x_{1} = 2, x_{2} = 6$ . Why? Finish it and land a number. $= (\frac{6}{2})^{3} \cdot \frac{e ^{2} - 1}{e ^{6} - 1} = 27 \cdot \frac{6.389}{402.4} = 27 \times 0.01588 = 0.4287.$

Verify: the $ν^{3}$ factor gave a healthy $27 \times$ boost, but the exponential $e^{x}$ knocked it down to $0.43$ — the $ν^{3}$ -up/ $e^{x}$ -down tug-of-war made explicit. Every physical constant cancelled ✓, exactly what the problem demanded. The final answer $0.4287$ is dimensionless, as it must be for a ratio. Cell I done.

Recall Self-test: name the cell, then solve

Given $x = 0.01$ , is $⟨ ϵ ⟩$ closer to $k_{B} T$ or to $0$ ? ::: Closer to $k_{B} T$ — Cell A, $⟨ ϵ ⟩ \approx k_{B} T (1 - x /2) = 0.995 k_{B} T$ . Which peak needs $x = 4.965$ , the $ν$ -peak or the $λ$ -peak? ::: The $λ$ -peak (Cell F); the $ν$ -peak uses $x = 2.821$ . What single dimensionless number should you compute first, always? ::: $x = h ν / k_{B} T$ . Why does the same star give two different peak wavelengths? ::: The Jacobian $c / λ^{2}$ from $u (ν) d ν = u (λ) d λ$ shifts the maximum (Cell F).

Related building blocks: Bose-Einstein statistics (why $1/ (e^{x} - 1)$ ), Equipartition theorem (the classical $k_{B} T$ ), Density of states (the $ν^{2}$ counting), and the failed Rayleigh-Jeans law this whole page repairs.