Worked examples — Blackbody radiation — Planck's quantum hypothesis

---

The scenario matrix

Every problem in blackbody radiation is really a question about one dimensionless number: $x=\frac{h\nu }{k_{B}T}\text{("how many thermal units does one quantum cost?")}$

• $x$ small ($\ll 1$): a quantum is cheap → classical behaviour returns.
• $x$ large ($\gg 1$): a quantum is expensive → the mode is frozen out.
• $x$ around $2$–$5$: the interesting middle where the spectrum peaks.

Two spectral densities appear on this page, and it matters which one you use:

The table below lists every class of situation. Each worked example is tagged with the cell(s) it covers.

Cell	Situation	Physical regime	Example
A	$h\nu \ll k_{B}T$ (low frequency)	classical / Rayleigh–Jeans limit	Ex 1
B	$h\nu \gg k_{B}T$ (high frequency)	Wien / frozen-out limit	Ex 2
C	$h\nu \approx k_{B}T$ (comparable)	exact formula, no shortcut	Ex 3
D	Degenerate: $\nu \to 0$ and $T\to 0$	limiting values / boundaries	Ex 4
E	Peak-finding (turning point), $\lambda$-peak vs $\nu$-peak	Wien's displacement, transcendental	Ex 5 (figure)
F	Total power (integral over all $\nu$)	Stefan–Boltzmann, real object	Ex 6
G	Real-world word problem	astronomy: temperature from colour	Ex 7
H	Exam twist: "which curve is hotter?"	comparing two curves, ratios	Ex 8 (figure)

Constants used throughout (keep them handy):

---

Ex 1 — Cell A: the low-frequency (classical) limit

Step 1 — compute $x=h\nu /k_{B}T$. Why this step? $x$ is the only thing that decides the regime; everything follows from its size. $x=\frac{(6.626\times 10^{-34})(1.0\times 10^{10})}{(1.381\times 10^{-23})(300)}=\frac{6.626\times 10^{-24}}{4.143\times 10^{-21}}\approx 1.60\times 10^{-3}.$ This is tiny — one quantum costs less than a thousandth of a thermal unit. So we are deep in Cell A.

Step 2 — Planck's average energy. Why this step? We want the honest quantum answer to compare against. $⟨E⟩=\frac{h\nu }{e^x-1}=\frac{6.626\times 10^{-24}}{e^{0.00160}-1}.$ Since $x$ is small, $e^x-1\approx x=1.60\times 10^{-3}$, giving $⟨E⟩\approx \frac{6.626\times 10^{-24}}{1.60\times 10^{-3}}\approx 4.14\times 10^{-21}\text{J}.$

Step 3 — classical value. Why this step? The whole point is the comparison. $k_{B}T=(1.381\times 10^{-23})(300)=4.14\times 10^{-21}\text{J}.$

---

Ex 2 — Cell B: the high-frequency (frozen-out) limit

Step 1 — compute $x$. Why this step? Again, $x$ decides everything. $x=\frac{(6.626\times 10^{-34})(3.0\times 10^{15})}{4.143\times 10^{-21}}=\frac{1.988\times 10^{-18}}{4.143\times 10^{-21}}\approx \mathrm{479.8.}$ This is enormous — one quantum costs ~480 thermal units. Cell B.

Step 2 — Planck's average energy. Why this step? We want the actual quantum answer, which should be crushingly small. $⟨E⟩=\frac{h\nu }{e^x-1}\approx h\nu e^{-x}=(1.988\times 10^{-18})e^{-479.8}.$ To size $e^{-479.8}$, convert to base 10: ${\log }_{10}(e^{-479.8})=-479.8\times 0.4343\approx -208.4$, so $e^{-479.8}\approx 10^{-208.4}\approx 4\times 10^{-209}$ — an unimaginably small number. $⟨E⟩\approx (1.988\times 10^{-18})(4\times 10^{-209})\approx 8\times 10^{-227}\text{J}\approx 0.$

---

Ex 3 — Cell C: comparable energies (no shortcut allowed)

Step 1 — compute $x$. Why this step? To confirm we are in the dangerous "middle" where neither shortcut is valid. $x=\frac{(6.626\times 10^{-34})(5.45\times 10^{14})}{(1.381\times 10^{-23})(5800)}=\frac{3.611\times 10^{-19}}{8.010\times 10^{-20}}\approx \mathrm{4.51.}$ Neither small nor huge — we must use the full formula.

Step 2 — evaluate the full expression. Why this step? In Cell C the "$-1$" genuinely matters; dropping it would be wrong by a few percent. $e^{4.51}\approx 90.9,e^{4.51}-1\approx \mathrm{89.9.}$ $⟨E⟩=\frac{h\nu }{e^x-1}=\frac{3.611\times 10^{-19}}{89.9}\approx 4.02\times 10^{-21}\text{J}.$

---

Ex 4 — Cell D: degenerate / boundary inputs

Step 1 — case (a): $\nu \to 0$. Why this step? This is the boundary of Cell A pushed to its extreme; we test that the formula stays sensible. As $\nu \to 0$, $x=h\nu /k_{B}T\to 0$, so $e^x-1\to x$: $⟨E⟩=\frac{h\nu }{e^x-1}\to \frac{h\nu }{x}=\frac{h\nu k_{B}T}{h\nu }=k_{B}T.$ So every mode approaches $k_{B}T$ at low frequency — a finite, non-zero value.

Step 2 — case (b): $T\to 0$. Why this step? This is Cell B pushed to the extreme; it checks the freeze-out is complete at absolute zero. As $T\to 0^{+}$, $x=h\nu /k_{B}T\to +\infty$, so $e^x\to \infty$ and $⟨E⟩=\frac{h\nu }{e^x-1}\to \frac{h\nu }{\infty }=0.$ At absolute zero no mode is excited — the cavity is dark.

---

Ex 5 — Cell E: finding the peak (Wien's displacement) + the $\lambda$-peak vs $\nu$-peak subtlety

Step 1 — read the figure. Why this step? Recall from the matrix that $u(\lambda ,T)$ is energy density per unit wavelength. In the figure the solid teal curve is $u(\lambda ,T)$; its maximum is marked by the orange dot, and the short horizontal plum segment through that dot is a flat tangent (slope $=0$). The dashed orange curve is the same radiation re-sliced as $u(\nu ,T)$ (per unit frequency, plotted against wavelength for comparison); its own peak — the plum dot — sits at a shorter wavelength. That visible gap between the two dots is the whole subtlety of part (b). A horizontal tangent means slope $=0$, so the maths task for part (a) is: set $du/d\lambda =0$.

Step 2 — write $u(\lambda ,T)$ and differentiate. Why this step? The peak is where $du/d\lambda =0$, exactly the flat-tangent point (orange dot) in the figure. $u(\lambda ,T)=\frac{8\pi hc}{{\lambda }^5}\frac{1}{e^{hc/\lambda k_{B}T}-1}.$ Write $u=C{\lambda }^{-5}g(\lambda )$ with $C=8\pi hc$ and $g=(e^{a/\lambda }-1{)}^{-1}$, where $a=hc/k_{B}T$. Product rule: $\frac{du}{d\lambda }=C[-5{\lambda }^{-6}g+{\lambda }^{-5}{g}^{\prime }].$

Step 3 — differentiate $g$ and simplify. Why this step? We need $g^{\prime }$ to make the bracket explicit, then factor. $g^{\prime }=\frac{d}{d\lambda }(e^{a/\lambda }-1{)}^{-1}=-(e^{a/\lambda }-1{)}^{-2}\cdot e^{a/\lambda }\cdot (-\frac{a}{{\lambda }^2})=\frac{a}{{\lambda }^2}\frac{e^{a/\lambda }}{(e^{a/\lambda }-1{)}^2}.$ Substitute back and set $du/d\lambda =0$. Dividing out the common non-zero factor $C{\lambda }^{-6}(e^{a/\lambda }-1{)}^{-1}$ leaves $-5+\frac{a}{\lambda }\frac{e^{a/\lambda }}{e^{a/\lambda }-1}=0.$

Step 4 — substitute $y=a/\lambda =hc/\lambda k_{B}T$. Why this step? $y$ packs all the constants into one dimensionless variable, collapsing the mess into a clean equation. $-5+y\frac{e^y}{e^y-1}=0\Rightarrow y\frac{e^y}{e^y-1}=5.$ Divide top and bottom of the fraction by $e^y$: $\frac{e^y}{e^y-1}=\frac{1}{1-e^{-y}}$, so $\frac{y}{1-e^{-y}}=5\Rightarrow \boxed{5(1-e^{-y})=y}.$

Step 5 — solve numerically. Why this step? Transcendental equations are solved by iteration/graphing.

• Try $y=5$: LHS $=5(1-e^{-5})=5(0.9933)=4.966$, RHS $=5$. Very close.
• Refine: the crossing is at $y\approx 4.965$.

Step 6 — get $b$. Why this step? $y=hc/{\lambda }_{\text{peak}}{k}_{B}T$ rearranges to ${\lambda }_{\text{peak}}T=hc/(y{k}_B)$, a constant. $b=\frac{hc}{4.965k_B}=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{(4.965)(1.381\times 10^{-23})}\approx 2.898\times 10^{-3}\text{m⋅K}.$

Step 7 — part (b): the frequency peak is a different equation. Why this step? This is the promised subtlety — students wrongly assume ${\lambda }_{\text{peak}}=c/{\nu }_{\text{peak}}$. Repeat the maximization on $u(\nu ,T)=\frac{8\pi {\nu }^2}{c^3}\frac{h\nu }{e^{h\nu /k_{B}T}-1}=\frac{8\pi h}{c^3}\frac{{\nu }^3}{e^{h\nu /k_{B}T}-1}$. The power out front is now ${\nu }^3$ (not ${\lambda }^{-5}$), so the "$5$" becomes a "$3$": $\boxed{3(1-e^{-z})=z},z=\frac{h\nu }{k_{B}T},z\approx \mathrm{2.821.}$ Because $y=4.965\ne z=2.821$, the two peaks pick out different colours. Converting the $\nu$-peak to a wavelength: ${\lambda }^{\prime }=c/{\nu }_{\text{peak}}=hc/(z{k}_{B}T)$, so ${\lambda }^{\prime }T=hc/(2.821k_B)\approx 5.10\times 10^{-3}\text{m⋅K}$ — a longer wavelength than $b=2.898\times 10^{-3}\text{m⋅K}$.

---

Ex 6 — Cell F: total emitted power (Stefan–Boltzmann, real object)

Step 1 — surface area. Why this step? Stefan–Boltzmann gives power per unit area; we need total, so multiply by $A$. $A=4\pi r^2=4\pi (0.10{)}^2=4\pi (0.01)\approx 0.1257{\text{m}}^2.$

Step 2 — apply $P=\sigma A{T}^4$. Why this step? This is the Stefan–Boltzmann law integrated over all frequencies — the $T^4$ comes from the four factors of $T$ the parent note tracked through the substitution. $T^4=(1000{)}^4=1.0\times 10^{12}{\text{K}}^4.$ $P=(5.67\times 10^{-8})(0.1257)(1.0\times 10^{12})\approx 7.13\times 10^3\text{W}.$ So about 7.1 kW.

---

Ex 7 — Cell G: real-world word problem (temperature from colour)

Step 1 — apply Wien's law. Why this step? Wien directly links the measurable peak colour to temperature — exactly what astronomers exploit, since you can't stick a thermometer in a star. (We use the wavelength peak $b$ here, consistent with a spectrum measured per unit wavelength — recall Ex 5's caution.) ${\lambda }_{\text{peak}}T=b\Rightarrow T=\frac{b}{{\lambda }_{\text{peak}}}.$

Step 2 — plug numbers. Why this step? Convert nm to metres first, then divide. $T=\frac{2.898\times 10^{-3}}{966\times 10^{-9}}=\frac{2.898\times 10^{-3}}{9.66\times 10^{-7}}\approx 3000\text{K}.$

Step 3 — interpret the colour. Why this step? A number is not an answer until it means something. At $3000\text{K}$ the peak is in the near-infrared, but the visible tail is strongest at the red end — so the star looks distinctly reddish-orange. Cooler blackbodies always peak at longer wavelengths (this is the same Wien's displacement law relationship, read in reverse).

---

Ex 8 — Cell H: exam twist ("which curve is hotter, and by how much?")

Step 1 — (a) which is hotter? Why this step? Wien says the peak shifts to shorter wavelength as $T$ rises. In the figure the teal curve peaks at 500 nm (bluer) and its whole body towers above the orange curve that peaks at 1000 nm. Both facts point the same way: Curve 2 (teal) is hotter.

Step 2 — (b) temperature ratio. Why this step? Wien's constant $b$ is the same for both bodies, so $\lambda T$ is a fixed number — the ratio of temperatures is just the inverse ratio of peaks. ${\lambda }_1{T}_1={\lambda }_2{T}_2=b\Rightarrow \frac{T_2}{T_1}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{1000}{500}=2.$ Curve 2 is exactly twice as hot. (This is visible in the figure: the two dotted peak-lines sit at 1000 nm and 500 nm, a factor of 2 apart.)

Step 3 — (c) power ratio. Why this step? Total power scales as $T^4$ (Stefan–Boltzmann), so a factor-2 temperature becomes a factor-$2^4$ in power. In the figure this is why the teal curve's area dwarfs the orange one's — the height gap is far bigger than the factor-2 temperature gap. $\frac{P_2}{P_1}={\left(\frac{T_2}{T_1}\right)}^4=2^4=16.$

Answers: (a) Curve 2 is hotter. (b) $T_2/T_1=2$. (c) $P_2/P_1=16$.

---

Recall Self-test: name the cell for each mini-scenario

A microwave photon in a warm oven ::: Cell A (low frequency, classical limit, $⟨E⟩\to k_{B}T$). An X-ray mode in a room-temperature cavity ::: Cell B (high frequency, frozen out, $⟨E⟩\approx 0$). The exact colour where the Sun's curve peaks ::: Cell C / Cell E (comparable energies, $x\approx 4.965$). Why does the frequency-peak name a different colour than the wavelength-peak? ::: Cell E part (b): the prefactor changes from ${\lambda }^{-5}$ to ${\nu }^3$, giving $z\approx 2.821$ instead of $y\approx 4.965$. Computing a star's temperature from its peak colour ::: Cell G (Wien word problem). "The peak halved — how much more power?" ::: Cell H ($T$ doubles, power $\times 16$).