Black-body radiation and Planck's quantum hypothesis E = hν: Worked examples

Everything we use is built from just three relations, each of which the parent already earned:

Let me name each symbol in plain words so line one is followable:

$ν$ (Greek "nu") = frequency = how many wave crests pass a point each second, in hertz (Hz = "per second").
$λ$ (Greek "lambda") = wavelength = the length of one full wave, in metres.
$c$ = the speed light travels.
$E$ = energy of one packet of light, in joules (J).
$⟨ E ⟩$ = the average energy of one mode (see the definition callout above).
$T$ = temperature in kelvin (K): absolute temperature, where $0 K$ is the coldest possible.
$k$ = Boltzmann's constant — the "exchange rate" between temperature and energy.

The scenario matrix

Every exam question on this topic is one of these cells. We will cover all of them.

Cell	Type of input / regime	What makes it tricky	Example
A	Given $ν$ → find $E$	direct plug-in	Ex 1
B	Given $λ$ → find $E$	must convert first, or use $E = h c / λ$	Ex 2
C	Given $E$ → find $ν$ and $λ$	run the formula backwards	Ex 3
D	Scale up to a mole of quanta	multiply by $N_{A}$ ; compare to bond energies	Ex 4
E	Count photons from a power/beam	energy ÷ energy-per-photon	Ex 5
F	Limiting case: $h ν ≪ k T$ (low freq / hot)	Planck → classical $k T$	Ex 6
G	Limiting case: $h ν ≫ k T$ (UV / cold)	mode "frozen out", $⟨ E ⟩ \to 0$	Ex 7
H	Degenerate input: $ν = 0$	check the formula doesn't break	Ex 8
I	Word problem: Wien peak → energy of peak photon	chain Wien's Displacement Law into $E = h ν$	Ex 9
J	Exam twist: ratio question (constants cancel)	two frequencies, compare $⟨ E ⟩$	Ex 10

The three "regimes" (F, G, H) are the heart of Planck's law, so we picture them once before working numbers.

Figure — Black-body radiation and Planck's quantum hypothesis E = hν

Read the figure like this. The horizontal axis is the dimensionless ratio $x = \frac{h ν}{k T}$ — "frequency measured in thermal units", small on the left (low frequency), large on the right (high frequency). The vertical axis is the average energy of the mode divided by $k T$ , i.e. $\frac{⟨ E ⟩}{k T}$ , so the whole plot is unit-free. The cyan curve is Planck's exact average $⟨ E ⟩ = \frac{h ν}{e ^{h ν / k T} - 1}$ . The dashed amber line sits at height $1$ and marks the classical value $⟨ E ⟩ = k T$ . On the left (low $ν$ , cell F) the cyan curve hugs the amber line — classical physics is fine there. On the right (high $ν$ , cell G) the curve dives toward zero — the mode is starved. At the origin (cell H) both top and bottom of the fraction vanish, and the limit works out to exactly $k T$ ; we prove that below.

Example 1 — Cell A: given frequency, find energy

Example 2 — Cell B: given wavelength, find energy

Worked example Convert, then plug

Ultraviolet light has $λ = 250 nm$ . Energy per quantum?

Forecast: UV is higher frequency than the 400 nm light in the parent's Example 2 ( $4.97 \times 1 0^{- 19} J$ ). Shorter $λ$ → higher $ν$ → more energy. Expect above $5 \times 1 0^{- 19}$ .

Step 1 — Convert nm to metres: $250 nm = 250 \times 1 0^{- 9} m = 2.5 \times 1 0^{- 7} m$ . Why this step? Our constants use metres; mixing units is the #1 error here.

Step 2 — Use the shortcut $E = \frac{h c}{λ}$ . Why this step? $E = h ν$ needs frequency, and $ν = c / λ$ ; substituting once gives the combined form, saving a step. $E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{2.5 \times 1 0 ^{- 7}}$

Step 3 — Compute numerator then divide. $numerator = 1.988 \times 1 0^{- 25} J m, E = \frac{1.988 \times 1 0 ^{- 25}}{2.5 \times 1 0 ^{- 7}} = 7.95 \times 1 0^{- 19} J$

Verify: Units: $\frac{( J s ) ( m/s )}{m} = J$ . ✓ And $7.95 \times 1 0^{- 19} > 4.97 \times 1 0^{- 19}$ — higher energy than 400 nm light, exactly as forecast.

Example 3 — Cell C: given energy, run it backwards

Example 4 — Cell D: energy of a whole mole

Example 5 — Cell E: counting photons from a beam

Worked example Power divided by packet size

A green laser pointer ( $ν = 5.0 \times 1 0^{14} Hz$ , so each photon $= 3.3 \times 1 0^{- 19} J$ from the parent's Example 1) outputs $1.0 mW$ . How many photons per second?

Forecast: a milliwatt is $1 0^{- 3} J$ each second, and each photon is a tiny $\sim 1 0^{- 19} J$ , so expect a huge number, around $1 0^{16}$ .

Step 1 — Note power means "joules per second": $1.0 mW = 1.0 \times 1 0^{- 3} J/s$ . Why this step? "How many per second" needs energy per second on top.

Step 2 — Divide total energy per second by energy per photon. Why this step? If each photon carries a fixed packet, the count is (total energy) ÷ (energy each) — like sharing a pile of money into fixed coins. $N = \frac{1.0 \times 1 0 ^{- 3} J/s}{3.3 \times 1 0 ^{- 19} J/photon}$

Step 3 — Compute. $N = 3.03 \times 1 0^{15} photons/s$

Verify: Units: $\frac{J/s}{J/photon} = photons/s$ ✓. About $3 \times 1 0^{15}$ per second — a colossal count, which is why laser light looks perfectly smooth even though it is grainy packets. This graininess is the seed of the Photoelectric Effect.

Example 6 — Cell F: the low-frequency (classical) limit

Worked example When Planck becomes classical

For a mode at $ν = 1.0 \times 1 0^{10} Hz$ (microwave) inside a cavity at $T = 300 K$ , compute the exact $⟨ E ⟩$ and compare to the classical $k T$ .

Forecast: microwaves are low frequency, so $h ν$ should be tiny next to $k T$ — expect $⟨ E ⟩ \approx k T$ .

Step 1 — Form the dimensionless ratio $\frac{h ν}{k T}$ . Why this step? Whether a mode is "classical" or "frozen" depends only on how $h ν$ compares to $k T$ ; that ratio is the deciding number. $\frac{h ν}{k T} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 1.0 \times 1 0 ^{10} )}{( 1.381 \times 1 0 ^{- 23} ) ( 300 )} = \frac{6.626 \times 1 0 ^{- 24}}{4.143 \times 1 0 ^{- 21}} = 1.60 \times 1 0^{- 3}$

Step 2 — Plug into $⟨ E ⟩ = \frac{h ν}{e ^{h ν / k T} - 1}$ . Why this step? This is the exact Planck average; we want to see it collapse to $k T$ . $⟨ E ⟩ = \frac{6.626 \times 1 0 ^{- 24}}{e ^{0.00160} - 1} = \frac{6.626 \times 1 0 ^{- 24}}{1.60128 \times 1 0 ^{- 3}} = 4.138 \times 1 0^{- 21} J$

Step 3 — Compare with $k T = 4.143 \times 1 0^{- 21} J$ . $\frac{⟨ E ⟩}{k T} = 0.9992 \approx 1$

Verify: They agree to $0.1%$ . This is the classical limit the parent note promised: for $h ν ≪ k T$ , $e^{x} - 1 \approx x$ , so $⟨ E ⟩ \approx \frac{h ν}{h ν / k T} = k T$ . Cell F is the left edge of the s01 figure. Uses the Boltzmann Distribution weighting underneath. ✓

Example 7 — Cell G: the high-frequency (frozen) limit

Worked example The mode that gets starved

Same cavity, $T = 300 K$ , but now an ultraviolet mode $ν = 3.0 \times 1 0^{15} Hz$ . Find $⟨ E ⟩$ and compare to $k T$ .

Forecast: UV is very high frequency, so $h ν ≫ k T$ — expect $⟨ E ⟩$ vastly below $k T$ , nearly zero. This is the ultraviolet-catastrophe cure in action.

Step 1 — The ratio again. $\frac{h ν}{k T} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{15} )}{4.143 \times 1 0 ^{- 21}} = \frac{1.988 \times 1 0 ^{- 18}}{4.143 \times 1 0 ^{- 21}} = 479.8$ Why this step? A ratio of $\sim 480$ means one quantum is $480 \times$ the thermal energy available — climbing even one step is nearly impossible.

Step 2 — Plug in. Why this step? We want the exact suppression. Since $e^{479.8}$ is astronomically large, $e^{479.8} - 1 \approx e^{479.8}$ . $⟨ E ⟩ = \frac{1.988 \times 1 0 ^{- 18}}{e ^{479.8} - 1} \approx 1.988 \times 1 0^{- 18} e^{- 479.8}$

Step 3 — Estimate the magnitude. $e^{- 479.8} \approx 1 0^{- 208}, ⟨ E ⟩ \approx 2 \times 1 0^{- 226} J$

Verify: Compared with $k T = 4.14 \times 1 0^{- 21} J$ , this is smaller by $\sim 200$ orders of magnitude — effectively zero. That is exactly why the intensity curve falls at high $ν$ instead of exploding: the mode is frozen out. Cell G is the right tail of s01. ✓

Example 8 — Cell H: the degenerate input $ν = 0$

Worked example Does the formula break at zero frequency?

What is $⟨ E ⟩$ as $ν \to 0$ ? Both numerator $h ν$ and denominator $e^{h ν / k T} - 1$ head to $0$ — a " $0/0$ " that looks dangerous.

Forecast: a zero-frequency "mode" is just a static field carrying thermal energy; guess it should give exactly $k T$ , the fully classical value.

Step 1 — Let $x = \frac{h ν}{k T}$ , so $⟨ E ⟩ = k T \cdot \frac{x}{e ^{x} - 1}$ , and study $x \to 0$ . Why this step? Rewriting in the single variable $x$ turns a two-symbol limit into one clean ratio.

Step 2 — Expand $e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots$ , so $e^{x} - 1 = x + \frac{x ^{2}}{2} + \dots$ Why this step? Near zero the exponential is almost a straight line $1 + x$ ; the series makes the cancellation visible instead of a mysterious $0/0$ . $\frac{x}{e ^{x} - 1} = \frac{x}{x ( 1 + \frac{x}{2} + \dots )} = \frac{1}{1 + \frac{x}{2} + \dots} x \to 0 1$

Step 3 — Therefore $⟨ E ⟩ \to k T \cdot 1 = k T$ .

Verify: Numerically at $x = 0.001$ : $\frac{0.001}{e ^{0.001} - 1} = \frac{0.001}{0.0010005} = 0.99950$ , already essentially $1$ . So the formula is perfectly well-behaved at $ν \to 0$ : it smoothly hands back the classical $k T$ . No division-by-zero disaster. ✓

Example 9 — Cell I: word problem chaining Wien's law

Worked example From a star's peak colour to its peak-photon energy

The Sun's spectrum peaks at $λ_{ma x} = 500 nm$ . (a) Estimate its surface temperature using Wien's Displacement Law $λ_{ma x} T = b$ with $b = 2.898 \times 1 0^{- 3} m K$ . (b) Find the energy of one photon at that peak wavelength.

Forecast: the Sun is famously $\approx 5800 K$ , and a visible photon is $\sim 4 \times 1 0^{- 19} J$ — let's confirm both.

Step 1 — Solve Wien's law for $T$ : $T = \frac{b}{λ _{ma x}}$ . Why this step? We are given the peak wavelength and want temperature; Wien's law is the exact bridge between them. $T = \frac{2.898 \times 1 0 ^{- 3}}{500 \times 1 0 ^{- 9}} = \frac{2.898 \times 1 0 ^{- 3}}{5.0 \times 1 0 ^{- 7}} = 5.80 \times 1 0^{3} K = 5800 K$

Step 2 — Convert the peak wavelength to metres for part (b): $λ_{ma x} = 500 nm = 5.0 \times 1 0^{- 7} m$ . Why this step? The photon-energy formula $E = h c / λ$ needs metres, so we fix units before substituting.

Step 3 — Compute the peak-photon energy with $E = \frac{h c}{λ _{ma x}}$ . Why this step? Part (b) is a Cell-B calculation nested inside the word problem: given $λ$ , find one quantum's energy. $E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{5.0 \times 1 0 ^{- 7}} = \frac{1.988 \times 1 0 ^{- 25}}{5.0 \times 1 0 ^{- 7}} = 3.98 \times 1 0^{- 19} J$

Verify: (a) $T = 5800 K$ matches the known solar surface temperature ✓. (b) $E = 3.98 \times 1 0^{- 19} J$ sits between green light ( $3.3 \times 1 0^{- 19}$ ) and 400 nm UV ( $4.97 \times 1 0^{- 19}$ ), exactly where a 500 nm photon should land. Both parts pass. ✓

Example 10 — Cell J: exam-twist ratio question

Worked example Compare two modes' average energies — the constants cancel

At a fixed temperature $T$ , mode 1 has frequency $ν_{1}$ with $\frac{h ν _{1}}{k T} = 1$ , and mode 2 has frequency $ν_{2}$ with $\frac{h ν _{2}}{k T} = 2$ . Find the ratio $\frac{⟨ E ⟩ _{2}}{⟨ E ⟩ _{1}}$ of their average energies.

Forecast: mode 2 is the higher-frequency one, so it is more suppressed by the freezing-out effect — expect the ratio to come out less than 1.

Step 1 — Write each average with the shorthand $x = \frac{h ν}{k T}$ : $⟨ E ⟩ = k T \cdot \frac{x}{e ^{x} - 1}$ . Why this step? In a ratio the shared factor $k T$ cancels, so the answer depends only on the two $x$ values — the physical constants disappear, which is the whole point of this exam style. $\frac{⟨ E ⟩ _{2}}{⟨ E ⟩ _{1}} = \frac{k T \frac{x _{2}}{e ^{x_{2}} - 1}}{k T \frac{x _{1}}{e ^{x_{1}} - 1}} = \frac{x _{2}}{x _{1}} \cdot \frac{e ^{x_{1}} - 1}{e ^{x_{2}} - 1}$

Step 2 — Substitute $x_{1} = 1$ and $x_{2} = 2$ . Why this step? Now it is pure arithmetic; use $e^{1} = 2.71828$ and $e^{2} = 7.38906$ . $\frac{⟨ E ⟩ _{2}}{⟨ E ⟩ _{1}} = \frac{2}{1} \cdot \frac{e ^{1} - 1}{e ^{2} - 1} = 2 \cdot \frac{1.71828}{6.38906} = 2 \times 0.26893 = 0.5379$

Verify: The ratio $0.538 < 1$ , matching the forecast — the higher-frequency mode holds less average energy even though each of its packets ( $h ν_{2}$ ) is bigger. This is precisely the "starving" trend that bends the intensity curve down at high $ν$ , and the exam-favourite subtlety flagged in the parent note's second Common Mistake. ✓

Recall Quick self-test across the matrix

Which cell needs a $ν = c / λ$ conversion first? ::: Cell B (given wavelength), unless you use $E = h c / λ$ directly. In which regime does $⟨ E ⟩ \to k T$ ? ::: Low frequency, $h ν ≪ k T$ (Cell F / the $ν \to 0$ limit H). In which regime does $⟨ E ⟩ \to 0$ ? ::: High frequency, $h ν ≫ k T$ (Cell G) — the ultraviolet cure. How do you count photons from a beam power? ::: Divide power (J/s) by energy per photon (J), giving photons/s (Cell E). How do you get a mole of photons' energy? ::: Multiply single-quantum energy by $N_{A} = 6.022 \times 1 0^{23}$ (Cell D).

Connections

Parent: 2.1.01 Black-body radiation and Planck's quantum hypothesis E = hν (Hinglish) — the theory these examples exercise.
Photoelectric Effect — Cell E's photon-counting is the doorway to it.
Wien's Displacement Law — powers Example 9's word problem.
Boltzmann Distribution — the $e^{- h ν / k T}$ weighting behind Cells F, G, H, J.
Bohr Model of the Atom · de Broglie Wavelength · Stefan-Boltzmann Law — where these quantized-energy skills go next.