Part 5 · The old quantum theory

5.1 Line spectra and blackbody radiation

Source: Postorino, week 4 slides, pp. 12, 17–18 and 23

Objectives

Picture. A gas emits and absorbs at the same few discrete frequencies; for hydrogen they follow Rydberg’s formula. A hot body emits a continuous spectrum that depends only on its temperature.
Vocabulary. Emission and absorption lines, wavenumber, Rydberg constant, blackbody, cavity, Stefan–Boltzmann law, Wien’s displacement law, energy density, modes.
Theorem. $R (T, λ) = \frac{c}{4} ρ (T, λ)$ , and the number of modes per unit volume and wavelength is $n (λ) = 8 π / λ^{4}$ .
Key. Discrete lines and stable atoms are incompatible with classical physics.

This part follows the experiments classical physics could not explain, and the ad hoc hypotheses that fixed them one at a time, before quantum mechanics proper. Notation: the slides write “Plank” for Planck; $K$ and $K_{B}$ both mean Boltzmann’s constant, $K_{B} = 1.38 \times 1 0^{- 23}$ J/K $= 8.62 \times 1 0^{- 5}$ eV/K, and $β = 1/ K_{B} T$ . Two conversions used throughout beyond the lectures: $h c = 1240$ eV·nm, so a photon of wavelength $λ$ in nm has energy $1240/ λ$ eV; and 1 eV corresponds to 8065.5 cm⁻¹.

The hydrogen spectrum

Emission. Analyse the radiation emitted by an incandescent gas, that is, measure its power spectrum $∣ c (ω) ∣^{2}$ (Key observation 1.7.2). It is made of a few discrete frequencies: the emission lines.

Absorption. Send white radiation (all frequencies) through the gas and measure the power spectrum of what is transmitted. It crosses the gas at every frequency except a few discrete ones: the absorption lines.

Key observation 5.1.1Emission and absorption lines coincideslides w4, p. 12

For a given element the absorption and emission lines sit at the same frequencies.

The slide shows hydrogen’s emission spectrum, four bright lines on black at about 410, 434, 486 and 656 nm, and its absorption spectrum, the full rainbow with four dark lines at the same wavelengths: the visible Balmer series (The energy levels of hydrogen).

Regularity. The lines are very regular. It is convenient to use the wavenumber $\tilde{ν} = 1/ λ$ , usually in cm⁻¹; it is proportional to frequency and to photon energy, $\tilde{ν} = ν / c = E / h c$ beyond the lectures.

Theorem 5.1.2Rydberg’s formulaslides w4, p. 12, 23

For hydrogen gas the emission and absorption wavenumbers are, almost exactly, $\tilde{ν}_{a, b} = R_{H} [\frac{1}{n _{a}^{2}} - \frac{1}{n _{b}^{2}}],$ with $n_{a} = 1, 2, 3, \dots$ and $n_{b} = 2, 3, 4, \dots$ positive integers, $n_{a} < n_{b}$ , and the Rydberg constant $R_{H} = 109677.58$ cm⁻¹.

Example 5.1.3The visible linesbeyond the lectures

$n_{a} = 2$ , $n_{b} = 3$ : $\tilde{ν} = R_{H} (1/4 - 1/9) = 15233$ cm⁻¹, $λ = 656.5$ nm, the red line. $n_{b} = 4, 5, 6$ give 486.3, 434.2 and 410.3 nm, the other three.

The formula was found empirically (Balmer 1885, Rydberg 1888).

Key observation 5.1.4Classical physics can’t explain itslides w4, p. 12

This pattern, together with the stability of the atom, is incompatible with classical physics.

RemarkWhybeyond the lectures

Stability. In a classical planetary atom the electron orbits the proton with a centripetal acceleration. An accelerating charge radiates electromagnetic waves, so it would lose energy and spiral into the nucleus, in about $1 0^{- 11}$ s.
Discreteness. While spiralling in, its orbital frequency would change continuously, so it would emit a continuous spectrum, not sharp lines.

Bohr’s atom explains both.

Blackbody radiation

Definition 5.1.5Blackbodyslides w4, p. 17

Any body at $T \neq = 0$ absorbs and emits electromagnetic radiation; at thermal equilibrium it emits as much energy as it absorbs.
A blackbody absorbs all the radiation falling on it: an ideal absorber, and therefore also an ideal emitter.
The blackbody spectrum is universal: it depends only on the temperature, not on the material or the shape.

A good absorber at a wavelength is an equally good emitter there; otherwise a body in equilibrium with radiation would heat up or cool down on its own (Kirchhoff’s law) beyond the lectures.

The model blackbody: a cavity at thermal equilibrium at temperature $T$ with a small hole. A ray entering the hole bounces around inside and is absorbed long before it finds the hole again, so the hole absorbs everything: it is black. The radiation that comes out of the hole is the equilibrium radiation inside the cavity. An interactive simulation is given on the slide: PhET, blackbody spectrum.

The experimental laws

Call $R$ the power emitted per m².

Theorem 5.1.6Stefan–Boltzmann lawslides w4, p. 17

$R (T) = σ T^{4}, σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4} .$

WarningSlide slip

The slide prints the units of $σ$ as W m² K⁻⁴. It is a power per unit area per K⁴: W m⁻² K⁻⁴.

The spectral version: $R (T, λ)$ is defined by $R (T) = \int_{0}^{\infty} R (T, λ) d λ$ , so $R (T, λ) d λ$ is the power emitted per m² between $λ$ and $λ + d λ$ .

Theorem 5.1.7Wien’s displacement lawslides w4, p. 17

The peak of $R (T, λ)$ moves to shorter wavelengths as $T$ grows: $λ_{max} T = b, b = 2.898 \times 1 0^{- 3} m \cdot K .$

(The value of $b$ is not on the slide.) The slide’s curves for 6000, 5000, 4000 and 3000 K: each rises steeply from zero at short wavelengths, peaks, and falls off slowly at long ones; hotter bodies emit more at every wavelength, with area $σ T^{4}$ .

Loading the demo…

Figure 5.1.1 The blackbody spectrum

R (T, λ)

against wavelength, with the visible band. Drag the temperature. The

peak

slides along Wien’s law

λ_{max} T = b

, and the area,

σ T^{4}

, grows as the fourth power. At 5800 K (the Sun) the peak is in the visible; at 3000 K (a filament bulb) it is in the infrared; at 300 K (you) near 10 µm.

At 6000 K the peak is near 480 nm, in the visible: why the Sun (about 5800 K) looks white-yellow.
At 3000 K it is near 970 nm, in the infrared: an incandescent bulb gives mostly heat.
A body at room temperature peaks near 10 µm, far infrared: thermal cameras see it.

From the cavity to the oscillators

Key observation 5.1.8The physical pictureslides w4, p. 18

Energy is continually exchanged between the oscillators of the cavity’s inner surface (oscillating dipoles, as in §3.1 Dielectrics: the Lorentz model) and the radiation inside.
At thermal equilibrium the electromagnetic energy density in the cavity must equal the energy density of the harmonic oscillators interacting with the radiation.

Let $ρ (T, λ)$ be the energy density of the radiation inside the cavity: an energy per volume, strictly per volume and per unit wavelength.

Proposition 5.1.9Emitted power and energy densityslides w4, p. 18

$R (T, λ) = ρ (T, λ) \frac{c}{4} .$ $c /4$ is a geometric factor.

Proof of the c/4beyond the lectures

Inside the cavity the radiation is isotropic: a fraction $d Ω/4 π$ of it travels within any solid angle $d Ω$ . Take a small hole of area $A$ .

Radiation travelling at angle $θ$ to the normal crosses the hole with normal speed $c cos θ$ : in a time $d t$ it brings the energy in a slanted cylinder of volume $A c cos θ d t$ .
The energy leaving per unit area and time is $R = ρ c \int_{hemisphere} cos θ \frac{d Ω}{4 π} = \frac{ρ c}{4 π} \int_{0}^{2 π} d φ \int_{0}^{π /2} cos θ sin θ d θ = \frac{ρ c}{4 π} \cdot 2 π \cdot \frac{1}{2} = \frac{ρ c}{4} .$

Only the outgoing hemisphere counts, and the $cos θ$ weighting halves it again: $1/4$ .

The energy density of the radiation equals that of the oscillators on the surface: $ρ (T, λ) = n (λ) \overline{ε (T, λ)},$ with $n (λ)$ the number of vibration modes per unit volume (and per unit wavelength) and $\overline{ε (T, λ)}$ the mean energy of one oscillator. Each mode of the field behaves like one harmonic oscillator, so energy per mode × number of modes is the energy density.

Theorem 5.1.10Number of modesslides w4, p. 18

$n (λ) = \frac{8 π}{λ ^{4}}$ modes per unit volume and per unit wavelength (“from thermodynamics”, the slide says).

Counting the modesbeyond the lectures

Take a cubic cavity of side $L$ with conducting walls. The field must vanish at the walls, so it is a superposition of standing waves (Proposition 2.4.4) with $k_{x} = \frac{π n _{x}}{L}, k_{y} = \frac{π n _{y}}{L}, k_{z} = \frac{π n _{z}}{L}, n_{x}, n_{y}, n_{z} = 1, 2, 3, \dots$

Each allowed mode is a point of a cubic lattice in $k$ -space, one per volume $(π / L)^{3}$ .
The modes with $∣ k ∣ < k$ fill one octant (positive $n$ 's) of a sphere of radius $k$ . With two independent polarisations for each $k$ (Theorem 2.2.7): $N (k) = 2 \cdot \frac{1}{8} \cdot \frac{\frac{4}{3} π k ^{3}}{( π / L ) ^{3}} = \frac{k ^{3} L ^{3}}{3 π ^{2}} .$
Per unit volume and per unit $k$ : $\frac{1}{L ^{3}} \frac{d N}{d k} = \frac{k ^{2}}{π ^{2}}$ .
Change variable to $λ$ with $k = 2 π / λ$ , $∣ d k / d λ ∣ = 2 π / λ^{2}$ : $n (λ) = \frac{k ^{2}}{π ^{2}} \frac{d k}{d λ} = \frac{4}{λ ^{2}} \cdot \frac{2 π}{λ ^{2}} = \frac{8 π}{λ ^{4}} .$

Short wavelengths have enormously many modes: halving $λ$ multiplies $n (λ)$ by 16. This is what causes the catastrophe of §5.2 Rayleigh–Jeans and Planck.

The mean energy of an oscillator

To compute $\overline{ε (T, λ)}$ one needs the probability of each energy value, $E_{k}$ or $E$ . Given a probability $P$ , the mean of a quantity $A$ is (Definition 4.1.10) $\overline{A (E_{k})} = k = 0 \sum \infty A (E_{k}) P (E_{k}) (discrete), \overline{A (E)} = \int A (E) P (E) d E (continuous) .$

Proposition 5.1.11Boltzmann probabilitiesslides w4, p. 18–19

At equilibrium the probability is Boltzmann’s (Definition 4.2.2): from $\tilde{n}_{k} = N e^{- β E_{k}} / Z$ and $\tilde{n}_{k} = P (k) N$ , $P (E_{k}) = \frac{e ^{- β E_{k}}}{\sum _{k} e ^{- β E_{k}}} (discrete energy), P (E) = \frac{e ^{- βE}}{\int e ^{- βE} d E} (continuous energy) .$

The next section computes $\overset{ε}{ˉ}$ both ways, and the difference between the two is the birth of quantum physics.

Exercises

Exercise 5.1.1beyond the lectures

The Sun’s surface is at about 5778 K. Find the wavelength of its emission peak and its emitted power per m².

Solution

$λ_{max} = 2.898 \times 1 0^{- 3} /5778 = 502$ nm, in the visible. $R = σ T^{4} = 5.67 \times 1 0^{- 8} \times 577 8^{4} = 6.3 \times 1 0^{7}$ W/m².

Exercise 5.1.2beyond the lectures

Compute the wavelength of the H $_{β}$ line ( $n = 4 \to 2$ ) from $R_{H}$ .

Solution

$\tilde{ν} = 109677.58 (1/4 - 1/16) = 20564.5$ cm⁻¹, so $λ = 486.3$ nm (in vacuum; the 4861 Å of tables is the wavelength in air).

Exercise 5.1.3beyond the lectures

The cosmic microwave background is a blackbody at 2.725 K. Where is its peak, and in what region?

Solution

$λ_{max} = 2.898 \times 1 0^{- 3} /2.725 = 1.06$ mm: microwaves.

Exercise 5.1.4beyond the lectures

By what factor does the power radiated by a body grow if its absolute temperature doubles? And how many more modes per unit volume are there between 500 and 501 nm than between 1000 and 1001 nm?

Solution

$2^{4} = 16$ . Modes: $n (λ) d λ \propto λ^{- 4}$ , so $(1000/500)^{4} = 16$ times more.

Exercise 5.1.5beyond the lectures

What is the shortest wavelength in the Lyman series ( $n_{a} = 1$ ), and the longest?

Solution

Shortest: $n_{b} \to \infty$ , $\tilde{ν} = R_{H}$ , $λ = 91.2$ nm (the series limit). Longest: $n_{b} = 2$ , $\tilde{ν} = \frac{3}{4} R_{H}$ , $λ = 121.6$ nm. All in the ultraviolet.