Part 5 · The old quantum theory

5.5 Franck–Hertz, Stern–Gerlach and wave–particle duality

Source: Postorino, week 4 slides, pp. 29–33

Objectives

Picture. Franck–Hertz: atoms accept energy only in definite amounts. Stern–Gerlach: angular momentum along a field is quantised.
Theorem. $E = c P^{2} + m^{2} c^{2}$ ; for a massless particle $E = P c$ , so a photon has $P = h / λ$ .
Recipe. De Broglie: $λ = h / P$ , $ν = E / h$ for any particle; compare $λ$ with the size of the apparatus.
Key. Bohr’s condition is a standing wave: $2 π r = nλ$ .

Two more experiments, one for each new quantum feature, and then the idea that ties waves and particles together.

Franck–Hertz: discrete energy levels

Apparatus (with the lecturer’s handwritten labels):

a tube filled with mercury vapour;
a heated cathode (catodo) emitting electrons;
a positive grid, the anode (anodo), which accelerates them through the voltage $V$ ;
a collector (collettore) just beyond, held slightly negative with respect to the grid, so it weakly repels the electrons;
an ammeter measuring the collected current.

How it works. A collision between an electron and an Hg atom can be elastic, or involve an exchange of energy.

As the grid voltage grows, the current grows.
When the electrons’ kinetic energy reaches 4.9 eV, the energy of an electronic transition of Hg, they can give it up in an inelastic collision.
The collector is weakly repulsive, so the electrons that gave 4.9 eV to Hg atoms have too little energy left to reach it: the current drops.
A second drop comes at $2 \times 4.9$ eV, when an electron can excite two atoms in turn, and so on.

The slide’s data (“Franck-Hertz data for mercury”): collected current against accelerating voltage, with peaks near 4, 9 and 14.5 V, each followed by a drop. The current keeps rising overall, so each peak is higher than the last; their spacing is the 4.9 eV that matters. The whole pattern is shifted by about 1 V by contact potentials in the tube beyond the lectures.

Loading the demo…

Figure 5.5.1 A model of the Franck–Hertz curve. Drag the accelerating

voltage

: the readout counts how many mercury atoms an electron can excite on its way and what is left over. The current drops just past each

multiple of 4.9 eV

(plus the 1 V contact shift).

Key observation 5.5.1The atom accepts energy only in definite amountsslides w4, p. 29

The atom’s energy levels are discrete, as Bohr postulated, and shown here without any light. Interactive demonstration on the slide: Wolfram, Franck–Hertz experiment.

The excited Hg atoms decay by emitting photons of $h c /4.9 eV = 253$ nm, and this ultraviolet line is indeed seen from the tube beyond the lectures.

Stern–Gerlach: spin

The slide only names the experiment and what it shows: the magnetic moment, the angular momentum, and the spin of the electron.

RemarkThe experimentbeyond the lectures

A beam of silver atoms passes through a strongly inhomogeneous magnetic field. An atom with magnetic moment $μ$ feels a force $F_{z} \approx μ_{z} \partial B_{z} / \partial z$ , which deflects it according to the component of its moment along the field.
Classically the moments point every which way, so the beam should smear into a continuous band.
It splits into two discrete spots. So the component of the magnetic moment, and of the angular momentum it comes from, along the field is quantised (“space quantisation”).
Two values, not an odd number as orbital angular momentum would give: silver’s moment comes from one unpaired electron’s intrinsic angular momentum, its spin, with component $\pm ℏ/2$ .

Wave–particle duality

Waves and photons. Interference shows the wave nature of radiation (§2.4 Adding waves: interference, standing waves, beats); the photoelectric effect and the photon show a particle nature (§5.3 The photoelectric effect). A wave is intrinsically delocalised (a monochromatic wave completely so, Definition 1.6.9); a particle is intrinsically localised.
Particles (electrons, neutrons) collide and are detected at points. Yet a beam of neutrons sent onto a crystal gives intensity peaks exactly as in X-ray diffraction (Theorem 2.5.6).

De Broglie’s hypothesis formalises a symmetric approach to radiation and matter. First, back to the photon: the electron in hydrogen is non-relativistic, but the photon moves at $c$ , so a little relativity is needed.

Elements of relativistic mechanics

Definition 5.5.2Lorentz transformationslides w4, p. 31

It takes the coordinates of an event, the four-vector $R \equiv (x, y, z, t)$ in an inertial frame, to its coordinates $R^{'} \equiv (x^{'}, y^{'}, z^{'}, t^{'})$ in another inertial frame moving with uniform velocity $v$ along $x$ : $t^{'} = γ (t - \frac{v}{c ^{2}} x), x^{'} = γ (x - v t), y^{'} = y, z^{'} = z, γ = \frac{1}{1 - ( v / c ) ^{2}},$ with $γ$ the Lorentz factor.

N.B. For $v ≪ c$ , $v / c^{2} \to 0$ and $γ \to 1$ : the transformation becomes the classical Galilean one, $t^{'} = t$ , $x^{'} = x - v t$ , $y^{'} = y$ , $z^{'} = z$ .

Definition 5.5.3Relativistic energy and momentumslides w4, p. 31

$E = γm c^{2}, P = γm v,$ with $m$ the invariant (rest) mass. In the particle’s own frame $γ = 1$ , so $E = m c^{2}$ .

Proposition 5.5.4Low-speed limitslides w4, p. 31

For $v ≪ c$ , with $\frac{1}{1 - x ^{2}} \approx 1 + \frac{1}{2} x^{2}$ , $γ \approx 1 + \frac{1}{2} \frac{v ^{2}}{c ^{2}}, E \approx m c^{2} + \frac{1}{2} m v^{2}, P \approx m v .$ The first term of $E$ is the rest energy, a constant; the second is the classical kinetic energy.

Massless particles. If $m = 0$ then $v = c$ , $γ = \infty$ , and $E = γm c^{2}$ , $P = γm v$ lose their meaning: $0 \cdot \infty$ . Separate the massive part from the kinetic one.

Theorem 5.5.5Energy–momentum relationslides w4, p. 32

$E = c P^{2} + m^{2} c^{2} .$ It makes sense for $m = 0$ too: for a massless particle $E = P c (dispersion relation) .$

Proof

Square $P = γm v$ and solve for $v$ : $P^{2} (1 - v^{2} / c^{2}) = m^{2} v^{2}$ gives $v^{2} = \frac{P ^{2} c ^{2}}{P ^{2} + m ^{2} c ^{2}}$ .
Square $E = γm c^{2}$ and use this. $1 - v^{2} / c^{2} = \frac{m ^{2} c ^{2}}{P ^{2} + m ^{2} c ^{2}}$ , so $E^{2} = \frac{m ^{2} c ^{4}}{1 - v ^{2} / c ^{2}} = c^{2} (P^{2} + m^{2} c^{2})$ .

(The intermediate algebra is not on the slide.)

De Broglie’s hypothesis

For the photon, $m = 0$ , so

Proposition 5.5.6Photon energy and momentumslides w4, p. 33

$E = h ν, P = \frac{E}{c} = \frac{h ν}{c} = \frac{h}{λ} = ℏ k or, inverting, ν = \frac{E}{h}, λ = \frac{h}{P},$ with $k = 2 π / λ$ and $ℏ = h /2 π$ .

Definition 5.5.7De Broglie’s hypothesis (1924)slides w4, p. 33

Associate with every particle a wave, of wavelength and frequency given by the same relations as for the photon: $λ = \frac{h}{P}, ν = \frac{E}{h} .$ N.B. For a non-relativistic particle the (kinetic) energy is $E = \frac{1}{2} m v^{2}$ and the momentum $P = m v$ .

Example 5.5.8Why we don’t see a bullet’s waveslides w4, p. 33

A bullet, $m = 10$ g, $v = 300$ m/s: $λ = \frac{h}{m v} = \frac{6.63 \times 1 0 ^{- 34} J \cdot s}{1 0 ^{- 2} kg \cdot 3 \times 1 0 ^{2} m/s} \approx 2.2 \times 1 0^{- 34} m .$

To observe interference you need an apparatus the size of the wavelength (Theorem 2.5.1). Such a device does not and cannot exist: atomic sizes are of order $1 0^{- 10}$ m.
In principle you could use very low speeds, but not arbitrarily low: Heisenberg’s uncertainty principle and thermal agitation stop you (to be seen later).

Example 5.5.9For atomic particles it is visiblebeyond the lectures

An electron accelerated through $V$ volts has $P = 2 m e V$ , so $λ = 1.226/ V$ nm. At 100 V, $λ = 0.123$ nm: the spacing of atoms in a crystal. Electrons diffract off crystals (Davisson and Germer, 1927), which is how electron microscopes resolve atoms.
A thermal neutron at 300 K, kinetic energy about $K_{B} T$ , has $λ = h / 2 m K_{B} T \approx 0.18$ nm: hence the neutron diffraction of the slide.

Loading the demo…

Figure 5.5.2 De Broglie wavelengths on the same length ruler as the electromagnetic spectrum. Pick a particle and set its speed: an electron at a few

1 0^{6}

m/s has the size of an atom, so a crystal diffracts it; a bullet’s wavelength is off the scale,

1 0^{20}

times smaller than a nucleus.

Key observation 5.5.10Bohr’s condition, revisitedbeyond the lectures

With $λ = h / m v$ , Bohr’s $m v r = n ℏ$ becomes $2 π r = nλ .$ A stationary orbit is one whose circumference holds a whole number of electron wavelengths: a standing wave closed on itself, like the modes of a string (Proposition 2.4.4). This is the first hint of why quantisation happens.

Loading the demo…

Figure 5.5.3 An electron wave wrapped round an orbit. Set how many wavelengths fit in the circumference: for a whole number the wave closes on itself; otherwise the next turn (orange) is out of step, and turn after turn it would cancel itself out.

Exercises

Exercise 5.5.1beyond the lectures

Find the de Broglie wavelength of an electron accelerated through 100 V, and compare with the bullet.

Solution

$P = 2 m e V = 2 \cdot 9.11 \times 1 0^{- 31} \cdot 1.6 \times 1 0^{- 17} = 5.4 \times 1 0^{- 24}$ kg·m/s, $λ = h / P = 0.123$ nm: $1 0^{24}$ times the bullet’s, and the size of an atomic spacing, so a crystal diffracts it.

Exercise 5.5.2beyond the lectures

Find the momentum of a 500 nm photon, and the speed an electron would need to have the same momentum.

Solution

$P = h / λ = 6.63 \times 1 0^{- 34} /5 \times 1 0^{- 7} = 1.33 \times 1 0^{- 27}$ kg·m/s. $v = P / m = 1.33 \times 1 0^{- 27} /9.11 \times 1 0^{- 31} = 1.46 \times 1 0^{3}$ m/s.

Exercise 5.5.3beyond the lectures

Show that the electron in Bohr orbit $n = 1$ has de Broglie wavelength $2 π a_{0}$ .

Solution

$λ = h / (m v_{1}) = h / (m α c)$ , and $a_{0} = ℏ/ (m α c)$ (from $a_{0} = 4 π ε_{0} ℏ^{2} / m e^{2}$ and $α = e^{2} /4 π ε_{0} ℏ c$ ). So $λ = 2 π ℏ/ (m α c) = 2 π a_{0} = 0.332$ nm: exactly one wavelength round the orbit.

Exercise 5.5.4beyond the lectures

What wavelength of light do mercury atoms excited in the Franck–Hertz tube emit?

Solution

$λ = h c / E = 1240/4.9 = 253$ nm, in the ultraviolet.

Exercise 5.5.5beyond the lectures

An electron has kinetic energy 1 MeV. Is $λ = h / 2 m K$ right? Compute $λ$ properly.

Solution

No: $K = 1$ MeV is twice the rest energy, so the electron is relativistic. $E = K + m c^{2} = 1.511$ MeV, $P c = E^{2} - (m c^{2})^{2} = 1.51 1^{2} - 0.51 1^{2} = 1.422$ MeV, $λ = h c / P c = 1240 eV \cdot nm /1.422 \times 1 0^{6} eV = 8.7 \times 1 0^{- 4}$ nm. The non-relativistic formula would give $1.23 \times 1 0^{- 3}$ nm, 40 % too long.