1The idea in one paragraph

Shine ultraviolet light on a clean metal and electrons fly out. The wave picture of light, so successful for interference and diffraction, predicts the wrong answer for almost every detail of this. Einstein's explanation is that light arrives in photons, each carrying energy hf, and that one photon gives all its energy to one electron. Twenty years later came the reverse surprise: electrons, which leave tracks and carry charge like tiny balls, diffract through crystals like waves, with a wavelength λ = h/p. And X-ray photons bounce off electrons like billiard balls, losing momentum and coming away with a longer wavelength. Light and matter both behave as waves in some experiments and as particles in others. That is wave–particle duality.

2Photons, and what intensity means

A photon is a single quantum of electromagnetic energy. Its energy depends only on its frequency:

E = hf = hc/λ, with h = 6.63 × 10⁻³⁴ J s

A photon of violet light at 400 nm carries hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (4.00 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J, which is 3.11 eV. A red photon at 650 nm carries only 1.91 eV. Ultraviolet photons carry more still.

The intensity of a beam is the power per unit area. In the photon picture, a more intense beam of one colour is not made of stronger photons; it is made of more photons per second. A 5.0 mW laser at 405 nm sends out 5.0 × 10⁻³ ÷ 4.91 × 10⁻¹⁹ = 1.0 × 10¹⁶ photons every second. Double the power and you double the number; each photon's energy is unchanged. Keep this sentence in mind: it answers half the questions in this subtopic.

3The photoelectric effect: what is observed

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation falls on it. The electrons are called photoelectrons. Figure 1 shows the standard apparatus.

Light falls on a metal plate, the photocathode, inside an evacuated tube. Emitted electrons cross to a second electrode, the anode, and the microammeter measures the current. The variable supply sets the potential difference between the electrodes, and it can be reversed. With the anode positive, every emitted electron is pulled across. With the anode made negative, electrons are pushed back and only the most energetic ones get across. The reverse potential difference that just stops the fastest electrons, so that the current falls to zero, is the stopping potential V_s. At that point the work done against the field, eV_s, equals the maximum kinetic energy:

E_max = eV_s, so a stopping potential of 0.81 V means E_max = 0.81 eV

Figure 2 shows the current as the potential difference is varied, for three beams.

Four results, and the exam asks about all of them.

1. There is a threshold frequency. For each metal there is a minimum frequency, the threshold frequency f₀. Below it no electrons are emitted at all, however intense the light. Above it, electrons are emitted even when the light is very dim.
2. The maximum kinetic energy depends on frequency, not intensity. In Figure 2, doubling the intensity (A to B) doubles the current but leaves the stopping potential the same. Raising the frequency (A to C) increases the stopping potential.
3. Emission is instantaneous. Electrons appear as soon as the light is switched on, with no measurable delay, even at very low intensity.
4. The current is proportional to the intensity, as long as the frequency is above threshold.

4Why the wave theory fails

The guide asks you to discuss which features the classical wave theory cannot explain, so learn this argument. In the wave picture, light's energy is spread continuously across the wavefront and grows with the intensity (the square of the amplitude). An electron would soak up energy from the wave until it had enough to escape.

The time delay is worth an estimate. Suppose dim light of intensity 1.0 × 10⁻⁶ W m⁻² falls on a metal whose electrons need 2.1 eV = 3.4 × 10⁻¹⁹ J to escape, and suppose one electron can gather energy from an area about the size of one atom, 1 × 10⁻¹⁹ m². It receives 10⁻⁶ × 10⁻¹⁹ = 10⁻²⁵ W, so it needs 3.4 × 10⁻¹⁹ ÷ 10⁻²⁵ ≈ 3 × 10⁶ s: more than a month. In the experiment, electrons appear within a tiny fraction of a second.

The fourth result, current proportional to intensity, is the only one the wave theory gets right, and it is also explained by photons. So the evidence does not just favour the photon model; it rules out the wave model for this effect.

5Einstein's explanation

In 1905 Einstein proposed that light is absorbed in whole photons, and that one photon gives all its energy to one electron. Some of that energy is needed to pull the electron out of the metal. The minimum energy needed to remove an electron from the surface is the work function Φ of the metal. Whatever is left becomes the electron's kinetic energy. For the electrons that escape most easily, the ones at the surface, that leftover is the largest possible:

E_max = hf − Φ

Figure 3 shows the energy accounts for three photons hitting the same metal.

Each observation now follows in one line.

• Threshold frequency. If hf < Φ, no photon has enough energy to free an electron, and an electron cannot save up energy from several photons. At the threshold, hf₀ = Φ, so f₀ = Φ/h.
• E_max depends on f. Each electron receives one photon's energy, hf. Raise f and E_max rises; the relation is a straight line.
• Intensity. A more intense beam has more photons per second, not more energetic ones, so more electrons come out each second with the same E_max. The current rises; the stopping potential does not.
• No delay. The energy arrives in a packet. The first photon absorbed can free an electron at once.

Most electrons come out with less than E_max, because they start below the surface and lose energy in collisions on the way out. That is why the equation gives a maximum, and why the current in Figure 2 falls gradually as the reverse potential difference rises rather than all at once.

Worked example. Light of wavelength 400 nm falls on a metal with work function 2.30 eV. Find the maximum kinetic energy of the photoelectrons, their maximum speed, and the threshold wavelength.

Light of wavelength longer than 540 nm, such as red light at 650 nm, ejects nothing from this metal however bright it is. The stopping potential for the 400 nm light is 0.81 V.

6The E_max–f graph: finding h and Φ

Rewrite Einstein's equation as E_max = hf − Φ and compare it with y = mx + c. A graph of E_max against f is a straight line:

• its gradient is h, the same for every metal;
• it meets the frequency axis at the threshold frequency f₀;
• extended backwards, it meets the energy axis at −Φ.

Figure 4 shows two metals. The lines are parallel, because h is a universal constant, and the metal with the larger work function has the larger threshold.

In the laboratory you measure the stopping potential, so the graph you actually plot is V_s against f. Since eV_s = hf − Φ, V_s = (h/e)f − Φ/e: the gradient is h/e, and the intercept on the V_s axis is −Φ/e, which in volts is numerically −Φ in electronvolts.

Worked example (Paper 1B style, invented data). A student measures the stopping potential for five frequencies.

Figure 5 plots them. The line of best fit passes through (6.0 × 10¹⁴ Hz, 0.39 V) and (10.0 × 10¹⁴ Hz, 2.05 V).

Take the gradient from the line, not from two data points, and use points far apart. The value of h is within about 0.2% of the accepted value. The V_s intercept lies far off the measured range, so it needs a long extrapolation; if the question asks for Φ, it is usually quicker and more accurate to read f₀ from the frequency axis and use Φ = hf₀.

7Matter waves: the de Broglie wavelength

In 1924 Louis de Broglie turned the argument round. If waves of light can behave like particles, perhaps particles of matter can behave like waves. He proposed that every particle with momentum p has a wavelength

λ = h/p

called its de Broglie wavelength. For a photon the same relation holds, which is why a photon carries momentum p = h/λ even though it has no mass. That fact returns in Compton scattering.

The wavelength is tiny for everyday objects. A 58 g tennis ball at 50 m s⁻¹ has λ = 6.63 × 10⁻³⁴ ÷ (0.058 × 50) = 2.3 × 10⁻³⁴ m, far smaller than any gap it could pass through, so no diffraction is ever seen. For an electron the story is different, because its mass is so small.

For an electron accelerated from rest through a potential difference V, the work done eV becomes kinetic energy, and momentum follows from E_k = p²/2m:

Worked example. Electrons accelerated through 54 V:

That is about the spacing of atoms in a crystal. So a crystal should act as a diffraction grating for electrons, exactly as it does for X-rays.

8Diffraction of particles: the evidence

In 1927 Clinton Davisson and Lester Germer fired 54 eV electrons at a nickel crystal and found a strong reflected beam at one particular angle, just as X-rays of the same wavelength would give. In the same year G. P. Thomson passed electrons through thin metal films and photographed rings. The wavelengths matched λ = h/p.

The school version is the electron diffraction tube in Figure 6. An electron gun accelerates electrons through a few kilovolts, and they pass through a very thin film of graphite onto a fluorescent screen.

The screen shows concentric rings, not a single spot. The graphite is made of many tiny crystals at random orientations, and the rows of atoms in them, about 10⁻¹⁰ m apart, act like the lines of a grating. Each ring is a direction in which waves scattered by neighbouring rows arrive in phase: constructive interference. Only waves interfere, so the rings are evidence that electrons have a wave nature.

The test that settles it: raise the accelerating voltage and the rings get smaller. A higher voltage gives the electrons more momentum, so a shorter de Broglie wavelength, so less diffraction and a smaller angle for each ring. Particles that were simply bouncing off atoms would show no such dependence. At 5.0 kV the wavelength is h/√(2m_e eV) = 1.7 × 10⁻¹¹ m.

Neutrons, atoms and even molecules have since been diffracted. And electrons sent through a double slit one at a time each land at a single point on the screen, like a particle, yet thousands of them together build up an interference pattern, like a wave. Each electron is detected as a particle; where it is likely to land is governed by a wave.

Wave–particle duality is the name for this. Light and matter both show wave properties (diffraction, interference) and particle properties (being emitted, absorbed or detected in whole, localised amounts of energy and momentum). Which one you see depends on the experiment. No experiment shows both at once for a single event, and neither picture on its own describes everything.

9A scattering experiment: where the first minimum falls

The guide requires one particular use of the de Broglie wavelength: a beam of particles diffracting off a small target, with the first minimum of intensity predicted from λ. The most important case is high-energy electrons scattered by nuclei. Electrons are chosen because they do not feel the strong nuclear force, so they probe the nucleus through the electric force alone.

A narrow beam of electrons, with energies of hundreds of MeV, hits a thin foil. A detector is moved round the foil and counts the electrons arriving at each angle θ. The count falls with angle, but not smoothly: at one angle it drops to a clear minimum, as in Figure 7.

The nucleus diffracts the electron waves the way an obstacle diffracts light. Treat it like the single slit of C.3, with the nuclear diameter D in place of the slit width. The first minimum is where

sin θ ≈ λ/D

so the angle of the first minimum measures the size of the nucleus. For the minimum to fall at a measurable angle, λ must be comparable with D, around 10⁻¹⁵ m; that is why such high electron energies are needed.

Worked example (invented data). Electrons with momentum 2.1 × 10⁻¹⁹ kg m s⁻¹ are scattered from a thin carbon target. The first minimum is observed at 35°. Estimate the radius of a carbon nucleus.

That agrees with R = R₀∛A = 1.20 × 10⁻¹⁵ × ∛12 = 2.7 × 10⁻¹⁵ m from E.1. This is one of the ways that rule was established. Two checks when answering: higher-energy electrons have a shorter wavelength, so the minimum moves to a smaller angle; and a larger nucleus also moves the minimum to a smaller angle.

10Compton scattering

In 1923 Arthur Compton fired X-rays of a single wavelength at a block of graphite and measured the wavelength of the X-rays scattered at different angles. Figure 8 shows the arrangement and what he found.

At each angle the scattered X-rays had two wavelengths: the original one, and a longer one. The longer wavelength moved further from the original as the angle increased.

The wave theory cannot produce this. A wave falling on an electron makes it oscillate at the wave's own frequency, and an oscillating charge re-radiates at that same frequency. Scattered waves should have exactly the incident wavelength, at every angle.

The photon model explains it as a collision. A photon carries energy E = hf and momentum p = h/λ. It strikes an almost free outer electron in the graphite, as in Figure 9, and bounces off at an angle θ while the electron recoils. Both energy and momentum are conserved. The electron takes some of the photon's energy, so the scattered photon has less energy, a lower frequency and a longer wavelength. The bigger the scattering angle, the harder the collision and the more energy the electron takes.

Applying conservation of energy and momentum gives the Compton shift (you use it; you do not derive it):

Δλ = λ_f − λ_i = (h/m_ec)(1 − cos θ)

where θ is the angle through which the photon is scattered and m_e is the electron's mass. The combination h/m_ec = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 3.00 × 10⁸) = 2.43 × 10⁻¹² m.

Three things follow. The shift does not depend on the incident wavelength or on the material, only on the angle. The largest possible shift is 4.85 pm. And a shift of a few picometres is only noticeable if the wavelength itself is small, which is why X-rays are used: for visible light at 500 nm the shift is less than 0.001%, too small to see. The unshifted line comes from photons that hit a tightly bound inner electron; the whole atom recoils, and with thousands of times the electron's mass it takes almost no energy.

Worked example. X-rays of wavelength 71.0 pm are scattered at 90° by free electrons. Find the scattered wavelength and the kinetic energy given to the electron.

How is this like a collision of two solid balls? In both, energy and momentum are conserved, the collision is elastic, and the struck body recoils while the other is deflected; a more head-on collision transfers more energy. How is it different? The photon has no mass and never slows down: it still travels at c after the collision. It loses energy by changing its frequency, not its speed, and its momentum is h/λ, not mv.

Why Compton scattering is the more convincing evidence. The photoelectric effect shows that light is absorbed in packets of energy hf, but the electron is bound in a metal and the photon disappears, so some physicists argued that the "packets" might come from the metal rather than from light. Compton scattering involves a single, nearly free electron, and the photon survives the collision. The shift matches, in detail, a calculation that treats the photon as a particle with momentum h/λ colliding with the electron. Light carries momentum in packets, exactly as a particle would.