What the effect is

The photoelectric effect is the emission of electrons from a material surface when electromagnetic radiation of high enough frequency strikes it. The electrons that leave are called photoelectrons; they have escaped from the surface after gaining energy from the incoming radiation.

A clean metal surface can, in this sense, turn light into an electric current. Shine suitable radiation on it, electrons leave, and if a second electrode collects them, charge moves through the external circuit. That is the basic idea behind a photoelectric cell.

A photon is a quantum of electromagnetic radiation that transfers energy and momentum in discrete interactions. The energy of one photon is

For light in vacuum,

Why this is particle evidence

The main observations are awkward — genuinely awkward — for a purely classical wave model of light:

• Below a certain frequency, no electrons are emitted, however intense the radiation is.
• Above that frequency, emission begins without a measurable delay, even when the radiation is very weak.
• Increasing intensity increases the number of photoelectrons emitted per second, but it does not increase the maximum kinetic energy of each photoelectron.
• Increasing frequency increases the maximum kinetic energy of the photoelectrons.

Classical wave theory treats the energy of light as spread continuously across the wavefront, with intensity proportional to amplitude squared. On that view, a very weak beam should eventually give an electron enough energy if we wait long enough. A more intense beam should also give electrons more energy. The experiment does not behave like that. One photon interacts with one electron; if the photon has enough energy, the electron can leave at once.

This is a neat case of falsification in physics. The wave model of light was not thrown away — diffraction and interference still require it — but the claim that light is only a wave could no longer stand. Einstein’s explanation put the particle description at the centre of the story.

The Bohr model does not save a wave-only explanation here. The Bohr model is an atomic model that describes electrons in quantized states of a hydrogen-like atom. The photoelectric effect concerns electrons near the surface of a metal, where bonding and conduction behaviour matter, and the Bohr model was not built to describe that. It is a useful historical model, not a general theory of metal surfaces.

The frequency cut-off

The threshold frequency is the minimum frequency of incident electromagnetic radiation needed to release photoelectrons from a specified metal surface. It is written

Watch the word “minimum”: it refers to frequency, not intensity. A weak ultraviolet source can release electrons from some metals, while a very bright red source may release none. Examiners like this detail because it tests the photon model, rather than the loose idea that “more light means more energy”.

Work function and threshold condition

The work function is the minimum energy required for an electron to escape from a specified metal surface. It is written

At the threshold frequency,

$hf_0 = \Phi$.

If $f < f_0$, each photon has too little energy, so no photoelectrons are emitted. If $f = f_0$, the photon energy is just enough for escape, and an electron would leave with essentially no kinetic energy. If $f > f_0$, the extra photon energy becomes kinetic energy of the emitted electron.

Different metals have different threshold frequencies because their surface electrons are bound with different strengths. The surface condition matters too: oxide layers, impurities and crystal structure can change the effective work function.

Energy accounting, one photon at a time

Einstein explained the photoelectric effect by applying conservation of energy to one photon and one electron. The photon is absorbed. Some of its energy frees the electron from the metal; the rest, if there is any, becomes kinetic energy of the photoelectron.

The maximum kinetic energy of the emitted photoelectrons is given by

“Maximum” is doing real work here: electrons do not all start in equally favourable positions inside the metal. Some lose energy on the way out, while the least hindered ones can escape with $E_{\max}$.

Since $\Phi = hf_0$, the same equation can be written as

What the graph says

If $E_{\max}$ is plotted against $f$, the result is a straight line. The gradient is $h$, the frequency-axis intercept is $f_0$, and the energy-axis intercept is $-\Phi$. Changing the metal changes $\Phi$, so the line shifts, but its gradient is still the Planck constant.

The graph also shows why intensity does not appear in Einstein’s equation. At a fixed frequency above the threshold, increasing intensity gives more photons per second, so the emission rate increases. The energy of each photon is unchanged, so $E_{\max}$ is unchanged too.

Stopping potential as a measurement

In a photoelectric cell, the collector can be made negative enough to repel even the fastest photoelectrons. The stopping potential is the potential difference that just reduces the photoelectric current to zero. It is written $V_s$, where $V_s$ is the stopping potential (V).

The maximum kinetic energy is then

This makes the electronvolt useful here: a photoelectron stopped by 1 V has lost 1 eV of kinetic energy.

Particle diffraction

Diffraction is the spreading and redistribution of a wave after it passes through an aperture or around an obstacle. It shows up most clearly when the aperture or obstacle is about the same size as the wavelength. So if particles produce a diffraction pattern, they are acting like waves in that experiment.

The classic evidence comes from electron diffraction. In the Davisson–Germer experiment, electrons were fired at a nickel crystal, and the scattered intensity had strong maxima at particular angles. In school versions, electrons pass through a thin graphite film and make bright and dark rings on a fluorescent screen. Those rings are the clue, not just a central bright patch: the electrons have been scattered into directions fixed by wave interference.

A scattering experiment and minimum intensity

One clean way to picture the measurement is to send a collimated beam of electrons through a narrow slit or a thin crystalline target, then move a detector across different scattering angles. If the electrons behaved only as tiny pellets, the spread would be mainly geometrical. Instead, the detected intensity rises to maxima and drops to minima.

For a single slit of width $a$, minima occur approximately when

Put a detector at one of these $\theta_n$ positions and it records minimum intensity, because the probability amplitudes from different parts of the slit cancel.

That idea links wave optics to quantum physics: what is waving is not a little material ripple in space, but a probability amplitude whose interference controls where particles are likely to be detected.

The dual description

Wave–particle duality is a quantum model where something like light or an electron shows wave-like behaviour in some experiments, and particle-like behaviour in others. The object is not swapping back and forth between a tiny ball and a water wave. The point is that neither classical picture works fully by itself.

Matter looks particle-like when it is detected: an electron reaches one point on a screen and transfers a localized amount of charge and energy. It looks wave-like while it propagates through a diffraction or interference arrangement: after many electrons have been detected, their final distribution forms a pattern with maxima and minima.

What counts as wave evidence?

Waves show superposition, diffraction and interference. So evidence for wave behaviour is more than a beam simply spreading out; it means there are systematic regions of high and low intensity, produced by reinforcement and cancellation.

The unsettling part is that this pattern still builds up when particles pass through the apparatus one at a time. Each electron is detected as a localized event. After many detections, though, the same interference pattern appears. The wave model predicts the probability distribution; particle detection gives the individual dots.

If you try to find out which slit an electron or photon passes through, you change the experiment. A measurement that can give path information disturbs the state enough to remove the two-source interference pattern. In quantum physics, observation is not always passive.

De Broglie’s proposal

The de Broglie wavelength is the wavelength linked to a moving particle because of its momentum. It is given by

For a non-relativistic particle,

Large everyday objects therefore have absurdly tiny de Broglie wavelengths. A tennis ball does not diffract around a doorway in any observable way; its wavelength is fantastically smaller than atomic and nuclear scales.

Electrons are the useful case

Electrons have small mass, so their de Broglie wavelengths can be comparable with atomic spacings. For an electron accelerated from rest through a potential difference $V$,

Increasing the accelerating potential difference gives the electron more momentum and decreases the de Broglie wavelength. In an electron diffraction tube, the ring pattern contracts. Decreasing the potential difference does the opposite: the wavelength increases and the rings spread out.

This is a neat practical check on the idea. Change a particle quantity — accelerating voltage, hence momentum — and the wave pattern shifts just as the wavelength equation predicts.

The scattering idea

Compton scattering is an inelastic scattering interaction where a photon hits a loosely bound or free electron, then comes out in a new direction with lower energy. It gives extra evidence for the particle nature of light, since the result only makes sense when the photon is treated as something with energy and momentum.

For a photon, the momentum magnitude is

In Compton’s experiment, X-ray photons were aimed at a low atomic number target such as carbon. A crystal analyser and an ionization detector picked up the scattered X-rays at different angles. The measured spectrum included a wavelength-shifted peak, and the size of the shift depended on the scattering angle.

Why it is stronger than the photoelectric evidence

The photoelectric effect gives strong evidence, but the incoming photon is absorbed and no longer exists. Its particle nature has to be inferred from the electron that is emitted. In Compton scattering, the incoming photon and the outgoing photon can both be studied: the photon is not just absorbed; it leaves with a measurable change in wavelength and direction.

So the conservation-law model is more direct. You can compare energy and momentum before the interaction with energy and momentum after it, just like in a collision problem.

Similar to, but not the same as, solid-ball collisions

The link with two solid balls colliding is that total energy and total momentum are conserved, and the scattered energy depends on angle. The difference is that a photon has zero rest mass, travels at speed $c$, and has to be described using quantum and relativistic ideas. A solid ball collision involves many atoms and contact forces; Compton scattering is a single photon–electron interaction.

Increased wavelength means lost photon energy

In Compton scattering, the scattered photon leaves with a longer wavelength than the incoming photon. Since $E_\gamma = \frac{hc}{\lambda}$, that longer wavelength corresponds to less photon energy. The energy difference goes into the recoil electron.

The electron picks up momentum as well, so it recoils at an angle to the original photon direction. That’s why the interaction has to be handled as a two-dimensional conservation problem, rather than a simple straight-line rebound.

The two peaks in real data

Real Compton spectra often contain two peaks. One stays at the original wavelength, caused by scattering from the atom as a whole or from tightly bound electrons; because the atom has a much larger mass, the wavelength change is too small to detect. The shifted peak is the Compton peak, produced when the photon scatters from electrons that can be treated as free during the interaction.

As the photon scattering angle gets larger, the wavelength increase gets larger too. In the straight-through direction the shift is zero; at larger angles, the shifted peak moves to longer wavelength.

The equation you use, not derive

For Compton scattering, use

You do not need to derive this equation. You do need to know what it says.

• If $\theta = 0^\circ$, then $\cos\theta = 1$ and $\Delta\lambda = 0$: a photon travelling straight on has no Compton wavelength shift.
• As $\theta$ increases, $1 - \cos\theta$ increases, so the wavelength shift increases.
• The shift depends on the scattering angle and fundamental constants, not on the incident wavelength itself.

The factor $h / (m_e c)$ is called the Compton wavelength of the electron. It fixes the scale of the effect. That is why Compton scattering shows up most clearly for X-rays and gamma rays: their wavelengths are small enough that the shift is a measurable fraction of the original wavelength.