The Photoelectric Effect

In the photon model, light is emitted and absorbed in discrete packets of energy called $\text{photons}$. Each photon’s energy is proportional to the frequency: $E=hf$, where $h=6.63\times 10^{-34}\mathrm{J}\mathrm{s}$ is the Planck constant.

Combining with $c=f\lambda$ gives $E=hc/\lambda$ — shorter wavelengths carry more energy per photon, which is why ultraviolet can damage skin and visible light cannot.

Photon energies are tiny in joules, so the $\text{electronvolt}$ is used: $1\mathrm{eV}=1.60\times 10^{-19}\mathrm{J}$, the energy gained by an electron accelerated through 1 V.

For a given metal there is a $\text{threshold frequency}$ $f_0$. Below it, no electrons are emitted however intense the light. Above it, emission happens.

Emission is $\text{instantaneous}$ — there is no delay even for very dim light.

The $\text{maximum kinetic energy}$ of emitted electrons depends on the frequency, not the intensity. Increasing intensity increases only the $\text{number}$ of electrons emitted per second.

The wave model says energy arrives continuously and spread across the surface. So brighter light, of any frequency, should eventually give electrons enough energy to escape — there should be no threshold. And dim light should need time to build up enough energy — there should be a delay. Neither is seen.

The photon model explains all three. An electron absorbs $\text{one photon}$, all of its energy at once. If that single photon’s energy $hf$ is less than the energy needed to escape, the electron cannot leave — no matter how many photons arrive, since it cannot combine them. That is the threshold.

Absorption of one photon is immediate, so there is no delay. Higher intensity means more photons per second, so more electrons — but each photon still has energy $hf$, so the maximum kinetic energy is unchanged.

The $\text{work function}$ $\varphi$ is the minimum energy needed to free an electron from the metal surface. Energy conservation for one photon then gives $hf=\varphi +E_{k,\max }$: the photon’s energy pays the escape cost, and any excess becomes kinetic energy.

At threshold, $E_{k,\max }=0$, so $h{f}_0=\varphi$.

The maximum kinetic energy is measured using a $\text{stopping potential}$ $V_s$ — the reverse voltage just large enough to stop the fastest electrons: $e{V}_s=E_{k,\max }$. A graph of $E_{k,\max }$ against $f$ is a straight line with gradient $h$ and intercept $-\varphi$.