Describe the photoelectric effect, explain why it requires the photon model, and apply Einstein's photoelectric equation

What this dot point is asking

SEAB wants you to describe the photoelectric effect, explain why its features defeat the classical wave model and demand a photon model, and apply Einstein's photoelectric equation quantitatively. The photoelectric effect is the foundational evidence for the quantisation of light.

The answer

The effect and the key observations

When light of high enough frequency shines on a metal surface, electrons (photoelectrons) are emitted. Four observations are crucial:

1. Threshold frequency. Below a certain frequency $f_0$, no electrons are emitted, however intense the light.
2. Energy depends on frequency, not intensity. Above threshold, the maximum kinetic energy of the electrons increases with frequency but not with intensity.
3. Intensity controls the rate. Higher intensity emits more electrons per second, all with the same maximum kinetic energy.
4. Instantaneous emission. Electrons are emitted immediately, with no time delay, even at very low intensity.

Why classical wave theory fails

The wave model treats light energy as spread continuously and delivered at a rate set by intensity. It predicts that any frequency should eventually eject electrons given enough intensity or time, and that low-intensity light should show a delay while energy accumulates. Both predictions are wrong: there is a sharp threshold frequency and no time delay.

The photon model

Einstein proposed that light consists of discrete packets (photons), each carrying energy:

where $h = 6.63 \times 10^{-34}\ \text{J s}$ is the Planck constant. A single photon is absorbed by a single electron in one interaction, transferring all its energy at once. This explains the threshold (one photon must carry enough energy on its own) and the instantaneous emission (the transfer is a single event).

Einstein's photoelectric equation

The energy of an absorbed photon goes partly into freeing the electron from the metal (the work function $\phi$) and partly into the electron's kinetic energy:

so the maximum kinetic energy is $E_{k,\max} = hf - \phi$. The work function is the minimum energy needed to remove an electron, and the threshold frequency is $f_0 = \dfrac{\phi}{h}$. A graph of $E_{k,\max}$ against frequency is a straight line of gradient $h$ and intercept $-\phi$.

Examples in context

Example 1. Solar cells. A photovoltaic cell absorbs photons whose energy exceeds the band-gap energy of the semiconductor, freeing charge carriers and driving a current. Photons below the threshold energy pass through without effect, the same quantised-energy principle as the photoelectric effect applied to generating electricity from sunlight.

Example 2. Light meters in cameras. A photoelectric sensor produces a current proportional to the rate of photon arrival (the intensity), which the camera uses to set exposure. The current measures how many photons arrive per second, exploiting the observation that intensity controls the emission rate.

Try this

Q1. State Einstein's photoelectric equation and define each term. [2 marks]

• Cue. $hf = \phi + E_{k,\max}$: $hf$ photon energy, $\phi$ work function, $E_{k,\max}$ maximum kinetic energy of the emitted electron.

Q2. A metal has a threshold frequency of $5.5 \times 10^{14}\ \text{Hz}$. Find its work function in joules ($h = 6.63 \times 10^{-34}\ \text{J s}$). [2 marks]

• Cue. $\phi = h f_0 = 6.63 \times 10^{-34} \times 5.5 \times 10^{14} = 3.6 \times 10^{-19}\ \text{J}$.

Q3. Explain why increasing the intensity of light above the threshold frequency does not increase the maximum kinetic energy of the photoelectrons. [2 marks]

• Cue. Higher intensity means more photons per second (more electrons emitted), but each photon still carries energy $hf$, so the maximum kinetic energy $hf - \phi$ per electron is unchanged.