Photons and the photoelectric effectEdexcel A-Level Physics: Revision notes

By the 19th century, interference and diffraction experiments, such as Young's double slit, had shown that light behaves as a wave, and Maxwell's theory described it as an electromagnetic wave. Light of a given frequency was thought to carry energy that spreads continuously across the wavefront, so a brighter beam simply delivers more energy per second.

In 1900 Planck suggested that energy is exchanged in discrete amounts, and in 1905 Einstein proposed that electromagnetic radiation itself travels as photons: packets of energy. The energy of one photon is

$E = hf = \dfrac{hc}{\lambda}$

where $h = 6.63 \times 10^{-34}$ J s is the Planck constant. Both models are needed: the wave model explains propagation, interference and diffraction, while the photon model explains how radiation exchanges energy with matter.

The photoelectric effect is the emission of electrons, called photoelectrons, from a metal surface when electromagnetic radiation of high enough frequency shines on it. The key observations are:

• Below a threshold frequency $f_0$ no electrons are emitted, however intense the radiation.
• Above $f_0$ emission is instantaneous, even for very low intensity.
• The maximum kinetic energy of the electrons increases with frequency but does not depend on intensity.
• The number of photoelectrons emitted per second is proportional to the intensity of the radiation (above $f_0$).

The wave model predicts that any frequency should cause emission if the intensity is high enough, that low-intensity light should cause a delay while electrons store enough energy, and that the electron energy should increase with intensity. None of these is observed.

The photon model explains all of them. Each electron absorbs one photon and gains all of its energy in one interaction. If hf is less than the energy needed to escape, nothing happens, and electrons cannot add up energy from several photons. A more intense beam has more photons per second, so more electrons per second, but each has the same energy.

The work function $\phi$ is the minimum energy needed to remove an electron from the surface of a metal. The threshold frequency is the frequency whose photons have exactly this energy: $\phi = hf_0$.

By conservation of energy, a photon of energy $hf$ gives the electron energy $\phi$ to escape and the remainder becomes kinetic energy:

$hf = \phi + \tfrac{1}{2}mv_{\max}^2$

Electrons deeper in the metal lose extra energy, so $\tfrac{1}{2}mv_{\max}^2$ is the maximum kinetic energy, for electrons emitted from the surface.

Worked example. Light of frequency $6.0 \times 10^{14}$ Hz falls on a metal with $\phi = 3.0 \times 10^{-19}$ J. $hf = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.98 \times 10^{-19}$ J, so $\tfrac{1}{2}mv_{\max}^2 = 0.98 \times 10^{-19}$ J.

The electronvolt (eV) is a convenient unit of energy for atomic-scale processes. One electronvolt is the energy gained by an electron moving through a potential difference of 1 V:

$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$

To convert from eV to J, multiply by $1.60 \times 10^{-19}$. To convert from J to eV, divide by $1.60 \times 10^{-19}$.

Example: a work function of 2.0 eV is $2.0 \times 1.60 \times 10^{-19} = 3.2 \times 10^{-19}$ J. Use joules in $\tfrac{1}{2}mv^2$ calculations.

The photoelectric effect is evidence for the particle nature of electromagnetic radiation, because the wave model cannot explain the threshold frequency, the instant emission or the independence of kinetic energy from intensity, while the photon model explains all of them. Interference and diffraction still show that radiation has wave properties, so radiation shows both wave and particle behaviour.

In an exam answer, name the observation, say what the wave model predicts, and then say how photons (one photon to one electron, E = hf) account for what is seen.