7I

The photoelectric effect

The photoelectric effect, the observations that the wave model could not explain, Einstein's photon model and K = hf − φ, work function and threshold frequency, stopping voltage and the graph that measures Planck's constant, and photon momentum p = h/λ.

Shine the brightest red light you can find on a clean zinc plate and nothing happens. Shine a dim ultraviolet lamp on it and electrons fly off at once. Brightness doesn't matter; colour does. Classical wave physics could not explain this at all, and the explanation won Albert Einstein his Nobel Prize. It wasn't for relativity. It was for this.

What is the photoelectric effect?

The photoelectric effect is the emission of electrons from a surface, usually a metal, when light or other electromagnetic radiation shines on it. The emitted electrons are called photoelectrons.

  • Heinrich Hertz, 1887. While producing the radio waves that confirmed Maxwell's theory (7A), Hertz noticed sparks jumped his spark gap more easily when ultraviolet light shone on it.
  • Philipp Lenard, 1902. Using a vacuum tube like the one below, Lenard measured the energy of the emitted electrons and found a result nobody expected: their maximum energy depended on the light's frequency, not its intensity.
A photocell circuit. Light ejects electrons from the metal emitter; a variable voltage either helps them reach the collector or holds them back.

The observations, and why waves fail

ObservationWave model predicts
Below a threshold frequency $f_0$, no electrons are emitted, however intense the light.Any frequency should work if the light is bright enough or shines long enough.
The maximum kinetic energy of the electrons increases with frequency, and does not depend on intensity.Brighter light has bigger-amplitude waves, so electrons should come off with more energy.
Increasing intensity increases the number of electrons (the current) but not their maximum energy.Partly agrees, but more energy per electron was expected too.
Emission is instantaneous (within $10^{-9}$ s), even in very dim light.In dim light, an electron should need seconds or longer to soak up enough energy from the spread-out wave.

Einstein's photon model

In 1905, Einstein proposed that light itself is quantised. Light travels as a stream of particles of energy, later named photons, each with energy

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

A photon interacts with a single electron and gives it all its energy, or none. Electrons are held in the metal, and the minimum energy needed to remove one from the surface is the work function, $\phi$. By conservation of energy, the electron leaves with kinetic energy of at most

$$K_{max} = hf - \phi$$

(Electrons deeper in the metal lose some energy on the way out, so they leave with less than $K_{max}$.)

Every observation now makes sense:

  • Threshold. If $hf < \phi$, no single photon can free an electron, and adding more photons doesn't help because each acts alone. The threshold frequency is $f_0 = \phi/h$.
  • Energy depends on frequency. $K_{max} = hf - \phi$ rises in a straight line with $f$.
  • Intensity sets the number. Brighter light means more photons per second, so more electrons, but each photon still has the same $hf$.
  • Instant emission. The energy arrives in one packet; there is no waiting to accumulate it.

Work functions are a few electronvolts. Typical values: caesium 2.1 eV, sodium 2.3 eV, calcium 2.9 eV, zinc 4.3 eV, platinum 5.6 eV. Recall that $1 \text{ eV} = 1.602 \times 10^{-19}$ J.

Worked example: sodium in violet light

Light of wavelength 400 nm falls on sodium (work function 2.3 eV). Find (a) the photon energy, (b) the maximum kinetic energy of the photoelectrons, and (c) the threshold wavelength.

$E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} = 4.97 \times 10^{-19}$ J $= 3.10$ eV
photon
$K_{max} = 3.10 - 2.3 = 0.80$ eV $= 1.3 \times 10^{-19}$ J
Kmax
$\lambda_0 = \dfrac{hc}{\phi} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{2.3 \times 1.602 \times 10^{-19}} = 5.4 \times 10^{-7}$ m $= 540$ nm. Green light or shorter will work; red will not.
threshold

Stopping voltage

To measure $K_{max}$, make the collector negative relative to the emitter. The electric field now pushes electrons back. Only those with enough kinetic energy get across, so the current falls. At the stopping voltage $V_0$, even the fastest electrons are turned back and the current reaches zero. The work done by the field equals the maximum kinetic energy:

$$eV_0 = K_{max} = hf - \phi$$

Current against voltage. Brighter light raises the saturation current, but the stopping voltage depends only on frequency.

With a positive collector, all emitted electrons are collected and the current levels off at a saturation current, which is proportional to intensity.

Measuring Planck's constant

Dividing $eV_0 = hf - \phi$ by $e$ gives $V_0 = \dfrac{h}{e}f - \dfrac{\phi}{e}$. A graph of stopping voltage (or $K_{max}$) against frequency is a straight line:

  • gradient $= h/e$ (or $h$ for a $K_{max}$ graph in joules)
  • x-intercept $=$ threshold frequency $f_0$
  • y-intercept $= -\phi/e$ (or $-\phi$)

Every metal gives a line with the same gradient, shifted sideways by its different work function. Robert Millikan, who expected to disprove Einstein, spent ten years making these measurements with clean metal surfaces in vacuum. In 1916 he confirmed the equation exactly and measured $h$ to within 0.5%. Einstein received the 1921 Nobel Prize "especially for his discovery of the law of the photoelectric effect".

Maximum kinetic energy against frequency for two metals. The lines are parallel with gradient h.
Worked example: reading the graph

A student measures stopping voltages of 0.59 V at $7.5 \times 10^{14}$ Hz and 1.82 V at $10.5 \times 10^{14}$ Hz for one metal. Find Planck's constant, the threshold frequency and the work function.

$\dfrac{h}{e} = \dfrac{1.82 - 0.59}{(10.5 - 7.5) \times 10^{14}} = 4.10 \times 10^{-15}$ V s
gradient
$h = (4.10 \times 10^{-15})(1.602 \times 10^{-19}) = 6.57 \times 10^{-34}$ J s
h
$V_0 = 0$ when $f_0 = 7.5 \times 10^{14} - \dfrac{0.59}{4.10 \times 10^{-15}} = 6.06 \times 10^{14}$ Hz
threshold
$\phi = hf_0 = (6.57 \times 10^{-34})(6.06 \times 10^{14}) = 4.0 \times 10^{-19}$ J $= 2.5$ eV
work function

Photons carry momentum

Photons have no mass, but they do carry momentum:

$$p = \frac{h}{\lambda}$$

The momentum of one photon is tiny, but huge numbers of photons together exert a measurable radiation pressure. Comet tails point away from the Sun partly because sunlight pushes on the dust, and the Japanese spacecraft IKAROS (2010) sailed between planets on a 14 m wide solar sail. In 1923, Arthur Compton showed that X-ray photons bouncing off electrons lose momentum exactly as $p = h/\lambda$ predicts, which convinced most physicists that photons are real.

Worked example: photon momentum

Find the momentum of a photon of wavelength 500 nm.

$p = \dfrac{h}{\lambda} = \dfrac{6.626 \times 10^{-34}}{500 \times 10^{-9}} = 1.33 \times 10^{-27}$ kg m s⁻¹
formula

So what is light?

Interference, diffraction and polarisation (7E–7G) need waves. Black-body radiation and the photoelectric effect need quanta. Neither model alone explains everything. Light shows wave–particle duality: it travels and spreads like a wave, but exchanges energy and momentum in whole photons. Which behaviour you see depends on the experiment.

InteractivePhotocell lab
Try this
  • With sodium and 400 nm light, read the photon energy and $K_{max}$. Check the 3.10 eV and 0.80 eV with the worked example.
  • Increase the wavelength slowly until the current stops. Read the threshold wavelength in the notes. Then turn the intensity up to 100%. Does anything come out?
  • Back at 400 nm, change the intensity. What happens to the current and to $K_{max}$?
  • Make the voltage negative until the current just reaches zero. This is the stopping voltage. Press Record stopping voltage.
  • Record stopping voltages at four or more wavelengths. Read $h$ from the gradient. How close is it to $6.63 \times 10^{-34}$ J s?
  • Choose zinc. Why does visible light do nothing? Repeat the graph with UV and compare the gradient and intercept with sodium's.

Test yourself

  1. Which observation of the photoelectric effect could the wave model of light NOT explain?

  2. Photons of energy 4.5 eV fall on a metal with a work function of 2.9 eV. The stopping voltage is

  3. The intensity of light on a photocell is doubled while the frequency stays the same. Which is correct?

  4. On a graph of $K_{max}$ (in J) against frequency, the gradient is

Label the photocell circuit

evacuated tubemetal cathodecollector (anode)UV lightphotoelectronsAmicroammetervariable voltage

Light is already causing emission from sodium. Sort each change by its main effect.

The minimum energy needed to remove an electron from a metal is the . The lowest frequency that causes emission is the frequency. The voltage that just stops all photoelectrons is the voltage. In 1905 explained the effect using photons. Light that behaves both as a wave and as particles shows wave–particle .

7 marks

Ultraviolet light of wavelength 250 nm falls on a zinc plate (work function 4.3 eV). (a) Calculate the maximum kinetic energy of the photoelectrons in joules. (b) Calculate the stopping voltage. (c) Explain why red light of any intensity produces no photoelectrons from zinc, and how this supports Einstein's model rather than the wave model.