The photoelectric effect provided the first experimental evidence that light behaves as discrete packets of energy called photons. This chapter explores how the photon model explains the ejection of electrons from metal surfaces, and how atomic emission and absorption spectra reveal the quantised energy levels inside atoms.
11.1 The Photoelectric Effect
Definition
Photoelectric effect
The photoelectric effect is the process whereby an electron is emitted by a substance when light shines on it. Einstein received the 1921 Nobel Prize for his contribution to understanding the photoelectric effect.
Definition
Work function (W)
The minimum energy needed to knock an electron out of a metal is called the work function (symbol W) of the metal. As it is energy, it is measured in joules (J). Energy is conserved, so if the photon has a higher energy than W then the excess energy goes into the kinetic energy Ek of the electron that was emitted from the substance.
Definition
Threshold frequency (f₀)
The minimum frequency of light that will eject electrons from a particular metal surface. Below this frequency, no electrons are emitted regardless of the intensity of the light. f₀ = W/h
Classical wave theory predicted that any frequency of light, given enough intensity, could eject electrons. But experiments showed that below a certain threshold frequency f₀, no electrons were ejected — no matter how intense the light. Above f₀, electrons were emitted instantly, even in very dim light. Einstein explained this in 1905: light comes in discrete packets (photons), each with energy E = hf. A single photon must have enough energy to overcome the work function W before an electron can escape.
Formula
Photon energy
E = photon energy (J), h = Planck's constant = 6.63×10⁻³⁴ J·s, f = frequency (Hz)
SI unit: J
Formula
Maximum kinetic energy of photoelectrons
Ek_max = max kinetic energy of ejected electron (J), hf = photon energy (J), W = work function (J)
SI unit: J
Formula
Threshold frequency
f₀ = threshold frequency (Hz), W = work function (J), h = 6.63×10⁻³⁴ J·s
SI unit: Hz
Exam Tip
Work functions are often given in electron-volts (eV). Convert to joules: 1 eV = 1.6×10⁻¹⁹ J. Always check units before substituting into Ek_max = hf − W.
Worked Example
Ultraviolet light of frequency 1.5×10¹⁵ Hz strikes a zinc surface whose work function is 6.9×10⁻¹⁹ J. Calculate (a) the energy of each photon, (b) the maximum kinetic energy of the ejected electrons, and (c) the threshold frequency for zinc.
Given
- f = 1.5×10¹⁵ Hz
- W = 6.9×10⁻¹⁹ J
- h = 6.63×10⁻³⁴ J·s
Find
E_photon, Ek_max, f₀
Solution
- 1E = hf = 6.63×10⁻³⁴ × 1.5×10¹⁵ = 9.945×10⁻¹⁹ J ≈ 9.95×10⁻¹⁹ J
- 2Ek_max = hf − W = 9.95×10⁻¹⁹ − 6.9×10⁻¹⁹ = 3.05×10⁻¹⁹ J
- 3f₀ = W/h = 6.9×10⁻¹⁹ / 6.63×10⁻³⁴ = 1.04×10¹⁵ Hz
Practice Question
Light of frequency 1.0×10¹⁵ Hz strikes a metal surface with work function 4.0×10⁻¹⁹ J. (a) Will electrons be ejected? Show your reasoning. (b) If they are ejected, calculate their maximum kinetic energy.
(5 marks)
Watch Out
Common mistake: thinking brighter light ejects faster electrons. Intensity determines how many photons arrive per second (so more electrons are ejected), but the kinetic energy of each electron depends only on the frequency of the photons, not on the intensity.
11.2 Emission and Absorption Spectra
Atoms can only exist in discrete energy levels — they are quantised. When an electron in an excited atom falls from a higher energy level to a lower one, it releases a photon whose energy exactly equals the difference between the two levels: E = hf. Because atoms of each element have unique energy levels, each element emits a unique set of frequencies — its line emission spectrum. Conversely, cool gases absorb exactly those same frequencies, producing dark lines in an absorption spectrum.
Emission vs Absorption Spectra
| Property | Emission Spectrum | Absorption Spectrum |
|---|---|---|
| Produced by | Hot, low-pressure gas | Cool gas in front of a continuous source |
| Appearance | Bright coloured lines on dark background | Dark lines on continuous colour spectrum |
| Cause | Electrons falling to lower energy levels, releasing photons | Electrons absorbing photons and jumping to higher levels |
| Frequencies | Specific to each element (unique fingerprint) | Same frequencies as emission spectrum of that element |
| Example | Hydrogen: red, blue-green, blue, violet lines | Sun's spectrum: dark Fraunhofer lines |
Note
The frequencies in an element's emission spectrum exactly match the dark lines in its absorption spectrum. This is because both involve the same electron transitions — one in emission, one in absorption.
Worked Example
An electron in a hydrogen atom falls from an energy level of −1.51 eV to the ground state at −13.6 eV. Calculate (a) the energy of the photon emitted and (b) the frequency of the emitted radiation. (1 eV = 1.6×10⁻¹⁹ J)
Given
- E₂ = −1.51 eV
- E₁ = −13.6 eV
- h = 6.63×10⁻³⁴ J·s
Find
E_photon and f
Solution
- 1Energy difference: ΔE = E₂ − E₁ = −1.51 − (−13.6) = 12.09 eV
- 2Convert to joules: E = 12.09 × 1.6×10⁻¹⁹ = 1.934×10⁻¹⁸ J
- 3Frequency: f = E/h = 1.934×10⁻¹⁸ / 6.63×10⁻³⁴ = 2.92×10¹⁵ Hz
Practice Question
Explain why each element has a unique line spectrum.
(4 marks)
Real World
Spectroscopy is used to identify elements in distant stars. By analysing which frequencies are absorbed in starlight, astronomers can determine the chemical composition of a star's atmosphere without ever visiting it.
11.3 Photon Model Consolidation and Applications
The photoelectric effect was critical in establishing quantum physics. Classical wave theory could not explain the threshold frequency or the instantaneous emission of electrons at low intensities. The photon model (Einstein, 1905) explained all observations: photons are discrete quanta of energy E = hf. This earned Einstein the 1921 Nobel Prize in Physics — not for relativity, but for the photoelectric effect.
Why the wave model fails and the photon model succeeds
- Wave model: any frequency with enough intensity should eject electrons. Photon model: only photons with f ≥ f₀ have enough energy per photon.
- Wave model: electrons should take time to accumulate energy. Photon model: one photon → one electron, instantaneous.
- Wave model: brighter light → faster electrons. Photon model: brighter light → more electrons, same Ek_max per electron.
Exam Tip
Exam question pattern: 'Explain why increasing the intensity of light below the threshold frequency does not cause emission.' Answer: Increasing intensity increases the number of photons but not the energy per photon. Each photon still has energy hf < W, so no single photon can eject an electron.
Worked Example
A metal has a threshold frequency of 8.0×10¹⁴ Hz. (a) Calculate the work function in joules and in eV. (b) Light of wavelength 250 nm strikes the surface. Calculate the maximum kinetic energy of the emitted electrons. (c = 3×10⁸ m·s⁻¹)
Given
- f₀ = 8.0×10¹⁴ Hz
- λ = 250 nm = 2.5×10⁻⁷ m
- h = 6.63×10⁻³⁴ J·s
- c = 3×10⁸ m·s⁻¹
Find
Work function W (J and eV), Ek_max
Solution
- 1W = hf₀ = 6.63×10⁻³⁴ × 8.0×10¹⁴ = 5.30×10⁻¹⁹ J
- 2W in eV = 5.30×10⁻¹⁹ / 1.6×10⁻¹⁹ = 3.31 eV
- 3Frequency of incident light: f = c/λ = 3×10⁸ / 2.5×10⁻⁷ = 1.2×10¹⁵ Hz
- 4Ek_max = hf − W = (6.63×10⁻³⁴ × 1.2×10¹⁵) − 5.30×10⁻¹⁹
- 5Ek_max = 7.956×10⁻¹⁹ − 5.30×10⁻¹⁹ = 2.66×10⁻¹⁹ J
Practice Question
The work function of sodium is 3.65×10⁻¹⁹ J. (a) Calculate the threshold frequency. (b) Light of frequency 7.5×10¹⁴ Hz shines on sodium. State with a reason whether electrons will be emitted.
(5 marks)
Practice Question
Describe ONE practical application of the photoelectric effect.
(2 marks)
Watch Out
Do not say 'the photon hits the electron'. Say 'the photon is absorbed by the electron' or 'the photon transfers its energy to the electron'. The photon is destroyed in the process.