The Photoelectric Effect — OCR A Level Physics Revision
Module 4 · Electrons, Waves and Photons

The Photoelectric Effect

Specification: OCR A H556  |  Section: 4.5.2  |  Teaching time: ~3 hours

By the end of this topic you should be able to…

What is the Photoelectric Effect?

The photoelectric effect is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

This phenomenon cannot be explained using classical wave theory — it was one of the key experiments that led to the development of quantum physics.

The Gold-Leaf Electroscope Demonstration

A classic demonstration uses a zinc plate mounted on a gold-leaf electroscope:

  1. Charge the electroscope negatively: The gold leaf rises due to repulsion between like charges
  2. Shine UV light on the zinc plate: The leaf collapses — electrons are being emitted from the zinc
  3. Shine visible light on the zinc plate: Nothing happens (even with very bright light)
  4. Charge the electroscope positively: UV light has no effect — positive charge holds electrons in the metal
💡 Key Observation

UV light (high frequency) ejects electrons from zinc, but visible light (lower frequency) does not — regardless of intensity. This was impossible to explain with wave theory!

The Photon Model of Light

To explain the photoelectric effect, Einstein proposed that light consists of discrete packets of energy called photons. Each photon has energy proportional to its frequency:

Photon Energy E = hf   or   E = hc/λ

Where:

  • h = 6.63 × 10⁻³⁴ J s — Planck's constant
  • f — frequency of the electromagnetic radiation (Hz)
  • c = 3.00 × 10⁸ m s⁻¹ — speed of light
  • λ — wavelength (m)

The electronvolt is often used as a convenient unit:

Unit Conversion 1 eV = 1.60 × 10⁻¹⁹ J

Experimental Observations

When light shines on a metal surface in a vacuum tube, several key observations emerge that challenge wave theory:

📋 Key Observations
  • Observation 1: Photoelectrons are emitted immediately when light strikes the surface — no time delay
  • Observation 2: No electrons are emitted if the light frequency is below a certain threshold frequency f₀, no matter how intense the light
  • Observation 3: The maximum kinetic energy of photoelectrons increases with light frequency but is independent of intensity
  • Observation 4: The number of photoelectrons per second is proportional to the light intensity (for f > f₀)

Why Wave Theory Fails

Classical wave theory predicts:

  • Energy depends on intensity (amplitude²), not frequency → contradicts observations 2 and 3
  • Electrons should accumulate energy over time → contradicts observation 1 (instantaneous emission)
  • Any frequency should work if intense enough → contradicts observation 2 (threshold frequency)
⚠️ The Wave Model Problem

In wave theory, a dim blue light should deliver less energy than a bright red light. But the photoelectric effect shows that blue light (higher frequency) can eject electrons while red light (lower frequency) cannot — even if the red light is much more intense. Wave theory cannot explain this.

Einstein's Explanation

Einstein explained the photoelectric effect using the photon model, for which he won the Nobel Prize in 1921.

Key Principles

  1. One-to-one interaction: Each photon interacts with a single surface electron. Either the photon has enough energy to eject the electron, or it doesn't.
  2. Energy quantisation: A photon's energy depends only on its frequency (E = hf), not on intensity.
  3. Work function: An electron needs a minimum energy to escape the metal surface. This is called the work function φ (phi) of the metal.
  4. Energy conservation: If hf > φ, the excess energy becomes the electron's kinetic energy.
Einstein's Photoelectric Equation hf = φ + KE_max

Where:

  • hf — energy of the incident photon
  • φ — work function of the metal (minimum energy to escape)
  • KE_max — maximum kinetic energy of emitted photoelectrons

Threshold Frequency

At the threshold frequency f₀, photoelectrons are emitted with zero kinetic energy:

Threshold Frequency hf₀ = φ   or   f₀ = φ/h

Below this frequency, no photoelectrons are emitted regardless of intensity.

Worked Example 1
The work function of caesium is 1.9 eV. Calculate the threshold frequency for caesium.
Convert φ to joules: φ = 1.9 × 1.60 × 10⁻¹⁹ = 3.04 × 10⁻¹⁹ J ✓
f₀ = φ/h = (3.04 × 10⁻¹⁹) / (6.63 × 10⁻³⁴) ✓
f₀ = 4.58 × 10¹⁴ Hz (this is in the red/infrared region) ✓

Why KE is Independent of Intensity

Each photon has energy hf. The intensity only affects how many photons arrive per second, not the energy of each photon. Since the interaction is one-to-one, a brighter light just means more photoelectrons (higher current), not faster ones.

Worked Example 2
UV light of wavelength 250 nm is incident on a metal with work function 4.5 eV. Calculate the maximum kinetic energy of the photoelectrons.
Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (250 × 10⁻⁹) ✓
E = 7.96 × 10⁻¹⁹ J = 4.97 eV ✓
KE_max = hf - φ = 4.97 - 4.5 = 0.47 eV (or 7.5 × 10⁻²⁰ J) ✓

Photoelectric Effect Simulator

Photoelectric Effect Visualisation
6.0×10¹⁴ Hz
5
Photon energy: eV
Work function φ: 4.5 eV
KE_max: eV
Status:

Adjust the frequency slider to see how photon energy affects electron emission. At low frequencies (red light), photons lack sufficient energy to overcome the work function. Above the threshold frequency, electrons are ejected with kinetic energy proportional to (hf - φ). Intensity only affects the number of photons/electrons, not their energy.

Stopping Potential

The stopping potential V_s is the minimum reverse voltage needed to stop the most energetic photoelectrons from reaching the collector plate.

How It Works

Photoelectrons are emitted from the metal surface with kinetic energy up to KE_max. If we apply a reverse voltage (collector negative relative to emitter), the photoelectrons must do work against the electric field. The stopping potential is reached when:

Stopping Potential eV_s = KE_max   or   V_s = KE_max / e

Where e = 1.60 × 10⁻¹⁹ C is the elementary charge.

Measuring Stopping Potential

  1. Shine monochromatic light on the metal surface
  2. Vary the reverse voltage until the photocurrent drops to zero
  3. This voltage equals the stopping potential V_s
  4. KE_max can then be calculated: KE_max = eV_s
Worked Example 3
When light of wavelength 400 nm illuminates a potassium surface, the stopping potential is 0.85 V. Calculate the work function of potassium.
Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J = 3.11 eV ✓
KE_max = eV_s = 0.85 eV ✓
φ = hf - KE_max = 3.11 - 0.85 = 2.26 eV

Graphs and Relationships

Understanding the graphical relationships is crucial for exam questions.

KE_max vs Frequency

Plotting maximum kinetic energy against frequency gives a straight line:

  • Equation: KE_max = hf - φ (compare to y = mx + c)
  • Gradient: h (Planck's constant)
  • y-intercept: -φ (negative of work function)
  • x-intercept: f₀ (threshold frequency)
📋 Exam Tip

The gradient of the KE_max vs f graph is always h — it's the same for all metals. Different metals have different work functions, so they give parallel lines shifted vertically. The line for caesium (low φ) sits above the line for zinc (high φ).

Photocurrent vs Intensity

For frequencies above the threshold:

  • Photocurrent is directly proportional to intensity
  • This is because intensity determines the number of photons per second
  • More photons = more photoelectrons = higher current

Photocurrent vs Stopping Potential

As the stopping potential is approached:

  • Current decreases gradually, then reaches zero at V_s
  • This is because electrons have a range of energies up to KE_max
  • At V_s, even the most energetic electrons cannot reach the collector
Worked Example 4 — Graph Analysis
A graph of KE_max against frequency for a metal gives a straight line with gradient 6.6 × 10⁻³⁴ J s and x-intercept at 8.0 × 10¹⁴ Hz. Determine the work function of the metal.
The gradient equals Planck's constant h = 6.6 × 10⁻³⁴ J s ✓
The x-intercept is the threshold frequency f₀ = 8.0 × 10¹⁴ Hz ✓
Work function: φ = hf₀ = 6.6 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.3 × 10⁻¹⁹ J ✓
φ = (5.3 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹) = 3.3 eV

Knowledge Check

1
State what is meant by the work function of a metal.
1 mark
  • The minimum energy required to remove an electron from the surface of the metal ✓
2
Explain why there is no time delay in the emission of photoelectrons, even for dim light.
2 marks
  • Energy is delivered in discrete packets (photons), not spread out as a wave ✓
  • Each electron interacts with a single photon; if hf > φ, emission is immediate ✓
3
Explain why increasing the intensity of light below the threshold frequency does not result in photoelectron emission.
2 marks
  • Each photon has energy hf which is less than φ (below threshold) ✓
  • Increasing intensity means more photons, but each still has insufficient energy; one-to-one interaction means electrons cannot accumulate energy from multiple photons ✓
4
Light of frequency 9.0 × 10¹⁴ Hz is incident on a metal with work function 2.3 eV. Calculate the maximum kinetic energy of the photoelectrons in eV.
3 marks
  • Photon energy: E = hf = 6.63 × 10⁻³⁴ × 9.0 × 10¹⁴ = 5.97 × 10⁻¹⁹ J = 3.73 eV ✓
  • KE_max = hf - φ = 3.73 - 2.3 ✓
  • KE_max = 1.43 eV

Work Functions of Common Metals

Different metals have different work functions, affecting their threshold frequencies and photoelectric behaviour:

MetalWork function φ (eV)Threshold frequency f₀ (×10¹⁴ Hz)Threshold wavelength (nm)
Caesium (Cs)1.94.6650 (red)
Potassium (K)2.35.6540 (green)
Sodium (Na)2.45.8520 (green)
Calcium (Ca)2.97.0430 (violet)
Zinc (Zn)4.310.4290 (UV)
Copper (Cu)4.711.4260 (UV)
Platinum (Pt)6.415.5190 (UV)
💡 Practical Significance

Caesium and potassium are used in photocells because their low work functions mean visible light can eject electrons. Zinc requires UV light — hence the gold-leaf electroscope demonstration uses UV on zinc, not visible light.

Exam-Style Questions

1
(a) Define the term threshold frequency. [1 mark]

(b) The threshold frequency of a metal is 6.0 × 10¹⁴ Hz. Calculate the work function of the metal in eV. [3 marks]

(c) Light of wavelength 450 nm is incident on this metal. Determine whether photoelectrons will be emitted. [3 marks]
7 marks

(a) [1]:

  • The minimum frequency of electromagnetic radiation required to eject photoelectrons from a metal surface ✓

(b) [3]:

  • φ = hf₀ = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ ✓
  • φ = 3.98 × 10⁻¹⁹ J ✓
  • φ = 3.98 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.49 eV

(c) [3]:

  • Frequency of light: f = c/λ = 3.00 × 10⁸ / (450 × 10⁻⁹) = 6.67 × 10¹⁴ Hz ✓
  • This is greater than the threshold frequency (6.0 × 10¹⁴ Hz) ✓
  • Yes, photoelectrons will be emitted
2
A student investigates the photoelectric effect using light of different frequencies. They measure the stopping potential V_s for each frequency and plot a graph of eV_s against frequency.

(a) Sketch the graph they would obtain, labelling the axes and any key points. [3 marks]

(b) The graph has gradient h. Explain why this gradient is the same for all metals. [2 marks]

(c) The graph crosses the frequency axis at 5.5 × 10¹⁴ Hz. Calculate the work function. [2 marks]
7 marks

(a) [3]:

  • Straight line with positive gradient ✓
  • x-intercept labelled as threshold frequency f₀ ✓
  • y-intercept labelled as -φ (or negative work function) ✓

(b) [2]:

  • The equation is KE_max = hf - φ, which has gradient h (Planck's constant) ✓
  • h is a universal constant, independent of the metal ✓

(c) [2]:

  • At x-intercept: hf₀ = φ, so φ = 6.63 × 10⁻³⁴ × 5.5 × 10¹⁴ ✓
  • φ = 3.65 × 10⁻¹⁹ J = 2.28 eV
3
(a) Explain what is meant by the statement "light exhibits wave-particle duality". [2 marks]

(b) Describe one phenomenon that provides evidence for the wave nature of light, and one that provides evidence for the particle nature of light. [2 marks]

(c) Light of wavelength 300 nm is incident on a metal surface. The stopping potential is 1.8 V. Calculate the work function of the metal in eV. [4 marks]
8 marks

(a) [2]:

  • Light can behave as a wave in some situations and as a particle (photon) in others ✓
  • Both models are needed to fully describe electromagnetic radiation ✓

(b) [2]:

  • Wave evidence: diffraction, interference, polarisation (any one) ✓
  • Particle evidence: photoelectric effect (must state) ✓

(c) [4]:

  • Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (300 × 10⁻⁹) = 6.63 × 10⁻¹⁹ J = 4.14 eV ✓
  • KE_max = eV_s = 1.8 eV ✓
  • hf = φ + KE_max ✓
  • φ = 4.14 - 1.8 = 2.34 eV
4
A photocell is illuminated with green light of wavelength 550 nm. The photocurrent is 2.0 μA.

(a) Calculate the energy of each photon in eV. [2 marks]

(b) Calculate the number of photons incident on the metal surface per second. [3 marks]

(c) The light is replaced with blue light of the same intensity but wavelength 450 nm. Explain the effect on: (i) the energy of each photon, (ii) the number of photons per second, (iii) the maximum kinetic energy of photoelectrons. [3 marks]
8 marks

(a) [2]:

  • E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (550 × 10⁻⁹) = 3.62 × 10⁻¹⁹ J ✓
  • E = 3.62 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.26 eV

(b) [3]:

  • Current I = 2.0 μA = 2.0 × 10⁻⁶ C/s ✓
  • Number of electrons per second = I/e = 2.0 × 10⁻⁶ / 1.60 × 10⁻¹⁹ = 1.25 × 10¹³ ✓
  • Assuming one photon per electron: photons per second = 1.25 × 10¹³ s⁻¹

(c) [3]:

  • (i) Blue light has shorter wavelength, so each photon has more energy
  • (ii) Same intensity means fewer photons per second (since each carries more energy) ✓
  • (iii) KE_max increases because hf increases while φ stays constant ✓
5
(a) State two observations from the photoelectric effect that cannot be explained by the wave model of light. [2 marks]

(b) Explain how Einstein's photon model accounts for the existence of a threshold frequency. [3 marks]

(c) A metal has work function 3.0 eV. Calculate the maximum wavelength of electromagnetic radiation that can cause photoelectric emission from this metal. [3 marks]
8 marks

(a) [2]: Any two:

  • Threshold frequency exists (below f₀, no emission regardless of intensity) ✓
  • KE_max depends on frequency, not intensity ✓
  • Instantaneous emission (no time delay) ✓

(b) [3]:

  • Each photon has energy E = hf ✓
  • An electron needs minimum energy φ to escape; it can only absorb one photon ✓
  • If hf < φ, the photon lacks sufficient energy; this defines threshold: hf₀ = φ ✓

(c) [3]:

  • At threshold: λ_max = hc/φ = hc/(3.0 × 1.60 × 10⁻¹⁹) ✓
  • λ_max = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.8 × 10⁻¹⁹) ✓
  • λ_max = 414 nm (violet light) ✓

Topic Summary

Photoelectric Effect

Emission of electrons from metal surfaces when EM radiation of sufficient frequency is incident. Evidence for particle nature of light.

Einstein's Equation

hf = φ + KE_max. Photon energy = work function + maximum kinetic energy of photoelectrons.

Threshold Frequency

f₀ = φ/h. Below this frequency, no emission occurs regardless of intensity. Each metal has a unique f₀.

Key Observations

Instantaneous emission; KE_max independent of intensity; current proportional to intensity; threshold frequency exists.

Equations to Know

E = hf = hc/λ
hf = φ + KE_max
f₀ = φ/h
eV_s = KE_max
1 eV = 1.60 × 10⁻¹⁹ J
h = 6.63 × 10⁻³⁴ J s