The Photoelectric Effect
Specification: OCR A H556 | Section: 4.5.2 | Teaching time: ~3 hours
- Describe the photoelectric effect and explain the experimental observations
- Explain why the wave model fails to account for the observations
- Use Einstein's photoelectric equation: hf = φ + KE_max
- Define and use work function φ and threshold frequency f₀
- Explain the significance of the stopping potential and how it relates to KE_max
- Interpret and sketch graphs of KE_max vs frequency and photoelectric current vs intensity
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:
- Charge the electroscope negatively: The gold leaf rises due to repulsion between like charges
- Shine UV light on the zinc plate: The leaf collapses — electrons are being emitted from the zinc
- Shine visible light on the zinc plate: Nothing happens (even with very bright light)
- Charge the electroscope positively: UV light has no effect — positive charge holds electrons in the metal
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:
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:
Experimental Observations
When light shines on a metal surface in a vacuum tube, several key observations emerge that challenge wave theory:
- 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)
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
- 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.
- Energy quantisation: A photon's energy depends only on its frequency (E = hf), not on intensity.
- Work function: An electron needs a minimum energy to escape the metal surface. This is called the work function φ (phi) of the metal.
- Energy conservation: If hf > φ, the excess energy becomes the electron's kinetic energy.
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:
Below this frequency, no photoelectrons are emitted regardless of intensity.
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.
Photoelectric Effect Simulator
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:
Where e = 1.60 × 10⁻¹⁹ C is the elementary charge.
Measuring Stopping Potential
- Shine monochromatic light on the metal surface
- Vary the reverse voltage until the photocurrent drops to zero
- This voltage equals the stopping potential V_s
- KE_max can then be calculated: KE_max = eV_s
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)
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
Knowledge Check
- The minimum energy required to remove an electron from the surface of the metal ✓
- 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 ✓
- 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 ✓
- 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:
| Metal | Work function φ (eV) | Threshold frequency f₀ (×10¹⁴ Hz) | Threshold wavelength (nm) |
|---|---|---|---|
| Caesium (Cs) | 1.9 | 4.6 | 650 (red) |
| Potassium (K) | 2.3 | 5.6 | 540 (green) |
| Sodium (Na) | 2.4 | 5.8 | 520 (green) |
| Calcium (Ca) | 2.9 | 7.0 | 430 (violet) |
| Zinc (Zn) | 4.3 | 10.4 | 290 (UV) |
| Copper (Cu) | 4.7 | 11.4 | 260 (UV) |
| Platinum (Pt) | 6.4 | 15.5 | 190 (UV) |
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
(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]
(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 ✓
(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]
(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 ✓
(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]
(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 ✓
(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]
(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 ✓
(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]
(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.