Photoelectric Effect and Wave-Particle Duality

The photoelectric effect and wave-particle duality represent a profound revolution in physics — the realisation that light and matter have both wave-like and particle-like properties. Einstein's explanation of the photoelectric effect earned him the Nobel Prize in 1921 and provided key evidence for quantum theory. This lesson covers the experimental observations, Einstein's photon model, the photoelectric equation, and de Broglie's hypothesis. This is assessed in AQA section 3.2.2.

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The Photoelectric Effect — Observations

When electromagnetic radiation (typically ultraviolet light) is shone onto a clean metal surface, electrons may be emitted from the surface. These emitted electrons are called photoelectrons. The key experimental observations are:

Observation 1: Threshold Frequency

• For a given metal, there is a minimum frequency of radiation (called the threshold frequency, f₀) below which no electrons are emitted, regardless of the intensity of the light.
• Above this threshold frequency, electrons are emitted immediately (within ~10⁻⁹ s).

Observation 2: Intensity and Rate

• Increasing the intensity of the light (above the threshold frequency) increases the number of photoelectrons emitted per second (the photocurrent), but does not increase the maximum kinetic energy of the electrons.

Observation 3: Frequency and Energy

• Increasing the frequency of the light increases the maximum kinetic energy of the emitted photoelectrons.
• The maximum kinetic energy is independent of intensity.

Why Classical Wave Theory Fails

Classical wave theory predicts that:

• Any frequency of light should eventually cause emission, given enough time — the energy would accumulate gradually. This contradicts Observation 1.
• Higher intensity should give electrons more energy. This contradicts Observation 3.
• There should be a time delay for low-intensity light. This contradicts the observation of instantaneous emission.

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Einstein's Photon Explanation (1905)

Einstein proposed that light consists of discrete packets of energy called photons. Each photon carries energy:

E = hf

where h = 6.63 × 10⁻³⁴ J s is the Planck constant and f is the frequency of the light.

Einstein's key insight: the photoelectric effect is a one-to-one interaction — each photon interacts with exactly one electron. Either the photon has enough energy to liberate the electron, or it does not. There is no accumulation of energy from multiple photons.

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The Photoelectric Equation

hf = φ + E_k(max)

or equivalently:

E_k(max) = hf − φ

where:

• hf is the energy of the incident photon
• φ (phi) is the work function of the metal — the minimum energy needed to release an electron from the surface
• E_k(max) = ½mv²_max is the maximum kinetic energy of the emitted photoelectron

Key Relationships

At the threshold frequency f₀, the photon has just enough energy to release an electron with zero kinetic energy:

hf₀ = φ

Therefore: φ = hf₀ and f₀ = φ/h

The Stopping Potential

The maximum kinetic energy of the photoelectrons can be measured using a stopping potential V_s:

E_k(max) = eV_s

where e = 1.60 × 10⁻¹⁹ C. Combining with the photoelectric equation:

eV_s = hf − φ

A graph of V_s against f gives a straight line with gradient h/e and x-intercept equal to the threshold frequency f₀.

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Worked Examples

Worked Example 1 — Threshold frequency and work function

The work function of sodium is 2.28 eV. Find the threshold frequency.

Convert to joules: φ = 2.28 × 1.60 × 10⁻¹⁹ = 3.648 × 10⁻¹⁹ J

f₀ = φ / h = 3.648 × 10⁻¹⁹ / 6.63 × 10⁻³⁴

f₀ = 5.50 × 10¹⁴ Hz

This is in the visible/UV range — sodium has a relatively low work function.

Worked Example 2 — Maximum kinetic energy

Ultraviolet light of frequency 8.0 × 10¹⁴ Hz is incident on a sodium surface (φ = 2.28 eV). Find the maximum kinetic energy of the emitted photoelectrons in eV and in joules.

Energy of photon: E = hf = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.304 × 10⁻¹⁹ J

Convert to eV: 5.304 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.315 eV

E_k(max) = hf − φ = 3.315 − 2.28 = 1.04 eV = 1.04 × 1.60 × 10⁻¹⁹ = 1.66 × 10⁻¹⁹ J

Worked Example 3 — Stopping potential

Find the stopping potential for the electrons in Worked Example 2.

eV_s = E_k(max)

V_s = E_k(max) / e = 1.66 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹

V_s = 1.04 V

(When working in eV, the stopping potential in volts is numerically equal to the maximum kinetic energy in eV.)

Worked Example 4 — Using the graph

A graph of stopping potential V_s against frequency f gives a straight line with gradient 4.14 × 10⁻¹⁵ V Hz⁻¹ and x-intercept 5.50 × 10¹⁴ Hz. Determine h and φ.