The Photoelectric Effect

The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency shines on it. It was one of the most important experiments in the history of physics because its results could not be explained by the classical wave theory of light. Einstein's explanation — that light consists of discrete packets of energy called photons — earned him the Nobel Prize and laid the foundation for quantum physics.

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The Observations

When ultraviolet light shines on a zinc plate connected to a negatively charged electroscope, the electroscope discharges — electrons are being emitted from the zinc surface. These emitted electrons are called photoelectrons.

Careful experiments revealed several observations that were deeply puzzling from the standpoint of classical wave theory:

Observation 1: Threshold frequency. For each metal, there is a minimum frequency of light (the threshold frequency, f₀) below which no electrons are emitted, no matter how intense the light. For zinc, visible light (even very bright visible light) cannot cause emission. Only ultraviolet light (above the threshold frequency) works.

Observation 2: Instantaneous emission. When light above the threshold frequency hits the metal, electrons are emitted virtually instantaneously (within about 10⁻⁹ s). There is no time delay, even at very low light intensities.

Observation 3: Maximum kinetic energy depends on frequency, not intensity. Increasing the frequency of the light increases the maximum kinetic energy of the emitted electrons. Increasing the intensity does not change the maximum kinetic energy — it only increases the number of electrons emitted per second (the photocurrent).

Observation 4: Intensity affects the rate. Increasing the light intensity increases the number of photoelectrons emitted per second, but does not change their maximum kinetic energy.

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Why the Wave Model Fails

Classical wave theory predicts:

• Any frequency of light should eventually cause emission if you wait long enough or use a bright enough source (because the wave energy would accumulate on the surface). This contradicts observation 1.
• There should be a significant time delay at low intensities while the electron accumulates enough energy. This contradicts observation 2.
• More intense light should give the electrons more energy (higher KE_max). This contradicts observation 3.

The classical wave model fails comprehensively. A completely new model was needed.

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Einstein's Photon Model

In 1905, Einstein proposed that light consists of discrete packets of energy called photons. Each photon carries energy:

$E = hf$E=hf

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

A single photon interacts with a single electron in the metal surface. If the photon has enough energy, it can liberate the electron. If it does not, no emission occurs — regardless of how many low-energy photons arrive.

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The Work Function

The work function (φ) is the minimum energy required to release an electron from the surface of a metal. It depends on the metal and is typically measured in electronvolts (eV) or joules.

Work Functions of Common Metals

Metal	Work function φ / eV	Work function φ / × 10⁻¹⁹ J	Threshold frequency f₀ / × 10¹⁴ Hz	Threshold wavelength λ₀ / nm
Caesium (Cs)	2.1	3.36	5.08	590 (visible — orange)
Potassium (K)	2.3	3.68	5.56	540 (visible — green)
Sodium (Na)	2.3	3.68	5.56	540 (visible — green)
Lithium (Li)	2.9	4.64	7.00	428 (visible — violet)
Zinc (Zn)	4.3	6.88	10.4	289 (UV)
Iron (Fe)	4.5	7.20	10.9	276 (UV)
Copper (Cu)	4.7	7.52	11.3	264 (UV)
Platinum (Pt)	5.6	8.96	13.5	222 (UV)

Notice that alkali metals (Cs, K, Na) have low work functions and can exhibit the photoelectric effect with visible light. Transition metals require UV light.

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The Photoelectric Equation

Einstein's photoelectric equation describes the energy balance:

$E_{\text{photon}} = \phi + KE_{\text{max}}$Ephoton​=ϕ+KEmax​

$hf = \phi + \frac{1}{2}mv_{\text{max}}^2$hf=ϕ+21​mvmax2​

• The photon energy hf is "spent" first on overcoming the work function φ
• Any remaining energy becomes the kinetic energy of the emitted electron
• KE_max is the maximum kinetic energy of the photoelectron (for electrons at the very surface)

Rearranging:

$KE_{\text{max}} = hf - \phi$KEmax​=hf−ϕ

At the threshold frequency (f₀), the maximum kinetic energy is zero — the photon has just enough energy to free the electron but none left over:

$hf_0 = \phi \quad \Rightarrow \quad f_0 = \frac{\phi}{h}$hf0​=ϕ⇒f0​=hϕ​

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Worked Calculations

Worked example 1: UV light of frequency 1.2 × 10¹⁵ Hz shines on a zinc surface (φ = 4.3 eV). Calculate the maximum kinetic energy of the emitted photoelectrons.

First convert φ to joules: φ = 4.3 × 1.60 × 10⁻¹⁹ = 6.88 × 10⁻¹⁹ J

E_photon = hf = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵ = 7.956 × 10⁻¹⁹ J

KE_max = hf − φ = 7.956 × 10⁻¹⁹ − 6.88 × 10⁻¹⁹ = 1.08 × 10⁻¹⁹ J

Converting to eV: KE_max = 1.08 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 0.67 eV

Worked example 2: Calculate the threshold frequency and threshold wavelength for potassium (φ = 2.3 eV).

f₀ = φ/h = (2.3 × 1.60 × 10⁻¹⁹) / (6.63 × 10⁻³⁴) = 3.68 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.55 × 10¹⁴ Hz

λ₀ = c/f₀ = 3.00 × 10⁸ / 5.55 × 10¹⁴ = 5.41 × 10⁻⁷ m = 541 nm (green light)

This means green, blue, violet, and UV light can all cause photoelectron emission from potassium, but red and orange light cannot.

Worked example 3: Light of wavelength 250 nm shines on a copper surface (φ = 4.7 eV). Calculate the maximum speed of the emitted photoelectrons. (Mass of electron = 9.11 × 10⁻³¹ kg)

E_photon = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (250 × 10⁻⁹) = 1.989 × 10⁻²⁵ / 2.50 × 10⁻⁷ = 7.96 × 10⁻¹⁹ J

φ = 4.7 × 1.60 × 10⁻¹⁹ = 7.52 × 10⁻¹⁹ J

KE_max = 7.96 × 10⁻¹⁹ − 7.52 × 10⁻¹⁹ = 4.4 × 10⁻²⁰ J

½mv² = 4.4 × 10⁻²⁰

v = √(2 × 4.4 × 10⁻²⁰ / 9.11 × 10⁻³¹) = √(9.66 × 10¹⁰) = 3.1 × 10⁵ m s⁻¹

This is about 0.1% of the speed of light, so non-relativistic treatment is valid.

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The Stopping Voltage

The maximum kinetic energy of photoelectrons can be measured using a stopping voltage (Vs). A reverse voltage is applied to slow down the photoelectrons. When the voltage is just large enough to stop even the fastest electrons:

$eV_s = KE_{\text{max}} = hf - \phi$eVs​=KEmax​=hf−ϕ

where e = 1.60 × 10⁻¹⁹ C is the electron charge.

Worked example 4: The stopping voltage for a metal illuminated with 400 nm light is 0.80 V. Calculate the work function of the metal.

E_photon = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J = 3.11 eV

KE_max = eVs = 0.80 eV

φ = hf − KE_max = 3.11 − 0.80 = 2.31 eV

This is consistent with sodium or potassium.

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The Photoelectric Effect Graph

A graph of KE_max against frequency f is a straight line:

$KE_{\text{max}} = hf - \phi$KEmax​=hf−ϕ

Comparing with y = mx + c:

• Gradient = h (the Planck constant) — the same for all metals
• y-intercept = −φ (different for each metal)
• x-intercept = f₀ (the threshold frequency — where KE_max = 0)

Different metals produce parallel lines (same gradient) shifted vertically according to their work function. A metal with a higher work function has a higher threshold frequency and a more negative y-intercept.

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Explaining Each Observation with the Photon Model

Observation	Wave model prediction	Photon model explanation
Threshold frequency	Any frequency should work given enough time	A photon must have energy ≥ φ; if hf < φ, the electron cannot be freed
Instantaneous emission	Time delay expected	One photon gives all its energy to one electron — no accumulation needed
KE_max depends on f, not intensity	Intensity should increase KE_max	KE_max = hf − φ; frequency determines photon energy
Intensity affects rate only	—	More photons per second = more electrons released per second

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Electronvolt Conversions

The electronvolt is a convenient energy unit in atomic and quantum physics:

$1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$1 eV=1.60×10−19 J

To convert eV to J: multiply by 1.60 × 10⁻¹⁹

To convert J to eV: divide by 1.60 × 10⁻¹⁹

Common exam mistake: Forgetting to convert eV to joules (or vice versa) when substituting into equations using SI units. The Planck constant is given in J s, so if you use it directly, energies must be in joules. Either convert φ to joules at the start or work entirely in eV (where h = 4.14 × 10⁻¹⁵ eV s).

Another common error: Claiming that "more intense light gives the electrons more kinetic energy." Intensity determines the number of photons per second (and hence the number of electrons emitted per second), not the energy per photon. Only frequency determines the energy per photon.

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A-Level Deep Dive: Photoelectric Effect

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in the sub-strands on the photon model, requires students to understand the particulate nature of electromagnetic radiation; describe the photoelectric effect as evidence for the photon model; use Einstein's photoelectric equation $hf = \phi + KE_{max}$hf=ϕ+KEmax​; define and use the work function $\phi$ϕ and threshold frequency $f_0$f0​; and explain the failure of the wave model to predict the observed phenomena (refer to the official Pearson Edexcel specification document for exact wording). Although the photoelectric effect sits within Topic 5, its conceptual reach extends across Topic 5's sub-strand on atomic energy levels and line spectra (the same $E = hf$E=hf links photon and atomic transition), into Topic 12 (turning points in physics) when discussing the historical wave-particle debate, and into Topic 12's de Broglie sub-strand where the symmetry $\lambda = h/p$λ=h/p extends the photon picture to massive particles. The Edexcel data-and-formulae booklet lists $E = hf$E=hf, $hf = \phi + \tfrac{1}{2}mv_{max}^2$hf=ϕ+21​mvmax2​ and $h$h to 3 sf, but does not list $\phi$ϕ values for individual metals — these are always given in the question stem.

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Worked example with full mark scheme

Question (8 marks):

A clean magnesium surface has a work function of $\phi = 3.7\ \text{eV}$ϕ=3.7 eV. Light of wavelength $\lambda = 250\ \text{nm}$λ=250 nm is incident on the surface.

(a) Calculate the threshold frequency $f_0$f0​ for magnesium. (2)

(b) Determine, with a calculation, whether photoemission occurs at $\lambda = 250\ \text{nm}$λ=250 nm. (3)

(c) Calculate the maximum kinetic energy, in joules, of the emitted photoelectrons. (3)

Use $h = 6.63 \times 10^{-34}\ \text{J s}$h=6.63×10−34 J s, $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$c=3.00×108 m s−1, $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$1 eV=1.60×10−19 J.

Solution with mark scheme:

(a) Step 1 — convert the work function to joules.

$\phi = 3.7 \times 1.60 \times 10^{-19} = 5.92 \times 10^{-19}\ \text{J}$ϕ=3.7×1.60×10−19=5.92×10−19 J

Step 2 — apply $hf_0 = \phi$hf0​=ϕ.

$f_0 = \frac{\phi}{h} = \frac{5.92 \times 10^{-19}}{6.63 \times 10^{-34}} = 8.93 \times 10^{14}\ \text{Hz}$f0​=hϕ​=6.63×10−345.92×10−19​=8.93×1014 Hz

M1 — correct rearrangement to $f_0 = \phi/h$f0​=ϕ/h with $\phi$ϕ converted to joules. The single most common slip here is leaving $\phi$ϕ in eV while $h$h is in J s — the units do not cancel and the answer comes out a factor of $1.6 \times 10^{-19}$1.6×10−19 wrong.

A1 — $f_0 = 8.9 \times 10^{14}\ \text{Hz}$f0​=8.9×1014 Hz (2 sf acceptable).

(b) Step 1 — find the frequency of the incident light.

$f = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}}{250 \times 10^{-9}} = 1.20 \times 10^{15}\ \text{Hz}$f=λc​=250×10−93.00×108​=1.20×1015 Hz

M1 — correct conversion of wavelength to frequency.

Step 2 — compare with $f_0$f0​.

Since $f = 1.20 \times 10^{15}\ \text{Hz} > f_0 = 8.93 \times 10^{14}\ \text{Hz}$f=1.20×1015 Hz>f0​=8.93×1014 Hz, the photon energy exceeds the work function and photoemission does occur.

M1 — explicit comparison of the calculated $f$f with $f_0$f0​ (or equivalently $E_{photon}$Ephoton​ with $\phi$ϕ).

A1 — clear conclusion ("yes, photoemission occurs because $f > f_0$f>f0​").

(c) Step 1 — calculate photon energy.

$E_{\text{photon}} = hf = 6.63 \times 10^{-34} \times 1.20 \times 10^{15} = 7.96 \times 10^{-19}\ \text{J}$Ephoton​=hf=6.63×10−34×1.20×1015=7.96×10−19 J

M1 — substitution into $E = hf$E=hf (or equivalently $E = hc/\lambda$E=hc/λ).

Step 2 — apply Einstein's equation.

$KE_{\text{max}} = hf - \phi = 7.96 \times 10^{-19} - 5.92 \times 10^{-19} = 2.0 \times 10^{-19}\ \text{J}$KEmax​=hf−ϕ=7.96×10−19−5.92×10−19=2.0×10−19 J

M1 — correct substitution into $KE_{max} = hf - \phi$KEmax​=hf−ϕ with both terms in joules.

A1 — $KE_{max} = 2.0 \times 10^{-19}\ \text{J}$KEmax​=2.0×10−19 J (or $2.04 \times 10^{-19}\ \text{J}$2.04×10−19 J to 3 sf, equivalently $1.3\ \text{eV}$1.3 eV).

Total: 8 marks (M5 A3).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A photocell uses a sodium cathode of work function $\phi = 2.3\ \text{eV}$ϕ=2.3 eV illuminated by monochromatic blue light of wavelength $\lambda = 450\ \text{nm}$λ=450 nm.

(a) Calculate the maximum kinetic energy, in eV, of the photoelectrons. (3)

(b) The intensity of the blue light is doubled at the same wavelength. State and explain how this affects (i) the maximum kinetic energy of the photoelectrons and (ii) the number of photoelectrons emitted per second. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1) — calculating photon energy $E = hc/\lambda = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(450 \times 10^{-9}) = 4.42 \times 10^{-19}\ \text{J}$E=hc/λ=(6.63×10−34)(3.00×108)/(450×10−9)=4.42×10−19 J.
• M1 (AO1) — converting to eV: $E = 4.42 \times 10^{-19}/1.60 \times 10^{-19} = 2.76\ \text{eV}$E=4.42×10−19/1.60×10−19=2.76 eV.
• A1 (AO2) — $KE_{max} = E - \phi = 2.76 - 2.30 = 0.46\ \text{eV}$KEmax​=E−ϕ=2.76−2.30=0.46 eV (or $0.5\ \text{eV}$0.5 eV).

(b)

• B1 (AO2) — $KE_{max}$KEmax​ is unchanged because it depends only on the photon frequency (and $\phi$ϕ), not the intensity.
• M1 (AO2) — doubling the intensity at fixed frequency means twice as many photons per second strike the cathode.
• A1 (AO3) — twice as many photoelectrons are emitted per second (so the photocurrent doubles), with explicit reasoning that one photon liberates at most one electron.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This question is typical of the Edexcel pattern of pairing a procedural calculation (a) with a qualitative explanation (b) that probes the conceptual core of the photon model.

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Synoptic links

Connects to:

1. Topic 5, sub-strand on the wave-particle duality of light: the photoelectric effect is the canonical particle-side evidence; diffraction and interference (the earlier sub-strands of Topic 5) are the canonical wave-side evidence. A coherent answer on duality must cite both — the photon model does not abolish the wave model, it complements it. The same electromagnetic radiation behaves as a wave when propagating and as a particle when interacting with matter.

2. Topic 5, sub-strand on photon energy $E = hf$E=hf: the photoelectric equation is just energy conservation applied to a photon-electron interaction. The same $E = hf$E=hf also governs the energy of a photon emitted in an atomic transition and the energy of an X-ray produced by deceleration. Once $E = hf$E=hf is internalised, the photoelectric equation reduces to "incoming energy = energy to escape + leftover kinetic energy".

3. Topic 5, sub-strand on atomic energy levels: the work function $\phi$ϕ is the energy required to lift an electron from the highest occupied level in the metal (the Fermi level) to the vacuum just outside the surface. It is the metallic analogue of the ionisation energy of an isolated atom — both describe escape of an electron from a binding potential.

4. Topic 12 (turning points — wave-particle duality and de Broglie): the de Broglie hypothesis $\lambda = h/p$λ=h/p extends the photon's wave-particle duality to massive particles. The photoelectric effect motivates the symmetry: if waves can be particles, particles can be waves. Electron diffraction (Davisson-Germer) is the experimental confirmation.