Photoelectric Effect: hf = φ + KE_max Formula & Calculations

The photoelectric equation is hf = φ + KE_max, where hf is the photon energy (h = 6.626 × 10⁻³⁴ J·s, f is photon frequency), φ (phi) is the work function of the metal (minimum energy to eject an electron), and KE_max = ½mv²_max is the maximum kinetic energy of ejected electrons. If the photon energy hf is less than the work function φ, no electrons are emitted regardless of intensity. This was Einstein's 1905 explanation of the photoelectric effect, for which he received the 1921 Nobel Prize.

The photoelectric effect was deeply puzzling to classical physics. Classical wave theory predicted that any frequency of light should eventually eject electrons if the intensity were high enough — just give the electrons time to absorb enough energy. But experiment showed that below a threshold frequency f₀ = φ/h, no electrons are ever emitted, no matter how intense the light. Einstein's insight was that light comes in discrete quanta (photons) each with energy hf — either a single photon has enough energy to eject an electron, or it doesn't. Intensity just determines how many photons arrive, not the energy of each.

The Photoelectric Equation

Rearranged:

Where:

• h = Planck's constant = 6.626 × 10⁻³⁴ J·s
• f = frequency of incident light (Hz)
• φ = work function of the metal (J or eV) — the minimum energy to free an electron from the surface
• KE_max = maximum kinetic energy of ejected electrons (J or eV)

The KE_max is a maximum because not all electrons are at the surface — some must spend additional energy escaping from below the surface, leaving them with less kinetic energy. The fastest electrons come from the surface layer.

Threshold Frequency

When the photon just barely has enough energy to eject an electron, KE_max = 0:

Below f₀: no photoelectric emission, regardless of intensity. Above f₀: electrons are emitted; their maximum KE increases linearly with frequency.

Work Functions of Common Metals

Metal	Work function φ (eV)	Threshold wavelength
Caesium	2.1	590 nm (visible)
Sodium	2.3	539 nm (visible)
Potassium	2.3	540 nm (visible)
Aluminium	4.1	302 nm (UV)
Copper	4.5	276 nm (UV)
Platinum	5.7	218 nm (UV)

Caesium requires only visible light to emit electrons — which is why it's used in photomultiplier tubes and light sensors. Most metals require UV light.

Stopping Potential

The stopping potential V₀ is the minimum reverse voltage needed to stop all emitted electrons (even the fastest ones). Since KE_max = eV₀:

V₀ = (hf − φ)/e. Measuring V₀ experimentally gives KE_max directly, and a graph of V₀ against frequency gives a straight line with gradient h/e and y-intercept −φ/e — allowing both Planck's constant and the work function to be measured.

4 Worked Examples

Example 1 — Maximum kinetic energy

Problem: UV light of frequency 1.5 × 10¹⁵ Hz hits sodium (φ = 2.3 eV = 3.68 × 10⁻¹⁹ J). Find the maximum KE of emitted electrons.

Solution:
Photon energy = hf = 6.626 × 10⁻³⁴ × 1.5 × 10¹⁵ = 9.94 × 10⁻¹⁹ J
KE_max = hf − φ = 9.94 × 10⁻¹⁹ − 3.68 × 10⁻¹⁹ = 6.26 × 10⁻¹⁹ J = 3.91 eV

Example 2 — Threshold frequency

Problem: Copper has a work function of 4.5 eV. Find the threshold frequency and wavelength for photoemission.

Solution:
φ = 4.5 × 1.6 × 10⁻¹⁹ = 7.2 × 10⁻¹⁹ J
f₀ = φ/h = 7.2 × 10⁻¹⁹ / 6.626 × 10⁻³⁴ = 1.087 × 10¹⁵ Hz
λ₀ = c/f₀ = 3 × 10⁸ / 1.087 × 10¹⁵ = 2.76 × 10⁻⁷ m = 276 nm (UV)

Example 3 — Stopping potential

Problem: Light of wavelength 200 nm hits a metal of work function 4.0 eV. Find the stopping potential.

Solution:
f = c/λ = 3 × 10⁸ / 200 × 10⁻⁹ = 1.5 × 10¹⁵ Hz
hf = 6.626 × 10⁻³⁴ × 1.5 × 10¹⁵ = 9.94 × 10⁻¹⁹ J = 6.21 eV
KE_max = 6.21 − 4.0 = 2.21 eV
V₀ = KE_max/e = 2.21 eV/e = 2.21 V

Example 4 — Identifying the metal from data

Problem: Light of frequency 8.0 × 10¹⁴ Hz ejects electrons with maximum KE of 1.2 × 10⁻¹⁹ J. Identify the work function and suggest the metal.

Solution:
φ = hf − KE_max = 6.626 × 10⁻³⁴ × 8.0 × 10¹⁴ − 1.2 × 10⁻¹⁹
= 5.301 × 10⁻¹⁹ − 1.2 × 10⁻¹⁹ = 4.101 × 10⁻¹⁹ J = 2.56 eV
This matches potassium (φ = 2.3 eV) reasonably well — likely potassium or a similar alkali metal.

Einstein's Explanation — Why Classical Physics Failed

The classical wave theory of light predicted: (1) below some intensity threshold, no electrons would be emitted (not observed); (2) with enough time, any frequency could supply enough energy to eject an electron (not observed — below threshold frequency, no emission at all, even after hours); (3) higher intensity should increase the kinetic energy of emitted electrons (not observed — intensity only increases the number of electrons, not their KE).

Einstein's 1905 explanation (for which he won the 1921 Nobel Prize — notably not for relativity): light is quantised into photons, each with energy E = hf. One photon interacts with one electron in an all-or-nothing transfer. If hf < φ (work function), the photon has insufficient energy to free any electron, regardless of intensity. If hf ≥ φ, one photon frees one electron with KE_max = hf − φ. Intensity determines how many photons arrive per second (how many electrons are emitted per second — the photocurrent), but not the energy of each photon or each emitted electron.

Stopping Potential — Direct Measurement of KE_max

The stopping potential V₀ is measured by connecting a battery in reverse — applying a retarding voltage that decelerates emitted electrons. At exactly V₀, even the fastest electrons (with KE_max) are stopped:

A graph of V₀ vs f is a straight line with gradient h/e and y-intercept −φ/e. Measuring this graph gives Planck's constant h and the work function φ simultaneously — and the x-intercept gives the threshold frequency f₀ = φ/h directly. This experiment, done by Millikan in 1916, gave h to within 0.5% of the modern value — a beautiful quantitative confirmation of Einstein's photon hypothesis.

Worked Example 5 — Determining Planck's constant from a V₀-f graph

Problem: Stopping potentials for zinc (φ = 4.31 eV) are measured using three UV frequencies: at f₁ = 1.2 × 10¹⁵ Hz, V₀ = 0.66 V; at f₂ = 1.5 × 10¹⁵ Hz, V₀ = 1.91 V; at f₃ = 1.8 × 10¹⁵ Hz, V₀ = 3.16 V. Find Planck's constant from the gradient.

Solution:
Gradient = ΔV₀/Δf = (3.16 − 0.66)/((1.8 − 1.2) × 10¹⁵) = 2.50/(0.6 × 10¹⁵) = 4.17 × 10⁻¹⁵ V·s
h = e × gradient = 1.6 × 10⁻¹⁹ × 4.17 × 10⁻¹⁵ = 6.67 × 10⁻³⁴ J·s
(Modern value: 6.626 × 10⁻³⁴ J·s — within 0.6% from three data points)

Applications of the Photoelectric Effect

Solar cells (photovoltaic): photons absorbed in a semiconductor create electron-hole pairs. These carriers are swept across a p-n junction by the built-in electric field, generating a current. Each photon with energy above the band gap (~1.1 eV for silicon) can create one electron-hole pair; excess energy above the band gap is wasted as heat. This limits single-junction silicon cell efficiency to ~33% (Shockley-Queisser limit). Multi-junction cells (using different semiconductor layers for different photon energies) reach ~47% efficiency in laboratory settings.

Photomultiplier tubes: a photon hits a photocathode (caesium-based, low work function ~2 eV), ejecting a photoelectron. The electron is accelerated to a dynode, where it ejects ~5 more electrons. These cascade through 10–14 dynodes, amplifying the signal by 10⁶–10⁸. Photomultipliers can detect single photons with nanosecond timing — used in particle physics detectors (scintillation counters), PET scanners, and astronomical photometers.

Photoelectric smoke detectors: a light source illuminates a chamber; normally no light reaches the detector. Smoke particles scatter light into the detector, triggering the alarm. (Ionisation smoke detectors use a different principle — alpha radiation ionises air; smoke interrupts the ion current.)

Wave-Particle Duality of Light

The photoelectric effect demonstrates the particle nature of light (photons). Yet light also shows wave behaviour — interference, diffraction, polarisation. How can light be both? This is the wave-particle duality, resolved by quantum mechanics: light (and all matter) has both wave-like and particle-like properties. Which aspect manifests depends on the experimental arrangement. Diffraction and interference experiments probe the wave nature; the photoelectric effect and Compton scattering probe the particle nature. Feynman's path-integral formulation of quantum mechanics treats all paths simultaneously and reproduces both kinds of behaviour from one mathematical framework. The duality is not a contradiction — it is the correct description of reality at the quantum scale.

Worked Example 6 — Photocurrent and intensity

Problem: A metal surface illuminated by 400 nm light produces a photocurrent of 2.5 μA. The light intensity is then doubled while keeping the wavelength the same. The voltage is kept at zero. Find: (a) the new photocurrent, (b) the new maximum kinetic energy of emitted electrons.

Solution:
(a) Doubling intensity doubles the photon flux → twice as many photoelectrons per second → photocurrent doubles to 5.0 μA
(b) KE_max = hf − φ = hc/λ − φ. This depends only on photon energy (fixed by λ = 400 nm) and work function (fixed by the material). Intensity does not affect photon energy. KE_max is unchanged.

Historical Significance

The photoelectric effect played a pivotal role in the development of quantum mechanics. By 1905, classical physics was in crisis: blackbody radiation (Planck's 1900 solution required energy quantisation), the Michelson-Morley experiment (no ether), and the photoelectric effect (classical wave theory fails) all demanded new physics. Einstein's 1905 paper on the photoelectric effect was the first to take Planck's quantisation seriously as a physical reality rather than a mathematical trick — light actually comes in discrete packets (photons), not as a continuous wave. This was the birth of quantum theory applied to light, and it paved the way for de Broglie's matter waves (1924), Schrödinger's equation (1926), and all of modern quantum mechanics. The photoelectric effect is thus not merely an interesting lab experiment — it is historically one of the most important discoveries in physics.

Exam Summary

Key equation: hf = φ + KE_max, where h = 6.626 × 10⁻³⁴ J·s, f is photon frequency, φ is work function (convert from eV to J: multiply by 1.6 × 10⁻¹⁹), KE_max = ½mv²_max = eV₀ (stopping potential times electron charge). Threshold frequency f₀ = φ/h — below this, no emission regardless of intensity. The effect of increasing intensity: more photons per second → more electrons emitted per second → larger photocurrent. No effect on KE_max. The effect of increasing frequency (above threshold): higher photon energy → larger KE_max = hf − φ. No effect on photocurrent if intensity is unchanged. The stopping potential V₀ = KE_max/e = (hf − φ)/e. Graph of V₀ vs f: straight line, gradient = h/e, x-intercept = f₀, y-intercept = −φ/e.

The photoelectric effect provides the clearest experimental evidence for the quantisation of light. The three key observations that classical theory cannot explain — the existence of a threshold frequency, the independence of KE_max from intensity, and the immediate onset of emission above the threshold — are all naturally explained by Einstein's photon model. Each observation maps directly to a testable prediction: threshold → hf₀ = φ; intensity independence → KE_max = hf − φ involves only f; instantaneous onset → one photon is absorbed by one electron in one quantum interaction. The clean agreement between Millikan's measurements and Einstein's predictions across a wide range of frequencies and materials remains one of the most compelling confirmations of quantum theory.

For exam calculations: convert all energies to the same unit before computing. If hf is in joules and φ is in eV, convert φ to joules first: φ(J) = φ(eV) × 1.6 × 10⁻¹⁹. Alternatively work entirely in eV: hf(eV) = hf(J)/(1.6 × 10⁻¹⁹), and then KE_max(eV) = hf(eV) − φ(eV). The stopping potential in volts numerically equals KE_max in eV (since eV₀ = KE_max, dividing both sides by e gives V₀ = KE_max/e, which in eV units means V₀ numerically equals KE_max in eV). This shortcut — V₀ (volts) = KE_max (eV) — saves one unit conversion step in stopping potential problems.

One of the most pedagogically powerful aspects of the photoelectric effect is that it demonstrates, in a single experiment, both the particle nature of light and the quantisation of energy. The threshold frequency proves energy is quantised (photon energy hf must exceed a threshold — a wave of any frequency could eventually supply enough energy). The linear KE_max vs frequency graph proves E = hf directly. And the independence of KE_max from intensity proves that one photon interacts with one electron. These three facts together make the photoelectric effect the most compact experimental proof of quantum theory available — which is why it appears on every physics syllabus and why understanding it deeply rewards students far beyond the exam marks it represents.