Analyse the photoelectric effect, including Einstein's photon equation hf = phi + KE_max, the role of Planck's constant, and the inability of the wave model to explain the threshold frequency and the kinetic-energy results

What this dot point is asking

NESA wants you to use Einstein's photon model to explain the photoelectric effect, write and apply $h f = \phi + KE_{\max}$, calculate the threshold frequency and stopping voltage, and clearly state which observations the classical wave model cannot account for.

The answer

The diagram below sketches the photoelectric effect. A photon of energy $hf$ above the metal's work function $\phi$ ejects an electron with kinetic energy $KE_{\max} = hf - \phi$. Below the threshold frequency $f_0 = \phi / h$, no electrons are ejected regardless of intensity.

Setting the scene

By the late 1800s the wave model of light was the standard. But in 1887 Heinrich Hertz noticed UV light striking metal electrodes increased the spark distance in his radio-wave apparatus. Lenard's careful experiments (1902) revealed three features the wave model could not explain:

1. Threshold frequency. Below a metal-specific frequency $f_0$, no electrons are ejected no matter how intense the light or how long it shines.
2. Frequency, not intensity, sets electron energy. $KE_{\max}$ depends linearly on frequency. Intensity sets the number of photoelectrons per second, not their energies.
3. No measurable time delay. Electrons are ejected effectively instantaneously when the light starts, even at very low intensity. A classical wave would need time to accumulate enough energy in one electron.

Einstein's photon hypothesis (1905)

Building on Planck's 1900 idea that energy is exchanged in discrete amounts $h f$, Einstein proposed that light itself is made of discrete energy packets:

A single photon is absorbed by a single electron all at once. If the electron is bound to the metal by an energy $\phi$ (the work function, the minimum energy to remove the least tightly bound electron), then:

This is Einstein's photoelectric equation. $KE_{\max}$ is the maximum kinetic energy of the ejected photoelectron; electrons deeper in the metal lose more energy before escape and emerge with less.

Threshold frequency

The minimum frequency that can eject any electron is the one where $KE_{\max} = 0$:

Below $f_0$, photons simply do not carry enough energy to free an electron, no matter how many arrive. This is the killer observation for the wave model: a classical wave of any frequency should eventually deliver enough energy if intense or sustained enough.

Stopping voltage

In the standard experiment, a positive collector electrode is gradually reverse-biased until the most energetic photoelectrons are turned back. The reversing voltage at which photocurrent vanishes is the stopping voltage $V_s$:

A plot of $V_s$ vs $f$ is a straight line with gradient $h / e$ (giving Planck's constant) and $x$-intercept $f_0$ (giving the threshold) and $y$-intercept $-\phi / e$ (giving the work function). This is Millikan's 1916 experiment, which confirmed Einstein's equation to high precision and helped earn both their Nobel Prizes.

Worked example: caesium photocell

Caesium has work function $\phi = 2.10$ eV. Calculate the maximum kinetic energy and stopping voltage when light of wavelength $400$ nm illuminates the cathode.

Photon energy: $E = h c / \lambda$. Using the shortcut $h c = 1240$ eV nm:

$E = 1240 / 400 = 3.10$ eV.

$KE_{\max} = 3.10 - 2.10 = 1.00$ eV.

Stopping voltage: $V_s = KE_{\max} / e = 1.00$ V.

If the wavelength is increased to $700$ nm: $E = 1240 / 700 = 1.77$ eV $< \phi$, so no photoelectrons are emitted regardless of intensity.

Why the wave model fails

Three predictions of the classical wave model are flatly contradicted:

The wave model is rescued for interference and diffraction, but for absorption and emission by atoms, the quantum (photon) model is needed. This duality is the heart of "wave-particle duality": light behaves as a wave in propagation and as a particle in interaction.

Planck's constant

$h = 6.626 \times 10^{-34}$ J s is the universal constant relating frequency to energy quantum. It also appears in:

• The energy of a photon: $E = h f$.
• The momentum of a photon: $p = h / \lambda$.
• De Broglie's matter-wave relation: $\lambda = h / p$.
• The Heisenberg uncertainty principle: $\Delta x \Delta p \geq \hbar / 2$.

Millikan's measurement of the slope $h / e$ in the photoelectric stopping-voltage plot gave $h = 6.57 \times 10^{-34}$ J s, agreeing with Planck's blackbody value.

Examples in context

Example 1. Photoelectric effect on a sodium photocathode at UNSW. Sodium has work function $\phi = 2.28 \text{ eV}$. UV light of $\lambda = 350 \text{ nm}$ has photon energy $E = h c / \lambda = 6.626 \times 10^{-34} \times 3.0 \times 10^8 / 3.5 \times 10^{-7} = 5.68 \times 10^{-19} \text{ J} = 3.55 \text{ eV}$. Maximum kinetic energy of ejected electrons is $KE_{\max} = h f - \phi = 3.55 - 2.28 = 1.27 \text{ eV}$. The stopping voltage is $V_s = KE_{\max} / e = 1.27 \text{ V}$. Doubling the light intensity doubles the number of photoelectrons but leaves $KE_{\max}$ unchanged, the key prediction of Einstein's model that classical wave theory cannot explain.

Example 2. Solar panels on a Sydney rooftop. A silicon solar cell has band gap $E_g = 1.12 \text{ eV}$, corresponding to threshold wavelength $\lambda_0 = h c / E_g = 1240 / 1.12 = 1107 \text{ nm}$ (near-IR). Visible photons at $\lambda = 600 \text{ nm}$ carry $E = 2.07 \text{ eV}$, so each one excites an electron across the band gap with $0.95 \text{ eV}$ "wasted" as heat. UV photons at $300 \text{ nm}$ ($E = 4.13 \text{ eV}$) still produce only one electron per photon, dumping $3.01 \text{ eV}$ as heat. This single-photon-single-electron limit is exactly the photoelectric ratio and explains the $\sim 33\%$ Shockley-Queisser efficiency cap for single-junction silicon.

Try this

Q1. State the photoelectric equation and explain what each term means. [2 marks]

• Cue. $h f = \phi + KE_{\max}$. Photon energy = work function + max KE of ejected electron.

Q2. A metal has work function $\phi = 4.5 \text{ eV}$. (a) Find the threshold wavelength. (b) When illuminated by $\lambda = 200 \text{ nm}$ light, calculate the maximum kinetic energy of photoelectrons. [4 marks]

• Cue. (a) $\lambda_0 = hc/\phi = 1240/4.5 = 276 \text{ nm}$. (b) $E_{\text{photon}} = 1240/200 = 6.20 \text{ eV}$; $KE_{\max} = 6.20 - 4.5 = 1.70 \text{ eV}$.

Q3. A photoelectric experiment plots $KE_{\max}$ versus frequency. (a) State the gradient and y-intercept of this graph. (b) Explain how the graph yields Planck's constant. (c) Identify two observations from photoelectric experiments that the wave model cannot explain. [2+2+2 marks]

• Cue. (a) Gradient $h$, intercept $-\phi$. (b) From slope; $h =$ rise/run. (c) Threshold frequency exists; $KE_{\max}$ depends on $f$ not intensity; no time delay.