Apply the photon model of light to the photoelectric effect using $E_{\text{photon}} = h f$ and $E_{k,\max} = h f - \phi$, where $\phi$ is the work function of the metal, and interpret the stopping voltage $V_0$ as $e V_0 = E_{k,\max}$

What this dot point is asking

VCAA wants you to describe the photoelectric effect, state why it cannot be explained by the classical wave model, apply Einstein's photoelectric equation $E_{k,\max} = h f - \phi$ to compute photon energy, maximum kinetic energy of ejected electrons, threshold frequency and stopping voltage. Cross-link: see the photoelectric effect calculator.

The photoelectric effect

When light shines on a clean metal surface, electrons can be ejected from the surface. The ejected electrons are called photoelectrons.

The phenomenon was first observed by Hertz (1887) and quantitatively studied by Lenard. Its behaviour cannot be explained by the classical wave model of light. Einstein's 1905 explanation (for which he won the 1921 Nobel Prize) introduced the photon and started the development of quantum mechanics.

The four classical-defying observations

The classical wave model of light predicted that:

• Light of any frequency, given enough intensity or enough time, should eventually eject electrons.
• Electron kinetic energy should depend on intensity (brighter light, faster electrons).
• There should be a measurable time lag for low intensities (waiting for enough energy to accumulate).

The experimental observations were the opposite:

1. There is a threshold frequency $f_0$ below which no electrons are emitted, regardless of intensity.
2. Electrons are emitted essentially instantaneously (within nanoseconds), even at very low intensity, provided $f > f_0$.
3. Intensity controls the current (number of electrons per second), not the maximum kinetic energy of each electron.
4. Maximum kinetic energy is determined by frequency, linearly above threshold.

These observations are inconsistent with the wave model and gave rise to the photon picture.

The photon model

Light consists of discrete energy quanta called photons. Each photon has energy:

where $h = 6.626 \times 10^{-34}$ J s $= 4.14 \times 10^{-15}$ eV s is Planck's constant.

A light beam of intensity $I$ consists of photons of energy $hf$, with the number of photons per second proportional to $I$.

Einstein's photoelectric equation

When a photon strikes the metal surface, it is absorbed by a single electron. The electron uses part of the photon's energy to escape the metal (overcoming the work function $\phi$), and the remainder becomes kinetic energy:

The work function $\phi$ is the minimum energy required to remove an electron from the metal's surface. It is a property of the metal:

• Sodium: $\phi \approx 2.3$ eV
• Caesium: $\phi \approx 2.1$ eV (lowest of common metals; used for photocathodes)
• Copper: $\phi \approx 4.7$ eV
• Platinum: $\phi \approx 6.4$ eV

If $h f < \phi$, no electron can escape and no photoelectric current is produced. If $h f \geq \phi$, electrons are emitted with maximum kinetic energy $h f - \phi$ (electrons originating from deeper in the metal lose extra energy to lattice scattering and emerge with less KE).

Threshold frequency

The threshold frequency $f_0$ is the minimum frequency at which photoemission occurs:

Equivalently, the threshold wavelength is $\lambda_0 = h c / \phi = c / f_0$.

Below threshold: no current at any intensity.

Above threshold: current proportional to intensity, and $E_{k,\max}$ increases linearly with $f$.

The stopping voltage

In a photoelectric experiment, the ejected electrons can be decelerated by applying a reverse voltage. The stopping voltage $V_0$ is the minimum reverse voltage that stops all photoelectrons (current drops to zero).

By energy conservation, the work done by the reverse voltage on an electron equals the maximum kinetic energy of the ejected electron:

So:

In the eV system, if you compute $E_{k,\max}$ in eV, $V_0$ in volts is numerically equal: $E_{k,\max} = 0.7$ eV means $V_0 = 0.7$ V.

The $V_0$ vs $f$ graph

Plotting stopping voltage against frequency for a given metal gives a straight line:

• Gradient: $h / e$, the same for all metals. Millikan's 1916 photoelectric experiment measured $h$ this way.
• $y$-intercept: $-\phi / e$ (negative).
• $x$-intercept: $f_0 = \phi / h$, the threshold frequency.

Different metals give parallel lines (same gradient $h / e$) shifted by their different work functions.

What changes if we...

• Increase the light intensity (more photons per second, same frequency). More electrons emitted per second (higher current), but each electron still has the same $E_{k,\max}$ because each photon still has the same energy $hf$. $V_0$ unchanged.
• Increase the frequency (higher-energy photons, same intensity). $E_{k,\max}$ increases linearly. $V_0$ increases linearly. Current may decrease slightly (fewer photons for the same intensity at higher energy).
• Change the metal (different work function, same light). Threshold frequency shifts. If light frequency is above the new threshold, $E_{k,\max}$ shifts (different intercept on the linear graph).
• Drop the frequency below $f_0$. No emission, no current, no $V_0$ measurement possible. Increasing intensity does not help.

Examples in context

Example 1. Australian Synchrotron photoelectron spectroscopy. The Synchrotron's soft X-ray spectroscopy beamline shines $50$ eV photons onto metal samples to measure binding energies of core electrons. For a clean copper sample with work function $\phi = 4.65$ eV, the maximum photoelectron kinetic energy is $E_{k,\max} = hf - \phi = 50 - 4.65 = 45.4$ eV. Stopping voltage is therefore $V_0 = E_{k,\max}/e = 45.4$ V. Sweeping the incident photon energy and measuring stopping voltage maps out core-level binding energies, identifying oxidation states and elemental composition with sub-nanometre surface sensitivity.

Example 2. Solar-panel photoelectric response at AGL Hornsdale. AGL's solar+battery hybrid sites use silicon PV cells with band gap $1.1$ eV, equivalent to a work-function-style threshold for photoelectron promotion across the junction. Light of wavelength $550$ nm has photon energy $E = hc/\lambda = 1.99 \times 10^{-25}/5.5 \times 10^{-7} = 3.61 \times 10^{-19}$ J = $2.26$ eV. Photoelectron kinetic energy after promotion is approximately $2.26 - 1.1 = 1.16$ eV. Infrared photons at $1200$ nm ($1.03$ eV) fall below the threshold and do not promote photoelectrons, which is the fundamental limit on silicon solar efficiency.

Try this

Q1. State the photoelectric equation and define each term. [2 marks]

• Cue. $E_{k,\max} = hf - \phi$. $hf$ is incident photon energy, $\phi$ is metal's work function.

Q2. Light of wavelength $400$ nm shines on a metal with work function $2.0$ eV. Calculate (a) the photon energy in eV, and (b) the stopping voltage. [4 marks]

• Cue. (a) $E = 1240/400 = 3.10$ eV. (b) $E_{k,\max} = 3.10 - 2.0 = 1.10$ eV; $V_0 = 1.10$ V.

Q3. Refer to the Synchrotron photoelectron spectroscopy example. (a) Calculate the maximum photoelectron KE for $50$ eV photons on copper ($\phi = 4.65$ eV). (b) Determine the stopping voltage. (c) Explain why stopping voltage is independent of light intensity. [2+2+2 marks]

• Cue. (a) $45.4$ eV. (b) $V_0 = 45.4$ V. (c) Stopping voltage depends only on max KE, which depends on photon energy not intensity; intensity changes photoelectron count rate only.