What is the stopping voltage?

In practice the maximum kinetic energy is measured by applying a reverse voltage that just stops the most energetic electrons. At the stopping voltage V_stop the work done against the field equals the maximum kinetic energy, so E_k,max = eV_stop. This gives a clean experimental route to E_k,max in electronvolts, as used in the calcium question above.

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Why does light behave like a stream of particles?

Use the photon model and Einstein's photoelectric equation to explain the photoelectric effect.

How the photoelectric effect reveals light as photons of energy E equals h f, Einstein's photoelectric equation, work function and threshold frequency.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

The photoelectric effect was one of the first experiments to reveal the quantum nature of light, and it earned Einstein his Nobel Prize. It is central to Unit 4's story of how classical physics gave way to quantum ideas.

The experiment

When light shines on a clean metal surface, electrons can be ejected. These ejected electrons are called photoelectrons. Careful measurements revealed three features that classical wave theory could not explain:

There is a threshold frequency below which no electrons are emitted, no matter how bright (intense) the light.
The maximum kinetic energy of the emitted electrons depends on the frequency of the light, not its intensity.
Emission is essentially instantaneous, even for very dim light above the threshold frequency.

Classical theory predicts that any frequency should work given enough intensity or time, so these results demanded a new model.

The photon model

Einstein proposed that light energy is carried in discrete quanta, now called photons. Each photon has energy proportional to the frequency $f$ of the light:

E = hf = \frac{hc}{\lambda}

where $h = 6.63 \times 10^{-34}\ \text{J s}$ is Planck's constant and $\lambda$ is the wavelength. One photon interacts with one electron in a single, all-or-nothing event, which is why emission is instantaneous and why the energy of each electron depends on the energy of a single photon.

Work function and threshold frequency

To escape the metal, an electron must be given at least a minimum energy called the work function $\phi$ , a property of the metal. The threshold frequency $f_0$ is the lowest frequency whose photon energy just equals the work function:

\phi = h f_0

If the photon energy $hf$ is less than $\phi$ , no electron escapes, however intense the beam. This neatly explains feature 1: it is the energy per photon, not the total energy delivered, that decides whether an electron escapes.

Einstein's photoelectric equation

Energy conservation for one photon ejecting one electron gives Einstein's photoelectric equation:

E_{k,\text{max}} = hf - \phi

The photon supplies energy $hf$ . Part of it ( $\phi$ ) is used to free the electron, and the rest becomes the maximum kinetic energy of the photoelectron. Increasing the frequency raises $E_{k,\text{max}}$ linearly, while increasing the intensity only increases the number of photoelectrons. A graph of $E_{k,\text{max}}$ against $f$ is a straight line of gradient $h$ and horizontal intercept $f_0$ , so plotting experimental data both confirms the theory and lets you measure Planck's constant.

The stopping voltage

In practice the maximum kinetic energy is measured by applying a reverse voltage that just stops the most energetic electrons. At the stopping voltage $V_\text{stop}$ the work done against the field equals the maximum kinetic energy, so $E_{k,\text{max}} = eV_\text{stop}$ . This gives a clean experimental route to $E_{k,\text{max}}$ in electronvolts, as used in the calcium question above.

Worked example: maximum kinetic energy

Light of frequency $1.2 \times 10^{15}\ \text{Hz}$ falls on a metal with work function $\phi = 3.0 \times 10^{-19}\ \text{J}$ . Find the maximum kinetic energy of the emitted photoelectrons.

Step 1: Calculate the energy carried by one photon

The energy of a single photon depends only on its frequency via $E = hf$ , where $h = 6.63 \times 10^{-34}\ \text{J s}$ is Planck's constant. This is the total energy that one photon brings to the interaction, and it sets an upper limit on how much energy the ejected electron can receive:

E = hf = (6.63 \times 10^{-34})(1.2 \times 10^{15}) = 7.96 \times 10^{-19}\ \text{J}.

Step 2: Apply Einstein's photoelectric equation

The work function $\phi$ is the minimum energy needed to free an electron from the metal surface. Whatever energy remains after paying this escape cost becomes the kinetic energy of the fastest electrons. Subtracting the work function from the photon energy gives the maximum kinetic energy directly:

E_{k,\text{max}} = hf - \phi = 7.96 \times 10^{-19} - 3.0 \times 10^{-19} = 5.0 \times 10^{-19}\ \text{J}.

Final answer: The maximum kinetic energy of the photoelectrons is about $5.0 \times 10^{-19}\ \text{J}$ .

In the exam, watch your units: energies are often quoted in electronvolts, where $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ . Convert to joules before substituting, and use the linear graph of $E_{k,\text{max}}$ against $f$ to extract Planck's constant or the work function when asked.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20225 marksZinc has a work function of

4.3\ \text{eV}

. Calculate the threshold frequency for zinc, then the maximum kinetic energy of photoelectrons emitted when UV photons of wavelength

1.85 \times 10^{-7}\ \text{m}

strike the zinc. State what increasing the intensity does to the photocurrent and to the threshold frequency.

Show worked answer →

Threshold frequency from $\phi = h f_0$ , with $\phi = 4.3\ \text{eV} = 4.3 \times 1.6 \times 10^{-19} = 6.88 \times 10^{-19}\ \text{J}$ :

f_0 = \frac{\phi}{h} = \frac{6.88\times10^{-19}}{6.63\times10^{-34}} = 1.04 \times 10^{15}\ \text{Hz}.

Photon energy of the UV light: $E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.0\times10^8)}{1.85\times10^{-7}} = 1.075 \times 10^{-18}\ \text{J}$ ( $= 6.72\ \text{eV}$ ).

Einstein's equation: $E_{k,\text{max}} = E - \phi = 1.075\times10^{-18} - 6.88\times10^{-19} = 3.87 \times 10^{-19}\ \text{J}$ ( $= 2.42\ \text{eV}$ ).

Increasing intensity increases the number of photons per second, so the photocurrent rises (more electrons ejected per second), but the threshold frequency is unchanged, since it depends only on the work function. Markers want $f_0 = \dfrac{\phi}{h}$ , $E_{k,\text{max}} = hf - \phi$ , and the distinction that intensity raises current but not threshold frequency.

TCE 20232 marksLight of wavelength

405\ \text{nm}

is shone on a clean calcium plate in a photoelectric tube. The photocurrent is stopped by a potential difference of

0.20\ \text{V}

. Calculate the photon energy in

\text{eV}

, then the work function of the calcium in

\text{eV}

Show worked answer →

Photon energy: $E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.0\times10^8)}{405\times10^{-9}} = 4.91 \times 10^{-19}\ \text{J}$ .
In electronvolts: $E = \dfrac{4.91\times10^{-19}}{1.6\times10^{-19}} = 3.07\ \text{eV}$ .

The stopping voltage gives the maximum kinetic energy directly: $E_{k,\text{max}} = eV_\text{stop} = 0.20\ \text{eV}$ .

Einstein's equation rearranged for the work function: $\phi = E - E_{k,\text{max}} = 3.07 - 0.20 = 2.87\ \text{eV}$ .

The calcium work function is about $2.87\ \text{eV}$ . Markers want $E = \dfrac{hc}{\lambda}$ converted to eV, $E_{k,\text{max}} = 0.20\ \text{eV}$ from the stopping voltage, then $\phi = hf - E_{k,\text{max}}$ .

TCE 20232 marksUsing Einstein's photoelectric equation and the photon model, explain why the statement 'brighter light should give faster photoelectrons' is wrong.

Show worked answer →

In the photon model each photoelectron is ejected by absorbing a single photon. The maximum kinetic energy is set by $E_{k,\text{max}} = hf - \phi$ , which depends only on the photon frequency $f$ (and the work function $\phi$ ), not on brightness.

Making the light brighter delivers more photons per second, but each photon still carries the same energy $hf$ . So brighter light ejects more electrons per second (a larger current) but gives no individual electron more energy.

Therefore brighter light does not produce faster photoelectrons; only increasing the frequency does. Markers want the one-photon-one-electron idea and $E_{k,\text{max}}$ depending on frequency, with intensity affecting only the number of electrons.