Calculate the energy of a photon of frequency 5.0 × 10^14\ Hz. [2 marks]

OCR OCR 2019-style practice: A clean metal surface has a work function of 2.8\ eV. Light of wavelength 4.0 × 10^-7\ m is shone onto it. Calculate the maximum kinetic energy of the emitted photoelectrons in joules. Take h = 6.63 × 10^-34\ J s, c = 3.0 × 10^8\ m s^-1 and e = 1.6 × 10^-19\ C.

Photon energy: E = (hc)/(λ) = (6.63 × 10^-34)(3.0 × 10^8)4.0 × 10^-7 = 4.97 × 10^-19\ J. Convert the work function to joules: = 2.8 × 1.6 × 10^-19 = 4.48 × 10^-19\ J. Einstein's equation: KE_ = hf - = 4.97 × 10^-19 - 4.48 × 10^-19 = 4.9 × 10^-20\ J. Markers reward the photon energy, converting the work function from electronvolts to joules, and the value about 4.9 × 10^-20\ J.

OCR OCR 2021-style practice: An electron is accelerated from rest, gaining a speed of 2.96 × 10^7\ m s^-1. Calculate the de Broglie wavelength of the electron. Take h = 6.63 × 10^-34\ J s and the electron mass m = 9.11 × 10^-31\ kg.

Use the de Broglie equation with the momentum: λ = (h)/(mv). Momentum: p = mv = (9.11 × 10^-31)(2.96 × 10^7) = 2.70 × 10^-23\ kg m s^-1. So λ = 6.63 × 10^-342.70 × 10^-23 = 2.5 × 10^-11\ m. Markers reward calculating the momentum, applying λ = (h)/(p), and the value about 2.5 × 10^-11\ m (comparable to atomic spacing, hence diffraction).

EnglandPhysicsSyllabus dot point

How does the photoelectric effect show that light is quantised, and how can particles behave as waves?

Quantum physics: the photon model and the photon energy equation, the photoelectric effect and Einstein's photoelectric equation, work function and threshold frequency, and wave-particle duality with the de Broglie wavelength.

A focused answer to the OCR H556 quantum physics content, covering the photon model and the energy of a photon, the photoelectric effect and Einstein's photoelectric equation, work function and threshold frequency, the electronvolt, and wave-particle duality with the de Broglie equation and electron diffraction.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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Quick answer

A photon is a discrete packet (quantum) of electromagnetic energy with energy $E = hf = \frac{hc}{\lambda}$ . The photoelectric effect is the emission of electrons from a metal surface when light above a threshold frequency is incident on it; it provides evidence that light is quantised because emission is instantaneous and depends on frequency, not intensity. Einstein's equation $hf = \phi + KE_{\max}$ states that a single photon's energy frees one electron (overcoming the work function $\phi$ ) and gives it kinetic energy. Below the threshold frequency no electrons are emitted however bright the light. Electrons also show wave behaviour: their de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$ , confirmed by electron diffraction.

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What this dot point is asking

OCR wants you to describe the photon model of electromagnetic radiation, use the photon energy equation, explain the photoelectric effect and apply Einstein's photoelectric equation with the work function and threshold frequency, use the electronvolt, and explain wave-particle duality using the de Broglie equation and electron diffraction.

The answer

The photon model

The electronvolt is a convenient energy unit at this scale: one electronvolt is the energy gained by an electron moving through a potential difference of one volt, $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ .

The photoelectric effect

The wave model predicts that any frequency, given enough time, should accumulate enough energy to release electrons. The observed threshold and instantaneous emission show energy arrives in discrete photons, one absorbed by one electron.

Einstein's photoelectric equation

A graph of $KE_{\max}$ against frequency is a straight line of gradient $h$ with an $x$ -intercept at the threshold frequency $f_0$ and a $y$ -intercept of $-\phi$ .

Wave-particle duality

Threshold frequency from the work function

A metal has a work function of $3.2 \times 10^{-19}\ \text{J}$ . Find its threshold frequency and the longest wavelength of light that will eject electrons. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ .

step 1: Threshold frequency

At the threshold all the photon energy frees the electron: $f_0 = \frac{\phi}{h} = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.83 \times 10^{14}\ \text{Hz}$ .

step 2: Longest wavelength

Longest wavelength means lowest frequency, the threshold: $\lambda_0 = \frac{c}{f_0} = \frac{3.0 \times 10^{8}}{4.83 \times 10^{14}} = 6.2 \times 10^{-7}\ \text{m}$ .

step 3: Interpret

Light of wavelength shorter than $6.2 \times 10^{-7}\ \text{m}$ (higher frequency) ejects electrons; longer wavelengths do not, regardless of intensity.

Examples in context

Solar cells and photodiodes rely on the photoelectric (photovoltaic) effect, releasing charge carriers when photons above the band-gap energy are absorbed. Photomultiplier tubes and image sensors detect single photons. The electron microscope exploits the tiny de Broglie wavelength of fast electrons to resolve detail far smaller than any optical microscope can, because resolution improves as wavelength falls. Light-emitting diodes emit photons of a fixed energy set by the material, demonstrating $E = hf$ in reverse.

Try this

Q1. State what is meant by the work function of a metal. [1 mark]

Cue. The minimum energy needed to free an electron from the surface of the metal.

Q2. Calculate the energy of a photon of frequency $5.0 \times 10^{14}\ \text{Hz}$ . [2 marks]

Cue. $E = hf = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.3 \times 10^{-19}\ \text{J}$ .

Q3. Explain why the photoelectric effect supports the photon model rather than the wave model of light. [3 marks]

Cue. Emission is instantaneous, there is a threshold frequency, and the maximum kinetic energy depends on frequency not intensity; the wave model cannot explain any of these, but a one-photon-one-electron model can.

Exam-style practice questions

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

OCR 20194 marksA clean metal surface has a work function of

2.8\ \text{eV}

. Light of wavelength

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

is shone onto it. Calculate the maximum kinetic energy of the emitted photoelectrons in joules. Take

h = 6.63 \times 10^{-34}\ \text{J s}

c = 3.0 \times 10^{8}\ \text{m s}^{-1}

and

e = 1.6 \times 10^{-19}\ \text{C}

Show worked answer →

Photon energy: $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{4.0 \times 10^{-7}} = 4.97 \times 10^{-19}\ \text{J}$ .

Convert the work function to joules: $\phi = 2.8 \times 1.6 \times 10^{-19} = 4.48 \times 10^{-19}\ \text{J}$ .

Einstein's equation: $KE_{\max} = hf - \phi = 4.97 \times 10^{-19} - 4.48 \times 10^{-19} = 4.9 \times 10^{-20}\ \text{J}$ .

Markers reward the photon energy, converting the work function from electronvolts to joules, and the value about $4.9 \times 10^{-20}\ \text{J}$ .

OCR 20213 marksAn electron is accelerated from rest, gaining a speed of

2.96 \times 10^{7}\ \text{m s}^{-1}

. Calculate the de Broglie wavelength of the electron. Take

h = 6.63 \times 10^{-34}\ \text{J s}

and the electron mass

m = 9.11 \times 10^{-31}\ \text{kg}

Show worked answer →

Use the de Broglie equation with the momentum: $\lambda = \frac{h}{mv}$ .

Momentum: $p = mv = (9.11 \times 10^{-31})(2.96 \times 10^{7}) = 2.70 \times 10^{-23}\ \text{kg m s}^{-1}$ .

So $\lambda = \frac{6.63 \times 10^{-34}}{2.70 \times 10^{-23}} = 2.5 \times 10^{-11}\ \text{m}$ .

Markers reward calculating the momentum, applying $\lambda = \frac{h}{p}$ , and the value about $2.5 \times 10^{-11}\ \text{m}$ (comparable to atomic spacing, hence diffraction).

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)