Photons — OCR A-Level Physics
Test yourself on Photons with OCR A-Level practice questions.
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Photons explained
Electromagnetic radiation can be modelled as a stream of discrete packets of energy called photons.
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This particulate model explains phenomena such as the photoelectric effect, where light below a threshold frequency cannot eject electrons no matter how intense it is. Each photon carries energy E = hf, where h is the Planck constant (6.63 × 10⁻³⁴ J s) and f is the frequency of the radiation. The photon model treats radiation as localised quanta rather than a continuous wave. For example, a beam of ultraviolet light consists of many photons, each with energy determined by its frequency. Increasing intensity increases the number of photons per second, not the energy of each photon. Students should understand that the photon model complements the wave model, and that both are needed to describe all properties of electromagnetic radiation.
(b) photon as a quantum of energy of electromagnetic radiation
A photon is a quantum, meaning a single indivisible unit, of electromagnetic radiation. Each photon carries a specific amount of energy determined by the frequency of the radiation: E = hf. This quantisation means that energy can only be transferred in whole multiples of hf. For example, a photon of red light (f ≈ 4.3 × 10¹⁴ Hz) has energy E = 6.63 × 10⁻³⁴ × 4.3 × 10¹⁴ ≈ 2.9 × 10⁻¹⁹ J. A photon of blue light has a higher frequency and therefore more energy. The concept of the photon as a quantum explains why energy exchanges in the photoelectric effect are discrete and why there is a minimum frequency for electron emission. Students should be able to calculate photon energy and explain that the photon is the smallest possible amount of electromagnetic energy at a given frequency.
(c) energy of a photon; E = hf and E = hc/λ
A photon is a quantum of electromagnetic radiation, and its energy depends only on frequency: E = hf, where h is the Planck constant (6.63 × 10⁻³⁴ J s) and f is frequency in hertz. Because c = fλ, substituting f = c/λ gives E = hc/λ, useful when wavelength is quoted. For example, red light of wavelength 6.5 × 10⁻⁷ m has E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (6.5 × 10⁻⁷) ≈ 3.1 × 10⁻¹⁹ J. Note that shorter wavelength means higher photon energy, and doubling frequency doubles energy. In MCQs, check whether the question gives f or λ, keep units consistent, and watch powers of ten.
(d) the electronvolt (eV) as a unit of energy
The electronvolt is a convenient energy unit for atomic and photon scales. One electronvolt is the kinetic energy gained by an electron accelerated through a potential difference of one volt, so 1 eV = 1.60 × 10⁻¹⁹ J. To convert, multiply eV by 1.60 × 10⁻¹⁹ to get joules, or divide joules by 1.60 × 10⁻¹⁹ to get eV. For example, a photon of energy 2.0 eV is 2.0 × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J. In MCQs, check whether the answer options are in eV or J and convert before comparing. The eV is a unit of energy, not potential difference or charge.
(e)
This row is a guided-reading pointer within section 4.5.1 Photons, not an assessed answer target. Use it to navigate the specification: read the surrounding statements on photon energy, the electronvolt and the LED experiment, and check how they connect. As you read, note that E = hf and E = hc/λ give photon energy, the eV provides a convenient atomic-scale energy unit, and the LED method uses eV = hc/λ to estimate h. Your task is to identify what each statement requires you to be able to do, then locate the relevant equations and practical context in your course materials. Record any gaps for revision.
(i) using LEDs and the equation eV = hc/λ to estimate the value of Planck constant h
In the LED experiment, a light-emitting diode emits a photon when an electron crosses the junction potential difference V. The energy lost by the electron is eV, and this equals the photon energy hc/λ, giving eV = hc/λ and hence h = eVλ/c. Measure the threshold voltage at which the LED just begins to emit light and the wavelength of the emitted light, then substitute. For example, V = 1.9 V and λ = 6.5 × 10⁻⁷ m give h = (1.60 × 10⁻¹⁹ × 1.9 × 6.5 × 10⁻⁷) ÷ (3.00 × 10⁸) ≈ 6.6 × 10⁻³⁴ J s. In MCQs, check the rearrangement and unit consistency.
(ii) Determine the Planck constant using different coloured LEDs.
In this practical determination, each LED emits photons when electrons in the junction recombine, and the minimum energy needed equals the photon energy hf. The threshold voltage V across the LED is measured for several colours, so that eV = hf. Since f = c/λ, a graph of V against 1/λ has gradient hc/e, so h = gradient × e ÷ c. You measure V just as each LED begins to conduct, using a voltmeter and a series protective resistor, and take λ from the LED specification. Using several colours gives a range of frequencies and reduces the effect of random uncertainty. The value obtained is approximate because the threshold is not perfectly sharp and the emission wavelength has a spread.
Your focus
- Describe the photon model of electromagnetic radiation.
- Calculate photon energy using E = hf.
- Explain how the photon model accounts for the threshold frequency in the photoelectric effect.
Show all 21 objectives
- Define a photon as a quantum of electromagnetic energy.
- Calculate the energy of a photon using E = hf.
- Explain how photon energy depends on the frequency of the radiation.
- Select and apply E = hf to find the energy of a photon from its frequency.
- Use c = fλ to convert between wavelength and frequency before applying E = hc/λ.
- Interpret how photon energy changes with frequency or wavelength in multiple-choice contexts.
- Define the electronvolt in terms of energy gained by an electron through 1 V.
- Convert between electronvolts and joules accurately in both directions.
- Select the appropriate energy unit when answering multiple-choice questions on photons.
- Identify the assessed statements surrounding this guided-reading pointer.
- Explain how photon energy, the electronvolt and the LED experiment connect.
- Produce a revision checklist covering each statement in section 4.5.1.
- Explain how the LED experiment links electron energy eV to photon energy hc/λ.
- Rearrange eV = hc/λ to estimate h from measured threshold voltage and wavelength.
- Evaluate the accuracy of an experimental value of h against the accepted value.
- Recall and apply E = hf and E = hc/λ to photons emitted by LEDs.
- Explain why the threshold voltage of an LED corresponds to the photon energy.
- Use a graph of V against 1/λ to determine the Planck constant and evaluate the result.
Photons exam tips
Marking Points
- States that electromagnetic radiation can be modelled as discrete packets of energy called photons.
- Recalls that the energy of a photon is given by E = hf.
- Explains that intensity relates to the number of photons per second, not the energy of individual photons.
- Describes how the photon model explains the threshold frequency in the photoelectric effect.
- Recognises that the photon model and wave model are complementary descriptions of electromagnetic radiation.
- Defines a photon as a quantum (discrete packet) of electromagnetic energy.
- States that photon energy is given by E = hf.
- Explains that energy is quantised in units of hf.
- Calculates photon energy using the Planck constant and frequency.
- Relates photon energy to frequency: higher frequency means higher energy per photon.
- States that photon energy is directly proportional to frequency, E = hf, with h = 6.63 × 10⁻³⁴ J s.
- Uses c = fλ to derive or apply E = hc/λ when wavelength rather than frequency is given.
- Recognises that shorter wavelength corresponds to higher photon energy for the same radiation speed c.
- Calculates correctly with powers of ten, for example E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (6.5 × 10⁻⁷) ≈ 3.1 × 10⁻¹⁹ J.
- Distinguishes photon energy from intensity: intensity depends on the number of photons per second, not on hf alone.
- Defines 1 eV as the energy transferred when an electron moves through a potential difference of 1 V.
- States the conversion 1 eV = 1.60 × 10⁻¹⁹ J and applies it in either direction.
- Converts photon energies between eV and J correctly, for example 2.0 eV = 3.2 × 10⁻¹⁹ J.
- Recognises that the eV is an energy unit, not a unit of potential difference or charge.
- Chooses eV for atomic and photon-scale energies to avoid unwieldy powers of ten.
- States that the energy lost by an electron crossing the LED junction is eV and equals the emitted photon energy hc/λ.
- Rearranges eV = hc/λ to h = eVλ/c and substitutes measured values correctly.
- Uses the threshold voltage at which the LED just begins to emit light as V in the equation.
- Converts wavelength to metres and uses e = 1.60 × 10⁻¹⁹ C and c = 3.00 × 10⁸ m s⁻¹.
- Evaluates h to an order of magnitude consistent with 6.63 × 10⁻³⁴ J s and comments on uncertainty in the estimate.
- Photon energy is related to frequency by E = hf, and to wavelength by E = hc/λ.
- The minimum energy supplied per electron is eV, where V is the threshold voltage of the LED.
- Equating eV = hc/λ gives V = (hc/e)(1/λ), so a graph of V against 1/λ is a straight line through the origin.
- The gradient of the V against 1/λ graph equals hc/e, so h = gradient × e ÷ c.
- Using several different coloured LEDs provides several frequencies and improves the reliability of the gradient.
- The threshold voltage is identified as the voltage at which the LED just begins to emit light or conduct appreciably.
Examiner Tips
- 💡Use E = hf and remember that f = c/λ if wavelength is given.
- 💡When explaining the photoelectric effect, state that one photon interacts with one electron.
- 💡Check that your answer distinguishes between intensity (photons per second) and photon energy (hf).
- 💡Quote the Planck constant to at least two significant figures when performing calculations.
- 💡Always write the equation E = hf before substituting values.
- 💡Convert all frequencies to Hz and wavelengths to metres before calculation.
- 💡Use standard form carefully, especially with powers of ten in the Planck constant.
- 💡When comparing photons, state that higher frequency corresponds to higher energy.
- 💡Write the equation you will use before substituting numbers, so the examiner can follow your method.
- 💡Convert all wavelengths to metres and keep values in standard form to reduce power-of-ten slips.
- 💡Sanity-check the answer: visible photons have energies of order 10⁻¹⁹ J, so a result near 10⁻¹⁵ J signals an error.
- 💡Write the conversion factor explicitly before calculating so the direction of conversion is clear.
- 💡Check the unit requested in the answer options and convert your value to match before selecting.
- 💡Use standard form throughout to keep track of the 10⁻¹⁹ factor.
- 💡Read the full section before making notes, so each statement is placed in context.
- 💡Turn each statement into a question you can answer, such as 'Can I convert between eV and J?'
- 💡Flag any statement you cannot explain aloud and revisit it with your teacher or textbook.
- 💡Write the full equation eV = hc/λ before rearranging, so the algebra is visible and checkable.
- 💡Keep wavelength in metres and use standard form to avoid power-of-ten errors.
- 💡Compare your result with the accepted value 6.63 × 10⁻³⁴ J s and comment briefly on the uncertainty of the estimate.
- 💡State the equation eV = hc/λ before rearranging, so the examiner can see the physical link between threshold voltage and photon energy.
- 💡Show the gradient calculation with coordinates read from the line of best fit, not from individual data points.
- 💡Give the final value of h with a sensible unit, J s, and compare it with the accepted value to comment on accuracy.
- 💡Mention one limitation, such as the spread of wavelengths emitted by an LED, when asked to evaluate the method.
Common Mistakes
- Believing that increasing the intensity of light increases the energy of each photon: it increases the number of photons.
- Thinking that the photon model replaces the wave model entirely: both models are needed for different phenomena.
- Confusing the Planck constant with other constants or using the wrong value.
- Assuming that photons have mass: photons are massless and travel at the speed of light in a vacuum.
- Thinking that a photon can have any energy: its energy is fixed by its frequency.
- Confusing the quantum of energy with the quantum of charge: the photon is the quantum of electromagnetic energy.
- Forgetting to convert frequency from kHz or MHz to Hz before calculating energy.
- Assuming that all photons have the same energy regardless of frequency.
- Using E = hf with wavelength substituted for f: correct by first finding f = c/λ, then applying E = hf.
- Forgetting to convert nanometres to metres: 500 nm must be written as 5.00 × 10⁻⁷ m before substitution.
- Treating photon energy as dependent on intensity: intensity changes the number of photons, while each photon's energy depends only on frequency.
- Confusing eV with volts: the volt is potential difference, while the electronvolt is energy; convert using 1 eV = 1.60 × 10⁻¹⁹ J.
- Multiplying by 1.60 × 10⁻¹⁹ when converting joules to eV: divide joules by 1.60 × 10⁻¹⁹ instead.
- Assuming the eV applies only to electrons: any particle or photon energy can be expressed in eV.
- Treating a guided-reading pointer as an examinable fact: it directs you to the surrounding statements, so study those instead.
- Reading statements in isolation: link photon energy, the electronvolt and the LED experiment as one coherent topic.
- Skipping the practical context: the LED method is a required experimental technique, not just a formula to memorise.
- Using the operating voltage rather than the threshold voltage: the threshold voltage corresponds to the minimum photon energy, so use that value.
- Forgetting to convert wavelength from nanometres to metres: 650 nm must be entered as 6.50 × 10⁻⁷ m.
- Rearranging incorrectly to h = eVc/λ: the correct rearrangement is h = eVλ/c.
- Using the normal operating voltage of the LED rather than the threshold voltage at which emission just begins; the threshold value is the one that corresponds to the minimum photon energy.
- Plotting V against λ instead of V against 1/λ; the linear relationship requires the reciprocal of wavelength.
- Forgetting to convert wavelength from nanometres to metres before calculating 1/λ, which changes the gradient by a factor of 10⁹.
- Treating the gradient itself as h rather than as hc/e, so the electron charge and speed of light are omitted from the final calculation.