9478 / 2027
Quantum Physics overview

Lesson 1 of 7

Evidence for photons

  • I can explain how photoelectric threshold evidence supports particulate radiation while interference and diffraction support its wave nature.Syllabus

    Syllabus 9478, 19(a). Explain how photoelectric threshold evidence supports particulate radiation while interference and diffraction support its wave nature.

  • I can define a photon as a quantum of electromagnetic radiation and use E = hf.Syllabus

    Syllabus 9478, 19(b). Define a photon as a quantum of electromagnetic radiation and use E = hf.

  • I can explain that a massless photon has momentum and use p = E/c = h/lambda.Syllabus

    Syllabus 9478, 19(c). Explain that a massless photon has momentum and use p = E/c = h/lambda.

Make a guess. You are not marked.

Ultraviolet light makes a clean zinc plate emit electrons. The intensity of the light is doubled at the same frequency. What happens to the maximum kinetic energy of the emitted electrons?

  1. It doubles, because twice as much energy arrives at the plate each second.
  2. It stays the same, because each photon still carries the same energy.
  3. It rises a little, because electrons can now absorb two photons together.
Show the answer

It stays the same, because each photon still carries the same energy.

Each electron gains its energy from one photon, of energy hf. The frequency is unchanged, so hf and the maximum kinetic energy hf - work function are unchanged. Twice the intensity gives about twice as many photons, so about twice as many electrons.

Light produces interference and diffraction patterns, but it also transfers energy in individual quanta. A successful explanation must account for both kinds of evidence.

In the photoelectric effect, electrons can leave a material when it absorbs suitable electromagnetic radiation. For a given surface there is a threshold frequency: below it, increasing ordinary illumination does not cause emission. Above it, emission is possible even with weak illumination.

A photon is one quantum of electromagnetic radiation. Its energy depends on frequency:

E = hf

Here h is the Planck constant, 6.63 × 10-34 J s. At fixed frequency, higher intensity means more incident photons per second for a fixed illuminated area. It does not increase each photon's energy.

Compare frequency and intensity separately

Consider a surface with supplied threshold f0 = 5.00 × 1014 Hz. Keep the surface, geometry and collection conditions unchanged. The comparisons below use the ordinary single-photon model and, above threshold, conditions without saturation.

Controlled photoelectric comparisons for the supplied surface
FrequencyIlluminationPhotoelectric result
4.00 × 1014 HzLower intensityNo emission: insufficient energy per photon.
4.00 × 1014 HzHigher intensityStill no emission in this model.
6.00 × 1014 HzLower intensityEmission is possible.
6.00 × 1014 HzHigher intensityThe emission rate can increase.

Below the threshold: greater intensity is insufficient

Below the threshold: greater intensity is insufficientThe same metal surface has threshold frequency five times ten to the fourteen hertz. Incident photons move right towards its exposed left surface at frequency four times ten to the fourteen hertz. No emitted electron or outgoing electron arrow is drawn. Increasing ordinary single-photon illumination at this unchanged frequency does not supply enough energy per photon. Photon and electron symbols and arrow lengths are schematic, not measured positions, sizes or rates.Incident f = 4.00 × 1014 HzSurface f0 = 5.00 × 1014 HzPhotonPhotonMetalNo photoelectric emission

Compare frequencies while holding the material and collection conditions fixed. The arrows show incoming photons and, where possible, outgoing electrons. Icon counts do not specify an efficiency or a measured rate.

Above the threshold: emission is possible

Above the threshold: emission is possibleThe same metal surface has threshold frequency five times ten to the fourteen hertz. Incident photons move right towards its exposed left surface at frequency six times ten to the fourteen hertz. One schematic electron is shown leaving the surface, moving up and left. Emission is possible even with weak ordinary illumination, but this icon count does not state an emission or collection efficiency. Photon and electron symbols and arrow lengths are schematic, not measured positions, sizes or rates.Incident f = 6.00 × 1014 HzSurface f0 = 5.00 × 1014 HzPhotonPhotonElectronMetalEmission is possible

Compare frequencies while holding the material and collection conditions fixed. The arrows show incoming photons and, where possible, outgoing electrons. Icon counts do not specify an efficiency or a measured rate.

Incoming photons and possible emitted electrons have different arrows. The supplied frequency determines whether the threshold condition is met; icon size, colour and number are schematic and do not specify an emission efficiency.

The threshold supports quantised energy exchange: adding more photons whose individual energies are too small does not meet the single-photon condition. This does not remove the wave evidence from interference and diffraction. Nor does it claim that one ordinary single-photon model covers arbitrarily intense fields.

The work function is an energy

The work function, written here as Φwork, is the minimum energy needed to remove an electron from the specified surface. At threshold:

Φwork = hf0
= (6.63 × 10-34)(5.00 × 1014)
= 3.315 × 10-19 J
= 2.07 eV

One electronvolt is the energy gained by a particle of charge magnitude e accelerated through a potential difference of 1 V. Thus 1 eV = e × 1 V = 1.60 × 10-19 J using the supplied elementary charge. An eV is an energy unit, not a voltage. Φwork is measured in joules; it is distinct from magnetic flux Φ measured in webers.

Worked photon calculation

Energy, wavelength and momentum

An optical frequency of order 5 × 1014 Hz with h of order 7 × 10-34 J s suggests a few times 10-19 J per photon: a few eV. Expect a wavelength of a few hundred nanometres and momentum of order 10-27 kg m/s.

For f = 6.00 × 1014 Hz, use c = 3.00 × 108 m/s in vacuum:

λ = c/f = 5.00 × 10-7 m = 500 nm
E = hf = 3.978 × 10-19 J
E = (3.978 × 10-19)/(1.60 × 10-19)
= 2.48625 eV ≈ 2.49 eV

A photon has zero rest mass but nonzero momentum. Its momentum points along propagation, with magnitude:

p = E/c = h/λ
= 1.326 × 10-27 kg m/s
≈ 1.33 × 10-27 N s

The non-relativistic formula p = mv for a massive particle is not a way to set photon momentum to zero. Check that energy divided by speed has momentum units.

A monochromatic beam with supplied power 1.989 µW at this frequency carries P/E = 5.00 × 1012 photons/s. That incident photon rate is not automatically a recorded electron rate: absorption, emission and collection efficiencies matter.

Try it step by step

Work it out in steps: Photon energy E = hc / lambda in J, then Photon energy in eV, then Maximum kinetic energy in eV.

  1. Problem 1. Light of wavelength 250 nm falls on a metal of work function 4.3 eV. Take h = 6.63 x 10-34 J s, c = 3.00 x 108 m/s and 1 eV = 1.60 x 10-19 J. Find the maximum kinetic energy of the emitted electrons in eV.
  2. Problem 2. Light of wavelength 200 nm falls on the same metal, work function 4.3 eV. Use the same constants. Find the maximum kinetic energy of the emitted electrons in eV.
  3. Problem 3. Light of wavelength 300 nm falls on a metal of work function 2.3 eV. Use the same constants. Find the maximum kinetic energy of the emitted electrons in eV.
Show the answer

Problem 1.

Photon energy E = hc / lambda in J: E = 6.63 x 10-34 x 3.00 x 108 / (250 x 10-9) = 7.96 x 10-19 J

Photon energy in eV: E = 7.96 x 10-19 / 1.60 x 10-19 = 4.97 eV

Maximum kinetic energy in eV: Maximum kinetic energy = 4.97 - 4.3 = 0.67 eV

Problem 2.

Photon energy E = hc / lambda in J: E = 6.63 x 10-34 x 3.00 x 108 / (200 x 10-9) = 9.95 x 10-19 J

Photon energy in eV: E = 9.95 x 10-19 / 1.60 x 10-19 = 6.22 eV

Maximum kinetic energy in eV: Maximum kinetic energy = 6.22 - 4.3 = 1.92 eV

Problem 3.

Photon energy E = hc / lambda in J: E = 6.63 x 10-34 x 3.00 x 108 / (300 x 10-9) = 6.63 x 10-19 J

Photon energy in eV: E = 6.63 x 10-19 / 1.60 x 10-19 = 4.14 eV

Maximum kinetic energy in eV: Maximum kinetic energy = 4.14 - 2.3 = 1.84 eV

Check your understanding

A surface has threshold frequency 5.00 x 1014 Hz. In the ordinary single-photon model, what happens if the intensity at 4.00 x 1014 Hz increases while frequency and other conditions stay fixed?

  1. Every incident photon must now eject one electron.
  2. The photon rate increases, but each photon still has insufficient energy for photoelectric emission.
  3. Emission must begin because the greater intensity lets one electron accumulate enough incident wave energy.
Show the answer

The photon rate increases, but each photon still has insufficient energy for photoelectric emission.

Intensity at fixed frequency changes incident energy per second and photon rate, not energy per photon. The frequency remains below the supplied threshold, so the ordinary single-photon condition is not met.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Define the work function of a metal and the threshold frequency. [2]

Saved on this browser only. Anyone using this browser can see it.

Exam-style question

Ultraviolet light of wavelength 250 nm falls on a metal with a work function of 2.30 eV. Calculate the maximum kinetic energy of the emitted electrons, in eV. Take h = 6.63 x 10-34 J s, c = 3.00 x 108 m s-1 and e = 1.60 x 10-19 C. [4]

Saved on this browser only. Anyone using this browser can see it.

Exam-style question

Explain how the photoelectric effect gives evidence that light consists of photons. [3]

Saved on this browser only. Anyone using this browser can see it.

Exam-style question

A graph of maximum kinetic energy of photoelectrons (in eV) against frequency of the light is a straight line. It crosses the frequency axis at 4.8 x 1014 Hz and passes through (8.0 x 1014 Hz, 1.32 eV). Use the graph to find the Planck constant and the work function in eV. Take e = 1.60 x 10-19 C. [3]

Saved on this browser only. Anyone using this browser can see it.

Exam-style question

Data-based question: Measuring the Planck constant with a photocell [20]

Light of a single wavelength shines on a clean metal surface inside an evacuated photocell. A reverse potential difference is increased until the photocurrent just falls to zero; this is the stopping potential Vs. The experiment is repeated for five wavelengths. Take c = 3.00 x 108 m s-1 and e = 1.60 x 10-19 C.

Table 1. Stopping potential at different wavelengths
wavelength / nm400450500550600
Vs / V1.110.760.490.260.07
  1. (a) Explain what the stopping potential tells you about the emitted electrons. [2]
  2. (b) Write down the photoelectric equation and show that eVs = hf - Φ, where Φ is the work function. [2]
  3. (c) Calculate the frequency of the light for each wavelength. [1]
  4. (d) Use the data to find the gradient of a graph of Vs against f, and hence a value for the Planck constant h. [3]
  5. (e) Find the threshold frequency, and the work function of the metal in eV. [3]
  6. (f) A student says: "Red light of wavelength 700 nm will release electrons from this metal if it is bright enough." Evaluate this claim. [2]
  7. (g) The intensity of the 400 nm light is doubled. State and explain the effect on the stopping potential and on the photocurrent before the reverse p.d. is applied. [3]
  8. (h) The 450 nm beam delivers 2.0 mW to the surface. Calculate the number of photons arriving each second and the photocurrent if 5.0% of them each release an electron. [3]
  9. (i) State one assumption you made in (h). [1]

Saved on this browser only. Anyone using this browser can see it.