AeroLab Reading room

Physics / A2 level

Quantum physics

Explain photons, threshold behaviour and discrete atomic energies.

Learning objectives

The big idea

Electromagnetic energy is exchanged in photons of energy hf. Photoelectric emission requires a photon with enough energy for an electron to escape. Increasing intensity at fixed frequency changes photon rate, not each photon’s energy.

Atomic transitions absorb or emit photons with particular energies because energy levels are discrete. Matter also exhibits wave behaviour: de Broglie wavelength relates wavelength to momentum. Neither classical particles nor classical waves alone explain every observation.

1. Photons carry discrete exchanges

A photon has energy hf and momentum E/c in vacuum. Increasing monochromatic beam intensity increases photon arrival rate rather than each photon’s energy. Frequency changes the energy of each photon. The electronvolt is an energy unit: one elementary charge moved through one volt changes energy by about 1.602 × 10⁻¹⁹ J. It is not a unit of voltage or charge.

2. Threshold and stopping potential answer different questions

For the simple photoelectric model, a photon must supply at least the work function to release an electron. Above threshold, maximum kinetic energy is hf − φ and is related to stopping potential by Kmax = eVs. Greater intensity can increase emitted-electron rate but does not increase maximum energy at fixed frequency. An emitted population can have a range of energies, so maximum kinetic energy is not the energy of every electron.

3. Atomic levels explain spectral lines

An atom can absorb a photon when its energy matches a permitted energy-level difference. An excited atom can emit photons as it moves to lower levels. Each photon satisfies hf = ΔE, so larger differences correspond to higher frequency and shorter wavelength. A line spectrum reveals discrete transitions rather than proving electrons orbit like tiny planets on visible paths. Different simplified atomic models explain different observations with different success.

4. Matter also shows wave behaviour

The de Broglie relation λ = h/p connects wavelength with momentum for particles. Increasing momentum shortens wavelength, explaining why diffraction can depend on accelerating voltage for electrons. Probability patterns emerge from many detections even though individual detections are localised. A wavefunction is not a visible material wave spread through an ordinary medium, and the classical particle/wave pictures each have limits.

A method you can reuse

  1. Identify whether the relevant quantity is per photon or per second.
  2. Convert electronvolts to joules when combining with SI constants.
  3. Apply energy conservation with the work function or level difference.
  4. Use the maximum-energy condition for stopping potential.
  5. For matter waves, calculate momentum using an appropriate non-relativistic or stated model.
Why does brighter light below threshold not emit photoelectrons in the simple one-photon model?

Each photon still lacks the energy needed to overcome the work function. More such photons increase their rate, not their individual energy, under the model’s stated ordinary-intensity conditions.

Equations and relationships

E=hf=hcλE=hf=\frac{hc}{\lambda}

Photon energy.

Kmax⁡=hf−ϕK_{\max}=hf-\phi

Photoelectric equation above threshold; φ is work function.

λ=hp\lambda=\frac hp

De Broglie wavelength.

pphoton=Ecp_{\text{photon}}=\frac Ec

Photon momentum magnitude.

1 eV=1.602×10−19 J1\,\mathrm{eV}=1.602\times10^{-19}\,\mathrm J

Energy gained by an elementary charge through 1 V.

Worked example

A photon has energy 6.0 × 10⁻¹⁹ J and a surface has work function 2.0 × 10⁻¹⁹ J. What is the maximum photoelectron kinetic energy?

Follow the solution

Kmax = 6.0 × 10⁻¹⁹ − 2.0 × 10⁻¹⁹ = 4.0 × 10⁻¹⁹ J.

Investigate and explain

Compare simulated atomic energy-level differences with emitted wavelengths. Do not operate high-voltage discharge equipment independently.

Compare atomic transitions, detection patterns and thermal spectra as distinct quantum ideas. These models are related, not interchangeable; a blackbody curve alone is not a direct photoelectric experiment.

Open the interactive investigation

Check your understanding

1. What determines the energy of one photon?

  1. The beam intensity alone
  2. How many other photons are nearby
  3. Its frequency
Check the answer and reasoning

Its frequency

E = hf applies to each photon.

2. A photon has energy 6.0 × 10⁻¹⁹ J and a surface has work function 2.0 × 10⁻¹⁹ J. What is the maximum photoelectron kinetic energy?

  1. 4.0 × 10⁻¹⁹ J
  2. 8.0 × 10⁻¹⁹ J
  3. 3.0 × 10⁻¹⁹ J
Check the answer and reasoning

4.0 × 10⁻¹⁹ J

Kmax = 6.0 × 10⁻¹⁹ − 2.0 × 10⁻¹⁹ = 4.0 × 10⁻¹⁹ J.

3. How does de Broglie wavelength change when momentum doubles?

  1. It doubles.
  2. It halves.
  3. It quadruples.
Check the answer and reasoning

It halves.

λ = h/p is inversely proportional to momentum.