A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Ultraviolet Catastrophe and Black-Body Radiation — mark scheme explained
The short answer
The ultraviolet catastrophe and black-body radiation are pivotal concepts in the development of quantum mechanics, marking a significant departure from classical physics. This section explores these phenomena, Planck's interpretation in terms of quanta, the failure of classical wave theory to explain photoelectricity, and Einstein’s revolutionary explanation.
The question
A metal surface has a work function of 4.2 eV. Calculate the minimum frequency of light required to eject electrons from this metal. [Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
What each mark is really for — in plain English — and the wording trap that loses it.
- S1
1. Convert the work function (φ) from electron volts to joules: φ = 4.2 eV × 1.602 × 10 -19 J/eV ≈ 6.73 × 10 -19 J.
- S2
2. Use the equation E = hν to find the minimum frequency (ν): ν = φ / h, where h is Planck's constant (6.626 × 10 -34 J·s).
- S3
3. Substitute the values: ν ≈ (6.73 × 10 -19 J) / (6.626 × 10 -34 J·s) ≈ 1.02 × 10 15 Hz.
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
1. Convert the work function (φ) from electron volts to joules: φ = 4.2 eV × 1.602 × 10 -19 J/eV ≈ 6.73 × 10 -19 J.
- S2
2. Use the equation E = hν to find the minimum frequency (ν): ν = φ / h, where h is Planck's constant (6.626 × 10 -34 J·s).
- S3
3. Substitute the values: ν ≈ (6.73 × 10 -19 J) / (6.626 × 10 -34 J·s) ≈ 1.02 × 10 15 Hz.
Final answer: The minimum frequency of light required to eject electrons from the metal is approximately 1.02 × 10 15 Hz.
Common mistakes
- Confusing the work function with the kinetic energy of ejected electrons. — Always remember that the kinetic energy of ejected electrons is given by K = hν - φ, where hν is the energy of the incident photon and φ is the work function.
- Using classical wave theory to explain the photoelectric effect. — Understand and use Einstein’s quantum explanation: the energy of a photon (E = hν) must exceed the work function to eject an electron, and any excess energy becomes the kinetic energy of the electron.
- Forgetting to convert units between joules and electron volts. — Always check the units of the given values and ensure they are consistent before performing calculations. Use the conversion factor 1 eV = 1.602 × 10 -19 J when necessary.
- Misapplying Wien's displacement law to calculate intensity instead of peak wavelength. — Use Wien's displacement law to find the peak wavelength of black-body radiation and other formulas for intensity calculations.
- Confusing Planck’s constant with Boltzmann’s constant. — Memorize the definitions and symbols for each constant: h = 6.626 × 10 -34 J·s and k B = 1.38 × 10 -23 J/K.
- Forgetting that the ultraviolet catastrophe is a classical prediction, not an experimental observation. — Understand that the ultraviolet catastrophe refers to the incorrect prediction by classical wave theory that energy density would diverge at short wavelengths, which contradicts experimental observations.
Where the marks go
- Full worked solution (all marking points)4 marks