A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Line Spectra and Energy Level Transitions in Atoms — mark scheme explained
The short answer
Line spectra, such as those observed for atomic hydrogen, provide strong evidence for the quantized nature of energy levels within atoms. This section delves into how these line spectra arise from transitions between discrete energy levels and how they can be observed using a diffraction grating.
The question
Calculate the wavelength of light emitted when an electron in a hydrogen atom transitions from the n = 3 level to the n = 2 level. [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. Identify the energy levels involved: n 1 = 2 and n 2 = 3.
- S2
2. Use the Rydberg formula to find the wavelength: 1/λ = R(1/n 1 2 - 1/n 2 2 ).
- S3
3. Substitute the values: 1/λ = 1.097 × 10 7 m -1 (1/2 2 - 1/3 2 ).
- S4
4. Simplify the expression: 1/λ = 1.097 × 10 7 m -1 (1/4 - 1/9).
- S5
5. Calculate the difference: 1/λ = 1.097 × 10 7 m -1 (0.25 - 0.1111) ≈ 1.097 × 10 7 m -1 × 0.1389.
- S6
6. Find the wavelength: λ = 1 / (1.097 × 10 7 m -1 × 0.1389) ≈ 6.56 × 10 -7 m or 656 nm.
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
1. Identify the energy levels involved: n 1 = 2 and n 2 = 3.
- S2
2. Use the Rydberg formula to find the wavelength: 1/λ = R(1/n 1 2 - 1/n 2 2 ).
- S3
3. Substitute the values: 1/λ = 1.097 × 10 7 m -1 (1/2 2 - 1/3 2 ).
- S4
4. Simplify the expression: 1/λ = 1.097 × 10 7 m -1 (1/4 - 1/9).
- S5
5. Calculate the difference: 1/λ = 1.097 × 10 7 m -1 (0.25 - 0.1111) ≈ 1.097 × 10 7 m -1 × 0.1389.
- S6
6. Find the wavelength: λ = 1 / (1.097 × 10 7 m -1 × 0.1389) ≈ 6.56 × 10 -7 m or 656 nm.
Final answer: 656 nm
Common mistakes
- Confusing the energy level equation with the Rydberg formula. — Review the specific equations: E n = -13.6 eV / n 2 for energy levels and 1/λ = R(1/n 1 2 - 1/n 2 2 ) for wavelengths.
- Using the wrong units for energy levels and photon energies. — Always check the units required in the question and use the conversion factor 1 eV = 1.602 × 10 -19 J when necessary.
- Forgetting to include the negative sign in energy level calculations. — Always include the negative sign when using the energy level equation E n = -13.6 eV / n 2 .
- Incorrectly identifying the higher and lower energy levels in transitions. — Always identify the higher (n 2 ) and lower (n 1 ) energy levels clearly before calculating the energy difference.
- Using the wrong value for Planck's constant or the Rydberg constant. — Memorize and double-check the values of important constants: h = 6.626 × 10 -34 J·s and R = 1.097 × 10 7 m -1 .
- Misinterpreting the diffraction grating equation. — Review the grating equation d sin(θ) = mλ and ensure you correctly identify d (grating spacing), θ (angle of diffraction), m (order), and λ (wavelength).
Where the marks go
- Full worked solution (all marking points)4 marks