OCR A Level Physics A: Stellar spectra, Wien's law, Stefan's law and stellar radius (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 5.5.2 (Electromagnetic radiation from stars), outcomes (a) to (k), of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. Use these constants unless a question gives others: h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, Wien constant 2.90 × 10⁻³ m K, radius of the Sun 6.96 × 10⁸ m.

Links: OCR A Level Physics hub, printable checklist, study guide, revision notes and the free diagnostics.

Questions

1. Explain why the energy levels of an electron in an isolated atom are given negative values. [2]

2. Two energy levels of a gas atom are −8.2 eV and −3.6 eV. Calculate the wavelength of the photon emitted when an electron falls from the higher level to the lower level. [3]

3. A photon emitted in a transition has energy 2.9 × 10⁻¹⁹ J. Calculate its frequency and its wavelength. [3]

4. The spectrum of a star is a bright continuous band crossed by dark lines. Explain how the dark lines form and how they can be used to identify an element in the star. [4]

5. An atom has four energy levels: −7.9 eV, −4.3 eV, −2.2 eV and −0.9 eV. Determine the number of different emission lines the gas could produce, and calculate the shortest wavelength emitted. [3]

6. Light of wavelength 610 nm is incident normally on a transmission grating marked 450 lines per mm. Calculate the angles of the first and second order maxima. [4]

7. A grating has 250 lines per mm and is used with light of wavelength 450 nm. Determine the total number of maxima that can be observed. [3]

8. Star A has a peak wavelength of 355 nm and star B has a peak wavelength of 1450 nm.

(a) Calculate the surface temperature of each star. [2] (b) Calculate the ratio of temperature of A to temperature of B. [1] (c) State, with a reason, which star appears bluer. [1]

9. A star has a radius of 2.1 × 10⁹ m and a surface temperature of 9200 K. Calculate its luminosity. [3]

10. The spectrum of a distant star peaks at a wavelength of 910 nm. The luminosity of the star is 3.0 × 10²⁹ W. Estimate the radius of the star, and state how many times larger it is than the Sun. [6]

11. The first order spectrum of a star is viewed through a grating with 600 lines per mm. Dark lines are seen at 23.2°, 18.1° and 17.0° to the straight-through direction. The table gives laboratory emission lines (supplied as data for this question).

Element	Wavelengths / nm
Hydrogen	656, 486
Sodium	589
Magnesium	518
Calcium	423

(a) Calculate the wavelength of each dark line. [3] (b) State which elements the star’s atmosphere contains, and what the angles tell you about sodium. [3] (c) The continuous spectrum of the star peaks at 450 nm. (i) Calculate the surface temperature. [2] (c) (ii) The star has luminosity 5.0 × 10²⁹ W. Calculate its radius. [2]

Answers

1. Zero energy is defined as the energy of an electron that has just escaped the atom, with no kinetic energy [1]. An electron held in the atom has lower energy than a free one, so its value is below zero [1]. [2] Examiner insight: Answers that only say “because the electron is attracted” explain the force but not why the zero is chosen where it is; both ideas are needed.

2. ΔE = −3.6 − (−8.2) = 4.6 eV [1]. ΔE = 4.6 × 1.60 × 10⁻¹⁹ = 7.36 × 10⁻¹⁹ J, and λ = hc/ΔE = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 7.36 × 10⁻¹⁹ [1]. λ = 2.70 × 10⁻⁷ m, which is 270 nm [1]. [3] Examiner insight: A common route to losing credit is putting 4.6 eV straight into hc/λ; the conversion to joules is part of the method, so show it.

3. f = ΔE/h = 2.9 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ [1]. f = 4.4 × 10¹⁴ Hz (4.37 × 10¹⁴ Hz unrounded) [1]. λ = c/f = 3.00 × 10⁸ / 4.37 × 10¹⁴ = 6.9 × 10⁻⁷ m, which is 690 nm [1]. [3] Examiner insight: Using the unrounded frequency in the second step avoids a rounding drift, and a clear unit on each answer helps your working to be followed.

4. The hot interior emits a continuous spectrum [1]. Atoms in the cooler outer gas absorb photons whose energy equals the gap between two of their energy levels [1]. The photons are re-emitted in all directions, so the intensity at those wavelengths in the beam is reduced and dark lines appear [1]. The wavelengths of the dark lines are compared with the laboratory line spectra of known elements; a match to the whole set of lines of an element identifies it [1]. [4] Examiner insight: Strong answers say “re-emitted in random directions” and link wavelength to a specific energy gap; “the photons are used up” is a frequent loose statement.

5. The number of drops is 3 + 2 + 1 = 6 lines [1]. The shortest wavelength is from the largest gap, −0.9 eV to −7.9 eV, which is 7.0 eV [1]. λ = hc/ΔE = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / (7.0 × 1.60 × 10⁻¹⁹) = 1.78 × 10⁻⁷ m, which is 178 nm [1]. [3] Examiner insight: Naming which two levels give the shortest wavelength shows the reasoning; stating only a final number leaves nothing to credit if it is wrong.

6. d = 1.00 × 10⁻³ / 450 = 2.22 × 10⁻⁶ m [1]. First order: sin θ = 610 × 10⁻⁹ / 2.22 × 10⁻⁶ = 0.2745, so θ = 15.9° [1]. Second order: sin θ = 2 × 0.2745 = 0.549 [1], so θ = 33.3° [1]. [4] Examiner insight: Take sin⁻¹ of the whole value 2λ/d. Doubling the first order angle (31.9°) is a typical slip and gives the wrong answer.

7. d = 1.00 × 10⁻³ / 250 = 4.00 × 10⁻⁶ m [1]. n must satisfy n ≤ d/λ = 4.00 × 10⁻⁶ / 450 × 10⁻⁹ = 8.89, so the highest order is 8 [1]. Maxima on both sides and the centre: 2 × 8 + 1 = 17 [1]. [3] Examiner insight: Round the highest order down, never up, because an order above d/λ would need sin θ greater than 1. The count of 16 omits the central maximum.

8. (a) T(A) = 2.90 × 10⁻³ / 355 × 10⁻⁹ = 8170 K [1]. T(B) = 2.90 × 10⁻³ / 1450 × 10⁻⁹ = 2000 K [1]. (b) T(A)/T(B) = 8170/2000 = 4.08 (or 1450/355 = 4.08) [1]. (c) Star A, because it has the shorter peak wavelength, so it is hotter and its peak lies nearer the blue/ultraviolet end [1]. [4] Examiner insight: Part (c) needs the reason as well as the star; saying “A is hotter” without the link to peak wavelength is only half of the argument.

9. Surface area = 4πr² = 4π × (2.1 × 10⁹)² = 5.54 × 10¹⁹ m² [1]. σT⁴ = 5.67 × 10⁻⁸ × (9200)⁴ = 4.06 × 10⁸ W m⁻² [1]. L = 5.54 × 10¹⁹ × 4.06 × 10⁸ = 2.25 × 10²⁸ W [1]. [3] Examiner insight: Raise the temperature to the fourth power before multiplying, and write the unit W. Squaring T instead of raising it to the fourth is the usual error.

10. T = 2.90 × 10⁻³ / λmax [1]. T = 2.90 × 10⁻³ / 910 × 10⁻⁹ = 3190 K (keep 3186.8 K unrounded) [1]. r² = L / (4πσT⁴) [1]. 4πσT⁴ = 4π × 5.67 × 10⁻⁸ × (3186.8)⁴ = 7.35 × 10⁷ W m⁻² [1]. r² = 3.0 × 10²⁹ / 7.35 × 10⁷ = 4.08 × 10²¹, so r = 6.4 × 10¹⁰ m [1]. r/r(Sun) = 6.39 × 10¹⁰ / 6.96 × 10⁸ = about 92 times the Sun’s radius [1]. [6] Examiner insight: Show the rearrangement before the numbers and keep the square root to the last line. Rounding T early to 3200 K gives a visibly different radius.

11. (a) d = 1.00 × 10⁻³ / 600 = 1.67 × 10⁻⁶ m [1]. λ = d sin θ with n = 1 [1]. 23.2° gives 657 nm, 18.1° gives 518 nm and 17.0° gives 487 nm [1]. (b) 657 nm and 487 nm match the two hydrogen lines (656 nm and 486 nm) within the precision of the angles, so hydrogen is present [1]. 518 nm matches magnesium [1]. A sodium line at 589 nm would appear at about 20.7°, and there is no dark line there, so there is no evidence for sodium [1]. (c)(i) T = 2.90 × 10⁻³ / 450 × 10⁻⁹ [1] = 6440 K [1]. (c)(ii) r² = L / (4πσT⁴) = 5.0 × 10²⁹ / (4π × 5.67 × 10⁻⁸ × 6444⁴) = 5.0 × 10²⁹ / 1.23 × 10⁹ [1]. r² = 4.07 × 10²⁰, so r = 2.0 × 10¹⁰ m [1]. [10] Examiner insight: Match all lines of an element, not just one, and give the angle you would expect for any element you rule out. In part (c) carry the unrounded temperature into the fourth power.

Where marks are usually lost

• Giving ΔE a negative sign or leaving the energy in eV when using h and c.
• Using d as lines per metre rather than its reciprocal.
• Doubling the angle for the second order instead of taking the inverse sine of nλ/d.
• Rounding the highest order up, or leaving out the central maximum or the other side in the total.
• Writing a temperature in degrees Celsius in either law.
• Rounding T before raising it to the fourth power.
• Putting r in place of r² when rearranging Stefan’s law, or forgetting the final square root.
• Saying absorbed photons “disappear” instead of being re-emitted in random directions.
• Stating an element from one matching line without checking the others.

Next steps

• Re-read the revision notes and the study guide.
• Visit the OCR A Level Physics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

This page follows section 5.5.2 of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR.