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

This guide covers 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. The specification says you also need knowledge of section 4.5 (photons) for the energy equations, and of 4.4.1 (wave motion, including intensity as power per unit area) for Stefan’s law.

Use the OCR A Level Physics hub and the printable checklist to track progress. Pair this page with the revision notes and the practice questions. The photon energy ideas are built up in the photons study guide. Test yourself with the free diagnostics.

What you need to be able to do

Outcome	You must be able to
5.5.2(a), (b)	Describe electron energy levels in isolated gas atoms and explain why the values are negative
5.5.2(c), (d)	Explain emission lines as photons from transitions; use hf = ΔE and hc/λ = ΔE
5.5.2(e)	Explain why different atoms give different lines, and use this to identify elements in stars
5.5.2(f)	Distinguish continuous, emission line and absorption line spectra
5.5.2(g), (h)	Use a transmission diffraction grating and d sin θ = nλ to find a wavelength
5.5.2(i)	Use Wien’s displacement law, λmax proportional to 1/T, to estimate a surface temperature
5.5.2(j)	Use luminosity and Stefan’s law, L = 4πr²σT⁴
5.5.2(k)	Combine Wien’s law and Stefan’s law to estimate a stellar radius

This guide uses 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 and a Wien constant of 2.90 × 10⁻³ m K.

Energy levels in isolated atoms (5.5.2(a), (b))

In a gas at low pressure the atoms are far apart, so each atom behaves as an isolated system. An electron in an isolated atom can only have certain discrete energies. These are the energy levels of the atom. The electron cannot have a value in between them.

Energy levels are given negative values. The zero of energy is defined as an electron that has just escaped the atom (it is free, with no kinetic energy). An electron that is bound to the nucleus has less energy than a free electron, so its energy is below zero. The lowest level is the ground state and is the most negative. The closer a level is to zero, the less tightly bound the electron is.

Example set of levels for a hypothetical atom, in electron volts: −1.2, −2.7, −5.1 and −9.4. The ground state is −9.4 eV. Removing an electron from the ground state needs 9.4 eV, the energy that takes it from −9.4 eV to 0.

Emission lines and photon energy (5.5.2(c), (d))

A hot gas has atoms that have been excited: their electrons are in higher levels. An electron falls to a lower level and one photon is emitted. The photon carries away the whole energy difference:

hf = ΔE, and since c = fλ, hc/λ = ΔE

where ΔE is the higher energy minus the lower energy, so it is positive. Convert from eV to J by multiplying by 1.60 × 10⁻¹⁹.

Each possible drop gives one wavelength, so a gas shows a set of separate bright lines. A bigger energy gap gives a higher frequency and a shorter wavelength.

Worked example 1: a photon from a transition

An electron in the hypothetical atom falls from −2.7 eV to −9.4 eV. Find the wavelength of the photon.

ΔE = −2.7 − (−9.4) = 6.7 eV ΔE = 6.7 × 1.60 × 10⁻¹⁹ = 1.07 × 10⁻¹⁸ J λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 1.07 × 10⁻¹⁸ = 1.86 × 10⁻⁷ m

The wavelength is about 186 nm, which is in the ultraviolet. The frequency is f = ΔE/h = 1.62 × 10¹⁵ Hz.

Worked example 2: counting lines

With four levels, an electron can end in any lower level, so the number of different drops is 3 + 2 + 1 = 6. The longest wavelength comes from the smallest gap: −1.2 eV to −2.7 eV, which is 1.5 eV, giving λ = 829 nm (infrared). Some lines may lie outside the visible region.

Identifying elements (5.5.2(e))

The energy levels depend on the nucleus and on the number and arrangement of electrons, so the set of levels is different for each element. Different elements therefore have different sets of allowed transitions and different line wavelengths. The pattern of lines works as a fingerprint.

To identify an element in a star, measure the wavelengths of lines in the star’s spectrum and compare them with the lines measured for known gases in the laboratory. If every line of an element is found at the right wavelengths, that element is present in the star’s outer layers. One wavelength on its own is weak evidence, because it may match more than one element; a full set is strong evidence.

Types of spectrum (5.5.2(f))

Spectrum	What you see	Produced by
Continuous	All wavelengths in a range, no gaps	A hot, dense source such as a filament lamp or the hot interior of a star
Emission line	Bright lines on a dark background	A hot, low pressure gas whose excited atoms emit photons
Absorption line	Dark lines across a continuous spectrum	Continuous light passing through a cooler gas

Absorption works in reverse to emission. A photon is absorbed only if its energy equals the gap between two levels in the atom, so those wavelengths are removed from the beam. The atoms re-emit photons, but in all directions, so the intensity in the original direction is reduced and a dark line appears. The dark lines of an element fall at the same wavelengths as its bright emission lines.

Stars show absorption spectra. The hot interior supplies a continuous spectrum and the cooler gas in the star’s atmosphere absorbs the wavelengths that match its atoms.

The transmission diffraction grating (5.5.2(g), (h))

A transmission diffraction grating is a plate with many equally spaced parallel slits. Light passing through the slits spreads out and the waves interfere. Bright maxima form where light from neighbouring slits arrives in phase, which means the path difference is a whole number of wavelengths.

For light that strikes the grating normally, the condition for maxima is

d sin θ = nλ

where d is the grating spacing (distance between neighbouring slits), θ is the angle between the straight-through direction and the maximum, n is the order (0, 1, 2 and so on) and λ is the wavelength. Proof is not required, and neither is the structure of an optical spectrometer. The practical link is PAG5, so be ready to use a grating to measure a wavelength.

The grating spacing is the reciprocal of the number of lines per unit length. A grating with 600 lines per mm has d = 1.00 × 10⁻³ / 600 = 1.67 × 10⁻⁶ m.

The largest possible order comes from sin θ ≤ 1, so n ≤ d/λ, and n must be a whole number. Maxima occur on both sides of the centre, so the total number is 2n + 1 (including the central maximum).

Worked example 3: angle and number of orders

Green light of wavelength 540 nm strikes a grating with 600 lines per mm. Find the angle of the first order and the number of maxima.

d = 1.67 × 10⁻⁶ m sin θ = λ/d = 540 × 10⁻⁹ / 1.67 × 10⁻⁶ = 0.324, so θ = 18.9° n ≤ d/λ = 3.09, so the highest order is 3 Total maxima = 2 × 3 + 1 = 7

Worked example 4: finding an unknown wavelength

Light falls on a grating with 300 lines per mm. A second order maximum is at 24.0° to the straight-through direction. Find the wavelength.

d = 1.00 × 10⁻³ / 300 = 3.33 × 10⁻⁶ m λ = d sin θ / n = 3.33 × 10⁻⁶ × sin 24.0° / 2 = 6.78 × 10⁻⁷ m

The wavelength is 678 nm. Measuring the angle on both sides of the centre and halving the difference reduces the error from the zero position.

Wien’s displacement law (5.5.2(i))

A hot object emits a spread of wavelengths. The wavelength where the emitted intensity is greatest is λmax. Wien’s displacement law says that λmax is inversely proportional to the thermodynamic temperature T of the surface:

λmax ∝ 1/T, or λmax T = constant

The constant is 2.90 × 10⁻³ m K. A hotter star has a shorter peak wavelength, which is why hot stars look blue-white and cool stars look red. Temperature must be in kelvin.

Worked example 5: surface temperature

A star’s spectrum peaks at 620 nm. Estimate its surface temperature.

T = (2.90 × 10⁻³) / (620 × 10⁻⁹) = 4.68 × 10³ K

The result is an estimate, because the peak of a real spectrum is broad.

You can also use the proportionality alone. If star A has a peak wavelength one third of star B’s, star A is three times hotter in kelvin.

Luminosity and Stefan’s law (5.5.2(j))

The luminosity L of a star is the total power it radiates, in watts, over all wavelengths. Stefan’s law states that for a star of radius r and surface temperature T,

L = 4πr²σT⁴

where σ is the Stefan constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴. The factor 4πr² is the surface area of a sphere. Each square metre of surface radiates σT⁴ watts, an intensity at the surface of power divided by area.

The fourth power matters. Doubling T at fixed radius multiplies L by 2⁴ = 16. Doubling r at fixed T multiplies L by 4.

Estimating stellar radius (5.5.2(k))

Wien’s law gives T from the observed spectrum. If you also know L, rearrange Stefan’s law:

r = √( L / (4πσT⁴) )

Worked example 6: radius of a star

A star has a spectrum that peaks at 620 nm and a luminosity of 1.9 × 10²⁷ W. Estimate its radius, and compare it with the Sun’s radius of 6.96 × 10⁸ m.

Step 1: T = 2.90 × 10⁻³ / 620 × 10⁻⁹ = 4680 K (3 s.f.) Step 2: T⁴ = 4.79 × 10¹⁴ K⁴ (using the unrounded T), so σT⁴ = 2.71 × 10⁷ W m⁻² and 4πσT⁴ = 3.41 × 10⁸ W m⁻² Step 3: r² = L / (4πσT⁴) = 1.9 × 10²⁷ / 3.41 × 10⁸ = 5.57 × 10¹⁸ m² Step 4: r = 2.36 × 10⁹ m

Compared with the Sun this is 2.36 × 10⁹ / 6.96 × 10⁸ = 3.4 times larger.

Ratio method

If a reference star has luminosity L₀, temperature T₀ and radius r₀, dividing the two Stefan’s law equations gives

r/r₀ = √(L/L₀) × (T₀/T)²

This avoids the constant σ and often reduces the arithmetic. For the star above, taking the Sun as reference (L₀ = 3.83 × 10²⁶ W, T₀ = 5780 K), L/L₀ = 4.96 and the ratio is about 3.4, matching the result above.

Common errors

• Writing energy levels as positive numbers, or giving ΔE as negative. The photon energy is the higher level minus the lower, which is positive.
• Forgetting to convert eV to J before using h and c.
• Using the number of lines per metre as d. The spacing d is the reciprocal.
• Counting only the maxima on one side of the centre when asked for the total.
• Using a temperature in degrees Celsius in Wien’s law or Stefan’s law.
• Taking the fourth root or the square root in the wrong place when rearranging for r or T.
• Saying absorbed photons disappear. They are re-emitted in random directions.

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.