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Inquiry Question 2: What is observed when light interacts with matter?

Investigate emission and absorption spectra, distinguish continuous, line emission and line absorption spectra, and analyse stellar spectra to identify chemical composition, surface temperature and motion

A focused answer to the HSC Physics Module 7 dot point on spectroscopy. Continuous, emission-line and absorption-line spectra explained by quantised atomic energy levels, plus how stellar spectra reveal chemical composition, surface temperature, rotation and radial velocity (Doppler shift).

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context

What this dot point is asking

NESA wants you to know the three types of spectrum, why atomic energy levels are quantised, and how spectroscopy is used to determine the composition, temperature and motion of stars. You should be able to read an absorption line in a stellar spectrum and explain what it tells you.

The answer

Quantised atomic energy levels

Electrons in atoms occupy discrete energy levels E1,E2,E3,E_1, E_2, E_3, \dots When an electron drops from a higher level EiE_i to a lower level EfE_f, a photon of energy:

hf=EiEfh f = E_i - E_f

is emitted. The reverse is absorption: an electron absorbs a photon of exactly the right energy and jumps to a higher level. Because the levels are discrete, only certain photon energies (and therefore wavelengths) appear in atomic spectra. Each element has a characteristic set of levels and therefore a unique spectral "fingerprint".

Continuous, line emission and line absorption spectra Three stacked rectangles representing spectra. Top: continuous spectrum, a smooth gradient from violet to red. Middle: line emission spectrum, a dark band with bright vertical lines at specific wavelengths. Bottom: line absorption spectrum, a continuous band crossed by a few dark vertical lines at the same wavelengths. Three types of spectrum Continuous Line emission Line absorption violet ← wavelength → red hf = Ei − Ef

Three types of spectrum (Kirchhoff's laws, 1859)

Continuous spectrum
A hot dense object (a glowing solid, liquid or high-pressure gas, or the interior of a star) emits a smooth distribution of wavelengths. The peak wavelength shifts with temperature (Wien's law); the total intensity follows the Stefan-Boltzmann law. The spectrum approximates a blackbody curve.
Line emission spectrum
A hot, low-density gas (a discharge lamp, the corona of a star, a nebula) emits only at specific wavelengths corresponding to its atoms' allowed downward transitions. The spectrum looks like bright lines on a dark background.
Line absorption spectrum
Continuum light passing through a cool gas loses the photons whose energies match the gas atoms' allowed upward transitions. The result is a continuous spectrum crossed by dark lines (Fraunhofer lines). Stellar spectra are predominantly of this type: the photosphere produces near-continuum light that is absorbed by the cooler outer atmosphere.

What stellar spectra reveal

A typical stellar spectrum is analysed for four things:

Chemical composition
Identify the absorption lines by wavelength and match to laboratory spectra. The most prominent lines in a Sun-like star are hydrogen Balmer lines (H-alpha at 656.3 nm656.3\ \text{nm}), neutral sodium, ionised calcium, magnesium and iron lines. Helium was discovered in 1868 from a solar absorption line that did not match any known terrestrial element.
Surface temperature
The relative strengths of different lines depend on temperature, because each transition has an optimal temperature for being populated. The shape of the underlying continuum (Wien's law, below) gives an independent temperature estimate. Together these classify stars into the spectral sequence O, B, A, F, G, K, M, from hottest (blue-white) to coolest (red).
Radial velocity (line of sight)
All lines are shifted from their laboratory wavelengths by the Doppler effect:

Δλλ0=vc(for vc)\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} \quad \text{(for } v \ll c\text{)}

A redshift (λobs>λ0\lambda_{\text{obs}} > \lambda_0) means the source is receding; a blueshift (λobs<λ0\lambda_{\text{obs}} < \lambda_0) means it is approaching. This is how we know about the expansion of the universe (Hubble) and detect orbiting exoplanets (the star wobbles).

Rotation. A rotating star has one limb moving toward us and the other away, so each spectral line is broadened symmetrically into a profile whose width measures the equatorial rotation speed.

Other inferences include surface gravity (from line widths sensitive to pressure broadening), magnetic field (Zeeman splitting of lines) and turbulent motion.

Wien's law and the blackbody curve

A hot object's continuous-spectrum intensity, plotted against wavelength, is a blackbody curve: it rises from zero, reaches a single peak, then falls away toward longer wavelengths. Wien's displacement law relates the wavelength of that peak, λmax\lambda_{\max}, to the object's absolute temperature TT:

λmax=bT,b=2.898×103 m K\lambda_{\max} = \frac{b}{T}, \qquad b = 2.898 \times 10^{-3}\ \text{m K}

A hotter object has a shorter peak wavelength (its curve shifts toward blue/violet) and, by the Stefan-Boltzmann law, radiates far more total power per unit area, so its curve is also taller and encloses a larger area. Applying Wien's law to a star's continuum gives a temperature estimate that is independent of, and should agree with, the temperature estimated from spectral line strengths.

Blackbody intensity versus wavelength for a 6000 K and a 4000 K star Two bell shaped intensity versus wavelength curves. The 6000 kelvin curve peaks higher and further left near 480 nanometres. The 4000 kelvin curve peaks lower and further right near 725 nanometres. Data points mark each curve including its peak, illustrating Wien's law: peak wavelength times temperature equals a constant, so the hotter curve peaks at a shorter wavelength and reaches a greater intensity. wavelength λ (nm) intensity 250480725 9601200 6000 K peak λmax ≈ 480 nm 4000 K peak λmax ≈ 725 nm Hotter star: shorter λmax, greater peak intensity (Wien's law).

Worked example: identifying composition

A stellar absorption spectrum shows strong lines at 588.99 nm588.99\ \text{nm} and 589.59 nm589.59\ \text{nm}. These match the laboratory sodium D doublet, so the star's atmosphere contains sodium. Comparing the doublet positions to laboratory values gives the radial velocity by the Doppler formula above.

Diffraction grating spectrometers

In practice, spectra are recorded by sending starlight through a slit, collimating it, dispersing it with a prism or diffraction grating, and imaging the result onto a CCD. A diffraction grating with line spacing dd produces principal maxima at:

dsinθ=mλd \sin \theta = m \lambda

so different wavelengths emerge at different angles and can be measured precisely. Gratings give much higher resolution than prisms and are standard in modern astrophysics.

Examples in context

Example 1. Mt Stromlo spectroscopy of a Cepheid variable. A spectrum taken at Mt Stromlo Observatory shows the calcium-K line at λobs=397.2 nm\lambda_{\text{obs}} = 397.2 \text{ nm} instead of the laboratory rest value λ0=393.4 nm\lambda_0 = 393.4 \text{ nm}. The redshift is z=Δλ/λ0=(397.2393.4)/393.4=9.66×103z = \Delta\lambda / \lambda_0 = (397.2 - 393.4)/393.4 = 9.66 \times 10^{-3}. For non-relativistic recession, v=zc=9.66×103×3.0×108=2.90×106 m/s=2900 km/sv = z c = 9.66 \times 10^{-3} \times 3.0 \times 10^8 = 2.90 \times 10^6 \text{ m/s} = 2900 \text{ km/s} away from us. Combined with Cepheid period-luminosity distance (d=40 Mpcd = 40 \text{ Mpc}), Hubble's constant H0=v/d=2900/40=72.5 km/s/MpcH_0 = v / d = 2900 / 40 = 72.5 \text{ km/s/Mpc}, matching modern values.

Example 2. Helium discovery in the solar spectrum, replicated at Sydney Observatory. During the 1868 solar eclipse, astronomers including those at Sydney Observatory recorded a yellow emission line at λ=587.6 nm\lambda = 587.6 \text{ nm} in the Sun's chromosphere that did not match any terrestrial element. Photon energy E=hc/λ=1240/587.6=2.11 eVE = h c / \lambda = 1240 / 587.6 = 2.11 \text{ eV}. This was the He I transition between the 1s2p1s2p and 1s2s1s2s levels in helium, an element not isolated on Earth until 1895. The discovery proved that astronomical spectra reveal compositions of objects we cannot sample, the cornerstone of all astrochemistry.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC4 marksDistinguish between continuous, line emission and line absorption spectra, and explain how an absorption line in the spectrum of a star can identify an element in the star's atmosphere.
Show worked answer →

Continuous spectrum: a smooth rainbow of all wavelengths, produced by a hot dense source (an incandescent solid, liquid or compressed gas) where the energy levels are smeared together by close packing.

Line emission spectrum: a series of bright lines on a dark background at specific wavelengths, produced by a hot, low-pressure gas. The lines correspond to photons released as electrons drop from higher to lower discrete energy levels in the atoms, Ephoton=EiEf=hfE_{\text{photon}} = E_i - E_f = h f.

Line absorption spectrum: a continuous spectrum crossed by dark lines at specific wavelengths, produced when continuum light passes through a cool gas. Electrons in the cool gas absorb only those photons whose energies exactly match the gas atoms' allowed transitions, removing those wavelengths from the transmitted beam.

A star produces continuum light from its hot dense interior, which passes through the cooler stellar atmosphere. Atoms in the atmosphere absorb the wavelengths matching their characteristic transitions, leaving dark lines in the stellar spectrum. Each element has a unique fingerprint, so matching the observed absorption-line wavelengths to laboratory spectra identifies the elements present (this is how helium was discovered in the Sun in 1868, before being found on Earth).

Markers reward all three spectrum types correctly described with examples, plus the absorption-by-atmosphere mechanism and the fingerprint-matching idea.

2019 HSC3 marksA hydrogen absorption line that has a laboratory wavelength of 656.3 nm is observed in a distant galaxy at 689.1 nm. Calculate the radial velocity of the galaxy and state whether it is moving toward or away from Earth.
Show worked answer →

Redshift fraction:

z=Δλ/λ0=(689.1656.3)/656.3=32.8/656.3=0.0500z = \Delta \lambda / \lambda_0 = (689.1 - 656.3) / 656.3 = 32.8 / 656.3 = 0.0500.

For vcv \ll c, the non-relativistic Doppler formula gives:

v=zc=0.0500×3.00×108=1.5×107v = z c = 0.0500 \times 3.00 \times 10^8 = 1.5 \times 10^7 m/s.

Sign and direction: the observed wavelength is longer than the laboratory wavelength (redshift), so the galaxy is receding from Earth at about 1.5×1071.5 \times 10^7 m/s.

Markers reward the formula, correct value of zz, conversion to velocity, and identification of "receding" because of the redshift. Some answers will use the relativistic Doppler formula; at v/c=0.05v / c = 0.05 the correction is small (0.1%\sim 0.1\%).

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksDistinguish between a line emission spectrum and a line absorption spectrum, and state one source of each.
Show worked solution →

Line emission spectrum. Bright lines on a dark background at specific wavelengths, produced by a hot, low-density gas (e.g. a neon discharge tube or a nebula), where excited electrons drop to lower energy levels and release photons.

Line absorption spectrum. A continuous spectrum crossed by dark lines at specific wavelengths, produced when continuum light passes through a cooler gas (e.g. the Sun's photosphere continuum passing through its cooler outer atmosphere), which removes photons matching its atoms' allowed transitions.

Marks: one for a correct description of each spectrum type, one for a correct source example for each.

foundation2 marksUse Wien's law to calculate the peak wavelength of the radiation emitted by a star with a surface temperature of 3000 K3000\ \text{K}. (b=2.898×103 m Kb = 2.898 \times 10^{-3}\ \text{m K}.)
Show worked solution →

Wien's law: λmax=bT\lambda_{\max} = \dfrac{b}{T}.

λmax=2.898×1033000=9.66×107 m=966 nm\lambda_{\max} = \dfrac{2.898 \times 10^{-3}}{3000} = 9.66 \times 10^{-7}\ \text{m} = 966\ \text{nm}.

Marks: one for the correct formula with values substituted, one for the answer to three significant figures with the unit (this falls in the infrared, consistent with a cool red star).

foundation3 marksState the type of source that produces (a) a continuous spectrum and (b) a line emission spectrum, and explain in one sentence why the two spectra look different even though both sources may contain the same element.
Show worked solution →

(a) A continuous spectrum comes from a hot, dense source (a glowing solid, liquid or high-pressure gas), where atoms are packed so closely that their energy levels blend into an effectively continuous range of emitted photon energies.

(b) A line emission spectrum comes from a hot, low-density gas, where widely separated atoms emit photons only at the discrete energies corresponding to their allowed transitions.

Why they differ. In a dense source, collisions and pressure broaden and smear the discrete atomic energy levels into a continuum; in a dilute gas the atoms are isolated, so only their sharp, discrete transition energies appear as lines.

Marks: one for the continuous-spectrum source, one for the line-emission-spectrum source, one for the pressure/density explanation of why the same element gives different-looking spectra.

core3 marksA hydrogen absorption line with a laboratory rest wavelength of 656.3 nm656.3\ \text{nm} is observed at 657.9 nm657.9\ \text{nm} in the spectrum of a distant galaxy. Calculate the galaxy's radial velocity and state whether it is approaching or receding. (c=3.00×108 m s1c = 3.00 \times 10^8\ \text{m s}^{-1}.)
Show worked solution →

Redshift fraction: z=Δλλ0=657.9656.3656.3=1.6656.3=2.44×103z = \dfrac{\Delta \lambda}{\lambda_0} = \dfrac{657.9 - 656.3}{656.3} = \dfrac{1.6}{656.3} = 2.44 \times 10^{-3}.

For vcv \ll c: v=zc=2.44×103×3.00×108=7.31×105 m s1v = zc = 2.44 \times 10^{-3} \times 3.00 \times 10^8 = 7.31 \times 10^5\ \text{m s}^{-1}.

Since λobs>λ0\lambda_{\text{obs}} > \lambda_0 this is a redshift, so the galaxy is receding from Earth at about 7.31×105 m s17.31 \times 10^5\ \text{m s}^{-1} (731 km s1731\ \text{km s}^{-1}).

Marks: one for the correct zz, one for v=zcv = zc correctly evaluated with unit, one for correctly identifying "receding" from the redshift.

core4 marksThe figure shows blackbody intensity-versus-wavelength curves for two stars, A (6000 K6000\ \text{K}) and B (4000 K4000\ \text{K}). **(a)** Read the peak wavelength of star A's curve from the graph. **(b)** Use Wien's law to confirm this peak wavelength is consistent with T=6000 KT = 6000\ \text{K}. **(c)** State which star is hotter and explain how both the peak position and the total area under the curve support your answer.
Show worked solution →

(a) Star A's curve peaks at approximately λmax480 nm\lambda_{\max} \approx 480\ \text{nm} (reading directly off the graph).

(b) λmax=bT=2.898×1036000=4.83×107 m=483 nm\lambda_{\max} = \dfrac{b}{T} = \dfrac{2.898 \times 10^{-3}}{6000} = 4.83 \times 10^{-7}\ \text{m} = 483\ \text{nm}, which matches the graph's peak within reading error.

(c) Star A (6000 K6000\ \text{K}) is hotter than star B (4000 K4000\ \text{K}). Its curve peaks at a shorter wavelength (Wien's law: λmax1/T\lambda_{\max} \propto 1/T, so a hotter object peaks further toward the blue/violet end), and its curve also sits higher and encloses a larger area (a hotter blackbody radiates more total power per unit area at every wavelength, consistent with the Stefan-Boltzmann law).

Marks: one for a peak-wavelength reading close to 480-485 nm480\text{-}485\ \text{nm}, one for the Wien's-law calculation confirming it, one for correctly identifying star A as hotter, one for citing both the shorter peak wavelength AND the greater area/height as evidence.

core3 marksExplain why the presence of a particular absorption line in a star's spectrum confirms the presence of an element in the star's atmosphere, but the ABSENCE of that line does not necessarily mean the element is absent.
Show worked solution →

Presence confirms. Each element has a unique set of allowed electron transitions, so a set of absorption lines at wavelengths matching an element's known laboratory spectrum can only be produced by that element being present in the light path - it is a unique spectral fingerprint.

Absence does not rule it out. A given transition line only appears strongly if the atmosphere's temperature and density populate the right starting energy level for that transition. An element may be present but at too low a temperature (or too fully ionised, or too rarefied) to produce a detectable line at that particular wavelength, so a missing line can reflect the physical conditions rather than the element's absence.

Marks: one for the fingerprint-matching argument for presence, one for identifying that line strength depends on temperature/level population, one for concluding this means absence of a line is not proof of absence of the element.

exam6 marksAnalyse how the analysis of a single stellar spectrum can reveal a star's chemical composition, surface temperature and radial velocity, explaining the physical origin of each piece of evidence.
Show worked solution →

Band-6 plan. Structure by the three quantities in the question: (1) composition - line identification against laboratory spectra; (2) temperature - Wien's law on the continuum plus relative line strengths; (3) radial velocity - Doppler shift of the line positions. For each, name the underlying physics (quantised energy levels, blackbody radiation, wave motion of light) and the observable feature used. Finish by noting all three are extracted from the SAME spectrum simultaneously.

Model answer. A star's photosphere produces a near-continuous blackbody spectrum that is then absorbed at specific wavelengths by cooler gas in the star's outer atmosphere, producing a line absorption spectrum. Three independent pieces of information can be read from this one spectrum.

Composition is determined from the wavelengths of the dark absorption lines. Because each element has a unique set of quantised electron energy levels, it absorbs photons only at the specific energies hf=EiEfhf = E_i - E_f matching its allowed transitions. Matching the observed line wavelengths against laboratory reference spectra identifies which elements are present in the star's atmosphere (this is how helium was first identified in the Sun in 1868, before being isolated on Earth).

Surface temperature is found in two independent ways from the same spectrum. First, the overall shape of the continuum approximates a blackbody curve, and its peak wavelength obeys Wien's law, λmax=b/T\lambda_{\max} = b/T; a shorter peak wavelength means a hotter star. Second, the relative strengths of different absorption lines depend on temperature, since each transition requires the lower energy level to be sufficiently populated, so the pattern of strong and weak lines places a star into a spectral class (O, B, A, F, G, K, M).

Radial velocity is found from a uniform shift of every absorption line's wavelength relative to its laboratory rest value. If the star is moving away from Earth along the line of sight, all lines are shifted to longer wavelengths (redshift); if approaching, to shorter wavelengths (blueshift), with Δλ/λ0=v/c\Delta\lambda/\lambda_0 = v/c for vcv \ll c. Because this shift affects every line by the same fractional amount, it can be distinguished cleanly from the fixed line positions used for composition.

All three results, composition, temperature and radial velocity, come from analysing a single captured spectrum: the line wavelengths (shifted by a common factor) give composition and velocity, and the continuum shape plus line-strength pattern gives temperature.

Marker's note: the top band explains the physical origin of each measurement (quantised levels for composition, blackbody/Wien's law and level population for temperature, wave-Doppler shift for velocity) rather than simply naming the three results, and explicitly notes that all three are extracted from one spectrum. A response that lists the three uses without the underlying physics caps in the middle band.

exam6 marksEvaluate the claim that a star's line absorption spectrum alone is sufficient evidence to fully characterise the star, addressing what it can and cannot determine.
Show worked solution →

Band-6 plan. Take a genuine evaluative stance: (1) what a line absorption spectrum CAN determine (composition, temperature, radial velocity, rotation) with brief physical justification for each; (2) what it CANNOT determine from itself alone (e.g. distance, luminosity, transverse motion, total mass) and why; (3) a concluding judgement (powerful but partial; needs combining with other data such as parallax or a light curve).

Model answer. A line absorption spectrum is a remarkably information-rich single measurement. The wavelengths of the absorption lines identify the elements present in the star's atmosphere by matching quantised-transition fingerprints against laboratory spectra. The shape of the underlying continuum, via Wien's law λmax=b/T\lambda_{\max} = b/T, together with which lines are strong or weak, gives the surface temperature and hence spectral type. A uniform shift of all the line wavelengths gives the radial velocity by the Doppler effect, and a symmetric broadening of the lines (from one limb approaching and the other receding) gives the star's rotation rate. In this sense the claim has real force: composition, temperature, radial velocity and rotation are all genuinely obtainable from spectroscopy alone.

However, a spectrum alone cannot determine everything about a star. It gives no direct measure of distance (needed for parallax or standard-candle methods) or of luminosity (total power output), since apparent brightness depends on both luminosity and distance, and a spectrum records neither in isolation. It also cannot determine transverse velocity (motion across the line of sight), because the Doppler effect is sensitive only to the radial component of velocity; a star moving purely sideways produces no line shift at all. Total mass likewise cannot be read from a single spectrum, though a binary system's spectrum can give some mass information via orbital Doppler variations combined with Kepler's laws.

On balance, the claim overstates what a single spectrum provides: it is an unusually powerful tool for atmospheric and kinematic (line-of-sight) properties, but a complete stellar characterisation, including distance, luminosity and full three-dimensional motion, requires combining spectroscopy with other observations such as parallax, proper motion and photometric brightness.

Marker's note: the top band gives specific correct physics for each "can" (fingerprint matching, Wien's law/line strengths, uniform Doppler shift, symmetric broadening) and each "cannot" (no distance/luminosity/transverse velocity from a spectrum alone, with a stated reason), then reaches an explicit, justified evaluative conclusion. A response that only lists what spectra reveal, without addressing the limitation, caps in the middle band.

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