Electromagnetic radiation from stars — OCR A-Level Physics

Energy levels in an atom are quoted as negative values because the zero of energy is defined as the electron being just free of the nucleus, at rest and infinitely far away. A bound electron is held by the electrostatic attraction, so removing it requires energy input; the bound state therefore sits below the zero line and its energy is negative. For example, the ground state of hydrogen is about −13.6 eV, and an electron excited to a higher level has a less negative value such as −3.4 eV. The energy needed to ionise the atom equals the difference between zero and the level, so a level of −13.6 eV needs 13.6 eV to free the electron. Transitions between levels involve differences, so the negative signs cancel correctly when you subtract.

A hot gas contains atoms whose electrons occupy discrete energy levels. When an electron drops from a higher level to a lower one, the atom loses energy and emits a single photon carrying that exact energy difference. Because only certain level gaps exist, only certain photon energies, and therefore certain frequencies and wavelengths, are emitted. A spectrometer spreads this light into a series of bright lines on a dark background, called an emission line spectrum. For example, excited hydrogen emits visible lines such as the red line near 656 nm and the blue-green line near 486 nm. Each line corresponds to one specific electron transition, so the pattern acts as a fingerprint of the element.

These two equations convert between the energy gap of a transition and the electromagnetic radiation produced or absorbed. ΔE = hf gives the photon energy from its frequency, where h is the Planck constant, about 6.63 × 10⁻³⁴ J s. Because for light in a vacuum c = fλ, substituting f = c/λ gives ΔE = hc/λ, which is useful when a wavelength is known. For example, a photon of wavelength 656 nm has energy hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (656 × 10⁻⁹), which is about 3.03 × 10⁻¹⁹ J. Convert nanometres to metres before substituting, and remember that ΔE is the difference between two energy levels, so it is positive for an emitted photon.

Every element has its own unique arrangement of electron energy levels, so the set of photon energies it can emit or absorb is unique. This produces a characteristic pattern of spectral lines, often called a spectral fingerprint. Astronomers record the spectrum of a star, identify the wavelengths of the lines present, and match them against laboratory spectra of known elements. For example, strong hydrogen lines and helium lines in a stellar spectrum show those elements are present in the star's outer layers. The same principle applies to absorption lines formed when cooler gas in the star's atmosphere absorbs specific wavelengths from the continuous spectrum beneath. Comparing line patterns therefore reveals composition without visiting the star.

A continuous spectrum contains all wavelengths of visible light, appearing as an unbroken band of colour. An emission line spectrum shows only bright lines at specific wavelengths on a dark background, produced when excited atoms in a low-pressure gas emit photons as electrons drop to lower energy levels. An absorption line spectrum shows a continuous spectrum crossed by dark lines at the same wavelengths as the emission lines of the same element, produced when white light passes through a cooler gas that absorbs photons matching its energy level gaps. Stars show absorption spectra because their hot dense core emits a continuous spectrum that passes through cooler outer gases.

A transmission diffraction grating has many equally spaced parallel slits. Monochromatic light passing through the grating produces sharp maxima at angles given by d sin θ = nλ, where d is the grating spacing, n is the order and λ is the wavelength. To determine wavelength, measure the angle θ for a known order n and calculate λ = d sin θ / n. The grating spacing d equals 1 divided by the number of lines per metre. Because the maxima are sharp, the angle can be measured precisely, giving an accurate wavelength. A spectrometer or optical bench with a protractor measures the angle between the zero-order and nth-order maxima.

For a diffraction grating, constructive interference produces maxima when the path difference between light from adjacent slits equals a whole number of wavelengths. This condition is d sin θ = nλ, where d is the grating spacing (distance between adjacent slit centres), θ is the angle from the zero-order direction, n is the order (0, 1, 2, ...) and λ is the wavelength. The zero-order maximum (n = 0) occurs at θ = 0° for all wavelengths. Higher orders spread out more, so the grating separates wavelengths, which is why it is used in spectrometry. The maximum possible order is limited because sin θ cannot exceed 1.

Wien’s displacement law states that the wavelength of peak emission λmax is inversely proportional to the surface temperature T: λmax T = 2.9 × 10⁻³ m K. To estimate a star’s surface temperature, measure the wavelength at which its continuous spectrum is most intense, then rearrange to T = 2.9 × 10⁻³ / λmax. For example, if λmax = 5.8 × 10⁻⁷ m, then T = 2.9 × 10⁻³ / 5.8 × 10⁻⁷ ≈ 5000 K. Hotter stars peak at shorter wavelengths (blue), cooler stars at longer wavelengths (red). The constant has units of metre kelvin, so λmax must be in metres and T in kelvin.

Luminosity L is the total electromagnetic power a star radiates in all directions, measured in watts (W). Stefan’s law models a star as a black-body sphere: L = 4πr²σT⁴, where r is the star’s radius in metres, T is its surface temperature in kelvin, and σ is the Stefan constant, about 5.67 × 10⁻⁸ W m⁻² K⁻⁴. The 4πr² factor is the star’s surface area, so L scales with surface area and with the fourth power of temperature. Doubling r multiplies L by 2² = 4; doubling T multiplies L by 2⁴ = 16. Rearranged, r = √(L / (4πσT⁴)). Check units: W m⁻² K⁻⁴ × m² × K⁴ gives W.

Wien’s displacement law, λmaxT = 2.9 × 10⁻³ m K, links a star’s peak-emission wavelength λmax to its surface temperature T. Stefan’s law, L = 4πr²σT⁴, links luminosity, radius and temperature. To estimate a star’s radius: measure λmax from its spectrum, use Wien’s law to find T = (2.9 × 10⁻³) / λmax, then rearrange Stefan’s law to r = √(L / (4πσT⁴)). Luminosity may be given or inferred from apparent brightness and distance. Work in kelvin and metres throughout. This two-step method is powerful because it needs only the spectrum and the luminosity.