Questions on black-body radiation and spectral classes

Question 1

State Stefan's law and Wien's displacement law, defining each symbol and giving the unit of Wien's constant.

Accepted Answer

Stefan: P = σAT⁴, with A the surface area and T the surface temperature in kelvin (1). Wien: λ_maxT = 2.9 × 10⁻³ m K, the constant carrying the product unit metre kelvin (1).

Question 2

Write the spectral classes in order from hottest to coolest, and state the class of the Sun.

Accepted Answer

O B A F G K M, hottest to coolest (1). The Sun, at about 5800 K, is a class G star (1).

Question 3

State the two assumptions made when a star's temperature, power and radius are deduced from its measured spectrum.

Accepted Answer

The star radiates as a black body (1), and the light travels to the telescope without absorption on the way, by dust or by atmosphere (1).

Question 4

State what is meant by a black body.

Accepted Answer

A body that absorbs all the electromagnetic radiation incident on it (1). It is therefore also the best possible emitter, radiating a spectrum that depends only on its temperature (1).

Question 5

State the colour of a class O star and of a class M star, and name the molecule prominent in class M absorption spectra.

Accepted Answer

Class O stars are blue; class M stars are red (1). Class M spectra show molecular absorption by titanium oxide, TiO (1).

Question 6

A star's surface temperature rises. Describe two changes to its black-body radiation curve.

Accepted Answer

The peak moves to a shorter wavelength (Wien's law) (1), and the curve rises at every wavelength, so the total power radiated (the area under the curve) increases (1).

Question 7

Calculate the peak wavelength of the radiation from a star with a surface temperature of 10 000 K, and state the region of the spectrum in which it lies.

Accepted Answer

λ_max = 2.9 × 10⁻³/10 000 (1) λ_max = 2.9 × 10⁻⁷ m (1) In the ultraviolet, so the star looks blue-white (1)

Question 8

A star's spectrum peaks at 9.7 × 10⁻⁷ m. Calculate its surface temperature and suggest its spectral class.

Accepted Answer

T = 2.9 × 10⁻³/(9.7 × 10⁻⁷) (1) T = 2990 K (1) Just below 3500 K, so a red class M star (1)

Question 9

A star has twice the Sun's radius (take the Sun's radius as 6.96 × 10⁸ m) and a surface temperature of 9000 K. Calculate its power output.

Accepted Answer

A = 4πr² = 4π × (1.39 × 10⁹)² (1) A = 2.43 × 10¹⁹ m² (1) P = σAT⁴ = 5.67 × 10⁻⁸ × 2.43 × 10¹⁹ × 9000⁴ (1) P = 9.1 × 10²⁷ W (1)

Question 10

Explain why the hydrogen Balmer absorption lines are strongest in class A stars, weaker in cooler stars, and weak again in the hottest stars.

Accepted Answer

Balmer absorption needs hydrogen atoms already excited to the n = 2 level (1). In cool stars almost all atoms sit in the ground state, so the lines are weak; in the hottest stars the hydrogen is ionised, so no bound electrons remain to absorb (1). Class A, 7500 to 11 000 K, is the productive middle (1).

Question 11

A star's surface temperature triples while its radius stays the same. State the factor by which its power output increases.

Accepted Answer

P ∝ T⁴ at fixed area (1), so the power rises by a factor of 3⁴ = 81 (1).

Question 12

Two stars have equal radii. One has its spectral peak at 400 nm, the other at 800 nm. Calculate the ratio of their power outputs.

Accepted Answer

λ_maxT is constant, so the 400 nm star is twice as hot as the 800 nm star (1). With equal areas P ∝ T⁴ (1), so the ratio is 2⁴ = 16, the hotter star radiating 16 times the power (1).

Question 13

A white dwarf of radius 7.0 × 10⁶ m radiates a total power of 1.1 × 10²⁴ W. Calculate its surface temperature. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

Accepted Answer

A = 4πr² = 4π × (7.0 × 10⁶)² = 6.2 × 10¹⁴ m² (1). T⁴ = P/σA (1) = 1.1 × 10²⁴/(5.67 × 10⁻⁸ × 6.2 × 10¹⁴) (1), so T = 1.3 × 10⁴ K (about 13 000 K) (1).

Question 14

Calculate the power radiated by each square metre of the Sun's surface, taking the surface temperature as 5800 K and σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

Accepted Answer

Power per unit area = σT⁴ = 5.67 × 10⁻⁸ × 5800⁴ (1) = 6.4 × 10⁷ W m⁻² (1).

Question 15

A red giant radiates 3.9 × 10²⁸ W, one hundred times the Sun's output, from a surface at 3480 K. Calculate its radius, and express it in solar radii (Sun's radius 6.96 × 10⁸ m).

Accepted Answer

A = P/σT⁴ = 3.9 × 10²⁸/(5.67 × 10⁻⁸ × 3480⁴) (1) A = 4.7 × 10²¹ m² (1) r = √(A/4π) (1) r = 1.9 × 10¹⁰ m (1) ≈ 28 solar radii (1)

Question 16

Two stars have the same radius. Star X is class M and star Y is class A. Compare, with reasons, their colours and their power outputs.

Accepted Answer

X, below 3500 K, peaks towards the infrared and looks red (1); Y, between 7500 and 11 000 K, peaks towards the ultraviolet and looks blue-white (1). With equal areas, P ∝ T⁴, so Y out-radiates X by a factor of between about 20 and a few hundred (1).

Question 17

A student says a star's colour tells you its temperature. State what measurement makes this precise, and the law that converts it.

Accepted Answer

Measure the wavelength at which the star's black-body spectrum peaks (1); Wien's law λ_maxT = 2.9 × 10⁻³ m K then gives the surface temperature directly (1).

Question 18

An astronomer must decide which of two stars is the larger. Star J radiates 4.6 × 10²⁶ W with its spectral peak at 690 nm; star K radiates 1.5 × 10²⁷ W with its peak at 230 nm. Deduce which star has the larger radius. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; Wien constant 2.9 × 10⁻³ m K.

Accepted Answer

Wien: T_J = 2.9 × 10⁻³/(6.9 × 10⁻⁷) = 4.2 × 10³ K and T_K = 1.3 × 10⁴ K (1). A = P/σT⁴ and r = √(A/4π) (1). r_J = 1.4 × 10⁹ m (1); r_K = 2.9 × 10⁸ m (1). Star J is about 5 times larger in radius, despite its smaller power output: its low temperature must be offset by area (1).

Question 19

Describe fully how astronomers determine the radius of a distant star from its spectrum and its measured brightness, naming each law used and stating the assumptions involved.

Accepted Answer

Measure the wavelength at which the star's spectrum peaks; Wien's law λ_maxT = 2.9 × 10⁻³ m K gives the surface temperature (1). Measure the radiant flux at Earth and, knowing the star's distance, use the inverse square law to find the total power radiated (1). Stefan's law P = σAT⁴ then gives the surface area (1), and A = 4πr² converts area to radius (1). This assumes the star radiates as a black body (1) and that no light is absorbed between the star and the telescope, by dust or atmosphere (1).