Black-Body Radiation and Spectral Classes — AQA A-level Physics

What is Black-Body Radiation?

Definition

Black-Body Radiator: A perfect emitter and absorber of all possible wavelengths of radiation. Stars can be approximated as black bodies.

The amount of radiation emitted by a body depends on: the surface area, the type of surface, and the temperature of the body.

Key Properties of Black-Body Radiation

The spectrum is continuous (all wavelengths present)

The shape of the curve depends only on temperature

Hotter objects emit more radiation at all wavelengths

The peak wavelength shifts to shorter wavelengths as temperature increases

Key Point

By analysing the black-body spectrum of a star, we can determine its surface temperature without ever visiting it.

Black-Body Curves

The graph shows the variation of intensity with wavelength for objects at different temperatures:

Key Facts from These Curves

Area under curve ∝ total energy emitted per unit area at that temperature

The maxima move towards shorter wavelengths at higher temperatures

Curves for lower temperatures lie completely inside those of higher temperature

Intensity and Distance

The intensity of light emitted by a star follows the inverse square law. This means intensity is inversely proportional to the square of the distance between the star and the observer.

Formula

I = P / (4πd²)I = intensity (W m⁻²) | P = power output of star (W) | d = distance from star (m)

Why 4πd²?

Light is emitted equally in all directions from a point source, so it spreads out in the shape of a sphere. The surface area of a sphere is 4πr², so at distance d, the power is spread over area 4πd².

Key Point

Double the distance → intensity drops to ¼. Triple the distance → intensity drops to ⅑.

Wien's Displacement Law

Wien's displacement law states that the peak wavelength (λₘₐₓ) of emitted radiation is inversely proportional to the absolute temperature (T) of the object:

Formula

λₘₐₓ T = constant = 2.9 × 10⁻³ m Kλₘₐₓ = wavelength of light released at maximum intensity (m) T = absolute temperature (K) | Unit: metres-Kelvin (not milliKelvin!)

This can be rearranged to find either quantity:

Find λₘₐₓ

λₘₐₓ = (2.9 × 10⁻³) / T

Find T

T = (2.9 × 10⁻³) / λₘₐₓ

Key Point

Wien's law shows that as a black body gets hotter, the peak wavelength decreases (shifts to blue), meaning the frequency and energy of the waves increases.

Worked Example

1

Finding Surface Temperature

A star has a peak emission wavelength of 4.83 × 10⁻⁷ m (blue light). Calculate the surface temperature of this star.

1

Identify: λₘₐₓ = 4.83 × 10⁻⁷ m, Wien's constant = 2.9 × 10⁻³ m K

2

Apply Wien's law: T = (2.9 × 10⁻³) ÷ λₘₐₓ

3

T = (2.9 × 10⁻³) ÷ (4.83 × 10⁻⁷)

4

T = 6004 K ≈ 6000 K

T = 6000 K

Worked Example

2

Finding Peak Wavelength

A red giant star has a surface temperature of 3500 K. Calculate the peak wavelength of its emission.

1

Identify: T = 3500 K, Wien's constant = 2.9 × 10⁻³ m K

2

Apply Wien's law: λₘₐₓ = (2.9 × 10⁻³) ÷ T

3

λₘₐₓ = (2.9 × 10⁻³) ÷ 3500

4

λₘₐₓ = 8.29 × 10⁻⁷ m = 829 nm (infrared)

λₘₐₓ = 8.3 × 10⁻⁷ m

Stefan's Law

Stefan's law states that the power output (luminosity) of a black-body radiator is directly proportional to its surface area (A) and its absolute temperature to the fourth power (T⁴):

Formula

P = σAT⁴For a spherical star: L = 4πr²σT⁴

Where σ (sigma) is the Stefan constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

Understanding the Equation

4πr² = surface area of a sphere

σT⁴ = power radiated per unit area

This law can be used to compare the power output, temperature and size of stars

Key Point

L ∝ r² — Double the radius → 4× the luminosity

L ∝ T⁴ — Double the temperature → 16× the luminosity

Worked Example

3

Finding Luminosity

Calculate the luminosity of a star of radius 10⁹ m and surface temperature 6000 K. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

1

Given: r = 10⁹ m, T = 6000 K, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

2

Use: L = σ × 4πr² × T⁴

3

L = 5.67 × 10⁻⁸ × 4 × π × (10⁹)² × (6000)⁴

4

L = 5.67 × 10⁻⁸ × 4π × 10¹⁸ × 1.296 × 10¹⁵

5

L = 9.23 × 10²⁶ W

L = 9.2 × 10²⁶ W

Worked Example

4

Comparing Two Stars

Star A has the same radius as Star B, but Star A has twice the surface temperature. What is the ratio of their luminosities L_A / L_B?

1

Given: r_A = r_B, T_A = 2T_B

2

For each star: L = 4πr²σT⁴

3

L_A = 4πr²σ(2T_B)⁴ = 4πr²σ × 16 × T_B⁴

4

L_A = 16 × (4πr²σT_B⁴) = 16 × L_B

L_A / L_B = 16

Stellar Spectral Classes

Stars can be classified into spectral classes based on the strength of absorption lines in their spectra. These absorption lines are dependent on the temperature of the star.

Spectral Class Sequence (Hot → Cool)

O B A F G K M

"Oh Be AFine Girl/Guy, Kiss Me"

By looking at the spectrum of a star, astronomers can determine:

The temperature: Found by measuring the variation in intensity across the spectrum and applying Wien's law

The composition: Determined by identifying which elements produce the absorption lines observed

Spectral Classes: Properties

ClassColourTemp (K)Absorption LinesH Balmer

OBlue25,000–50,000He⁺, He, HWeak

BBlue11,000–25,000He, HSlightly stronger

ABlue-white7,500–11,000H, ionised metalsStrongest

FWhite6,000–7,500Ionised metalsWeak

GYellow-white5,000–6,000Ionised + neutral metalsNone

KOrange3,500–5,000Neutral metalsNone

MRed<3,500Neutral atoms, TiONone

The Sun is a G-type star with a temperature of about 5,800 K.

Hydrogen Balmer Lines

Hydrogen Balmer lines are absorption lines found in the spectra of O, B, and A type stars. They are caused by the excitation of hydrogen atoms from the n = 2 state to higher energy levels.

Why H Balmer Lines Peak at A-Type

The intensity of hydrogen Balmer lines is dependent on temperature:

Temperature Too HIGH (O, B stars)

Hydrogen atoms become excited beyond n = 2, or electrons become ionised (removed entirely).

→ Balmer lines are weak

Temperature Too LOW (F, G, K, M stars)

Hydrogen atoms are not excited to the n = 2 level - electrons remain in ground state.

→ Balmer lines are absent

Conclusion

A-type stars (~10,000 K) have the optimal temperature where hydrogen atoms have electrons in the n = 2 level, ready to absorb photons and produce strong Balmer absorption lines.

Worked Example

5

Identifying Spectral Class

A star has a peak emission wavelength of 5.80 × 10⁻⁷ m. Determine its surface temperature and identify its spectral class.

1

Use Wien's law: T = (2.9 × 10⁻³) ÷ λₘₐₓ

2

T = (2.9 × 10⁻³) ÷ (5.80 × 10⁻⁷)

3

T = 5000 K

4

From table: 5000 K is at the boundary of G (5000–6000 K) and K (3500–5000 K)

T = 5000 K, Spectral Class G/K

Worked Example

6

Combined Problem

A B-type star has surface temperature 20,000 K and luminosity 1.5 × 10³⁰ W. Calculate: (a) its peak wavelength, (b) its radius.

1

Part (a): λₘₐₓ = (2.9 × 10⁻³) ÷ T = (2.9 × 10⁻³) ÷ 20000

2

λₘₐₓ = 1.45 × 10⁻⁷ m = 145 nm (UV)

3

Part (b): L = 4πr²σT⁴ → r = √(L ÷ 4πσT⁴)

4

T⁴ = (20000)⁴ = 1.6 × 10¹⁷ K⁴

5

4πσT⁴ = 4π × 5.67 × 10⁻⁸ × 1.6 × 10¹⁷ = 1.14 × 10¹¹ W m⁻²

6

r² = (1.5 × 10³⁰) ÷ (1.14 × 10¹¹) = 1.32 × 10¹⁹ m²

7

r = 3.6 × 10⁹ m

(a) λₘₐₓ = 1.45 × 10⁻⁷ m (b) r = 3.6 × 10⁹ m

Summary

Black-Body Radiation

A black body is a perfect emitter and absorber of all wavelengths

Area under curve ∝ total energy emitted; lower-T curves lie inside higher-T curves

Intensity follows inverse square law: I = P / (4πd²)

Key Formulae

Wien's Law

λₘₐₓT = 2.9 × 10⁻³ m K

Stefan's Law

P = σAT⁴ = 4πr²σT⁴

Spectral Classes (Hot → Cool): O B A F G K M

O: 25,000–50,000 K | B: 11,000–25,000 K | A: 7,500–11,000 K | F: 6,000–7,500 K | G: 5,000–6,000 K | K: 3,500–5,000 K | M: <3,500 K

Hydrogen Balmer Lines

Strongest in A-type stars (~10,000 K) where H atoms have electrons in n=2. Too hot → ionised (weak). Too cold → not excited (none).