Black body radiation A blackbody is a perfect absorber of radiation, able to absorb completely radiation of any wavelength that falls on it. A blackbody not only absorbs all wavelengths of radiation falling on it, but also radiates all wavelengths (for its particular temperature) too. The simplest possible type of blackbody is a hole in a box, painted black on the inside.

Atomic Spectra Line spectra are emitted by matter in the gaseous state, in which the atoms are so far apart that interactions between them are negligible. Hot matter in condensed states (solid or liquid) emit radiation with a continuous distribution of wavelengths rather than a line spectra.

Continuous Spectra Objects that emit a continuous spectra of radiation do not do so equally at all wavelengths. The y-axis shows the intensity of light emitted at each wavelength. Intensity is the power per unit area (energy per second per metre squared). Intensity is measured in Watts per square metre (Wm-2)

Wien’s displacement law The blackbody radiation curve is dependent on the temperature of the object. The diagram shows how the graph changes at different temperatures. The wavelength of maximum intensity is inversely proportional to the temperature of the object: Wien’s displacement law If you double the temperature you half the peak wavelength. Question: At what wavelength would the intensity peak at a temperature of…… (a) 300K (b) 10,000K

Light that appears white Light that appears blue Light that appears red Light that appears white Light that appears blue

When astronomers look at the light from distant stars they can measure the peak wavelength. From this they can calculate the stars temperature using Wien’s displacement law. Question: Astronomers observe a star to have a peak wavelength of 450nm. What is its temperature?

The total intensity of radiation emitted increases as the temperature rises. Stefan-Boltzmann law Stefan-Boltzmann constant = 5.7x10-8Wm-2K-4 If you double the temperature, the total intensity increases by 16 times Question: If a source emits 1000Wm-2 of radiation, what temperature is it at?

Inverse square law As you get further away from a source, the intensity will drop off. This is because the power is spread over a larger area. If you double the distance from the source then the power is spread over 4 times the area, this will cause the intensity to drop to ¼. Inverse square law Example: A lamp has a power output of 100W. What will be the intensity 4 metres away (area of a sphere = 4πr2). I = 100/(4πx42) = 201Wm-2 Question: At the surface of a star (radius = 9x105km ) there is an intensity of 2x1012Wm-2. What will its intensity be at 1.6x109km away?

Hertzsprung Russell Diagram Luminosity = power output (Watts) Power output of the sun = 4x1026W (solar constant) There is a relationship between the temperature of a star and its luminosity/power output. This relationship depends on the stage of life a star is in.

Wien’s displacement law Alpha Centauri is one of our closest stars. Astronomers have measured its peak wavelength to be 300nm. The intensity of light received on the Earth from Alpha Centauri is 0.01Wm-2. Use this information to calculate the diameter of Alpha Centauri. Wien’s displacement law Stefan-Boltzmann law

Wien’s displacement law Alpha Centauri is one of our closest stars. Astronomers have measured its peak wavelength to be 300nm. The intensity of light received on the Earth from Alpha Centauri is 0.01Wm-2. Use this information to calculate the diameter of Alpha Centauri. Wien’s displacement law Stefan-Boltzmann law Inverse square law From the HR diagram 9700K gives a luminosity of 4x1028W