Transcript File

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.
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.
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).
P E
I 
A At
Intensity is measured
in Watts per square
metre (Wm-2)
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
3
the object:
λ mT  2.9 x10 m.K
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
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?
I  σT 4
The total intensity of radiation emitted increases as the temperature
rises.
4 Stefan-Boltzmann law
I  σT
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?
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
Power
1

 2
area
r
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?
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.
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
λ mT  2.9 x103 m.K
Stefan-Boltzmann law
I  σT 4
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
λ mT  2.9 x103 m.K
2.9 x103 2.9 x103
T

 9700K
-9
λm
300 x10
Stefan-Boltzmann law
I  σT 4  5.7x10 -8 x (9700) 4  5.0 x108 Wm2
Inverse square law
From the HR diagram 9700K gives a luminosity of 4x1028W
P
Power
1
I


 2
4r 2
area
r
r
P

4I
4x1028
9

2.5x10
m
8
4 (5x10 )