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Intensity
of emitted
e/m
radiation
Wien’s Displacement
Law for black body
radiation
peak wavelength a 1 / Temp
(wavelength of
maximum intensity)
6000K
2000K
1500K
VIBGYOR
1000K wavelength
Wein’s law
max . T = constant (Wein’s constant)
Where Wein’s constant = 0.0029 m K
(metres Kelvin)
(not milli Kelvins !)
Use it to estimate the surface temp of a star if we approximate
the star to being a black body
E.g. peak wavelength of the sun is 490 nm  effective surface
temp of … 5920K
Stefan’s law
E = 
4
T
Total energy radiated by a black body
- in unit time
- per unit surface area
- is proportional to the fourth power of the
absolute temperature of the body
Where E is the energy radiated per second per square metre of surface
T is the Temp of the black body
 is Stephan’s constant = 5.67 x 10-8 W m-2 K-4
Estimate the intensity of the radiation emitted per unit
area from a star if it’s effective surface Temp is 6000K
Estimate the energy emitted from a star if its peak
wavelength is 600nm.
Total power radiated
Power output = E x surface area of star
Surface area of a star = 4  R 2
Where R is the radius of the star
Conclusion: a combination of Wein’s Law and Stephan’s Law can
lead to an estimation of the temp of a star and it’s power output. This
could allow us to estimate its absolute magnitude and using a
rearrangement of m – M = 5 log (d/10) could lead to an estimation of
its distance away from us in parsecs. Of course its power output
depends not only on the temp of the star but also its size  radius and
therefore surface area … more techniques are required …