 # Wavelength Intensity of emitted e/m radiation 6000K 2000K 1500K 1000K VIBGYOR Wien’s Displacement Law for black body radiation peak wavelength  1 / Temp.

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wavelength Intensity of emitted e/m radiation 6000K 2000K 1500K 1000K VIBGYOR Wien’s Displacement Law for black body radiation peak wavelength  1 / Temp (wavelength of maximum intensity)

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 =  T 4 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 …

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