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				   The temp of the furnance is 2324 degree celcius and intensity is maximum in the radiation spectrum nearly at 12000 angstrom.If the intensity in the spectrum of a star is maximum nearly at 4800 angstrom then surface temp of a star is.


7 years ago

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										There is an inverse relationship between the wavelength of the peak of the emission of a black body and its temperature, and this less powerful consequence is often also called Wien's displacement law
$\lambda_{\mathrm{max}} = \frac{b}{T}$where

λmax is the peak wavelength in meters,
T is the temperature of the blackbody in kelvins (K), and
b is a constant of proportionality

We can consider the star and the furnace both as blackbodies for this question.
(λmax)(T) = b in both cases.

Theresfore, (λmax)(T) for the furnace = (λmax)(T) for the star
Then, for the furnace, λmax = 12000 A = 1.2e-6 m and T =2324 C = 2597 K

For the star, (λmax) = 4800 A = 4.8e-7 m

Hence , Tstar = (1.2e-6)(2597)/(4.8e-7) = 6492.5 K


7 years ago

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