Question

The temperature of a furnace is 2324°C and the
intensity is maximum in its radiation spectrum ned
at 12000 Aº. If the intensity in the spectrum of a
star is maximum nearly at 4800 A', then the sur-
face temperature of star is
(1) 8400°C
(2) 7200°C
376219.5°C
(4) 5900°C​

Answer 1

Answer:

Explanation:

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}  

1.    λmax is the peak wavelength in meters,
2.    T is the temperature of the blackbody in kelvins (K), and
3.    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.

Therefore, (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