Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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.

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.

Grade:12

1 Answers

AskIITians Expert Hari Shankar IITD
17 Points
11 years ago

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

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free