Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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.
8 years ago

17 Points
										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


8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Thermal Physics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
Get extra Rs. 551 off
USE CODE: CHEM20
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details
Get extra Rs. 594 off
USE CODE: CHEM20