Question

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature T and 3.0 times the diameter of the hotter star.
(a) What is the temperature of the hotter star in terms of T?
(b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

Answer 1

Answer:

Explanation:

Let hotter star has surface area of A . The cooler star would have surface area 9 times that of hotter star ie 9A , because its radius is 3 times hot star. Let temperature of hot star be T ₁.

Total radiant energy is same for both the star

Using Stefan's formula of black body radiation,

For cold star E = 9A x σ T⁴

For hot star E = A x σ T₁⁴

A x σ T₁⁴ = 9A x σ T⁴

T₁⁴ = (√3)⁴T⁴

T₁ = √3T .

b )

Let the peak intensity wavelength be λ₁ and λ₂ for cold and hot star .

As per wein's law

for cold star , λ₁ T = b ( constant )

for hot star  λ₂ √3T = b

dividing

λ₁ T / λ₂ √3T = 1

λ₂  / λ₁ =  1 / √3