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
This question involves the concepts of Wein's displacement law and Stefan-Boltzmann law.
(a) The temperature of the hotter star in terms of T is "√3 T".
(b) The ratio of the peak-intensity wavelength of the hotter star to the peak intensity wavelength of the cool star is "1/√3".
(a)
It is given that the energy radiation per second for both the stars is the same.
[tex]E_1=E_2[/tex]
using Stefan-Boltzmann Law:
[tex]\sigma A_1T_1^4=\sigma A_2T_2^4\\\\A_1T_1^4=A_2T_2^4\\\\\pi \frac{d_1^2}{4}T_1^4=\pi \frac{d_2^2}{4}T_2^4\\\\d_1^2T_1^4=d_2^2T_2^4\\\\[/tex]
where,
d₁ = diameter of hotter star = d
d₂ = diameter of cooler star = 3d
T₁ = Temperature of hotter star = ?
T₂ = Temperature of cooler star = T
Therefore,
[tex]d^2T_1^4=(3d)^2T^4\\T_1^4=9T^4\\[/tex]
T₁ = √3 T
(b)
Now, for the ratio of peak-wavelengths, we will use Wein's Displacement Law:
[tex](\lambda_1)(\sqrt3\ T)=constant\\(\lambda_2)(T)=constant\\[/tex]
where,
[tex]\lambda_1[/tex] = peak-wavelength of hotter star
[tex]\lambda_2[/tex] = peak-wavelength of cooler star
Dividing both the equations, we get:
[tex]\frac{(\lambda_1)(\sqrt3\ T)}{(\lambda_2)(T)}=\frac{constant}{constant}\\\\\frac{\lambda_1}{\lambda_2}=\frac{1}{\sqrt3}[/tex]
Learn more about Stefan-Boltzmann Law here:
https://brainly.com/question/14919749?referrer=searchResults
The attached picture shows Wein's Displacement Law.
