Question 635: Thermodynamics & KTG

Question

The outer surface of star in the form of a sphere radiates heat as a black body at temperature ' T '. The total radiant energy per unit area, normal to the direction of incidence, received at a distance ' $R$ ' from the centre of a star of radius ' $r$ ' is $(R>r)(\sigma=$ Stefan's constant $)$

Accepted Answer

Step 1: Total power radiated by the star
The surface flux from a black body at temperature $T$ is
$ P_{	ext{per unit area}} = \sigma T^4 . $
The total surface area of the star:
$ A_{	ext{star}} = 4 \pi r^2 . $
So, total radiant power (luminosity) is:
$ L = 4 \pi r^{2} \sigma T^4 . $
Step 2: Power distributed over spherical surface at distance $R$
At distance $R$, this power is isotropically spread across a sphere of radius $R$:
$ A_{R} = 4 \pi R^2 . $
The intensity (power per unit area normal to incidence) is:
$ I = \frac{L}{4 \pi R^2}  = \frac{4 \pi r^2 \sigma T^4}{4 \pi R^2}  = \frac{\sigma r^2 T^4}{R^2}. $
Step 3: Match option
That corresponds to:
Option A:
$ \boxed{\dfrac{\sigma r^2 T^4}{R^2}} $
Ã¢ÂÂ Correct Answer:
Option A

Answer

$\frac{\sigma \mathrm{r}^2 \mathrm{~T}^4}{\mathrm{R}^2}$

Answer

$\frac{\sigma r^2 T^4}{4 \pi R^2}$

Answer

$\frac{\sigma \mathrm{r}^2 \mathrm{~T}^4}{\mathrm{R}^4}$

Answer

$\frac{4 \pi \sigma r^2 T^4}{R^2}$

Thermodynamics & KTG

The outer surface of star in the form of a sphere radiates heat as a black body at temperature ' T '. The total radiant energy per unit area, normal to the direction of incidence, received at a distance ' R ' from the centre of a star of radius ' r ' is (R>r)(\sigma= Stefan's constant )

Choose the correct answer: