Question

The total radiant energy per unit area, normal to the direction of incidence, received at a distance $R$ from the center of a star of radius $r$, whose outer surface radiates as a black body at a temperature $T K$ is given by (where $\sigma$ is Stefan's constant)

• A. $\frac{\sigma r^2T^4}{R^2}$
• B. $\frac{\sigma r^2T^4}{4\pi r^2}$
• C. $\frac{\sigma r^4T^4}{r^4}$
• D. $\frac{4\pi\sigma r^2T^4}{R^2}$

Answer 1

The Correct Option is A

The problem requires us to determine the total radiant energy per unit area, in the direction of incidence, received from a star at a given distance.

1. **Understanding Black Body Radiation:**
A star can be approximated as a black body that emits radiation according to Stefan-Boltzmann law. The power P emitted per unit area of the star's surface is given by:

P = \sigma T^4

where \sigma is the Stefan's constant and T is the temperature in Kelvin.

2. **Calculating Total Power Emitted by the Star:**
The total power emitted by the star can be calculated by multiplying the power per unit area by the surface area of the star (which is a sphere with radius r):

P_{\text{total}} = \sigma T^4 \times 4\pi r^2

3. **Power Received at a Distance R:**
The power is radiated uniformly in all directions, forming a sphere of radius R. The surface area of this sphere is 4\pi R^2.

4. **Deriving Radiant Energy Per Unit Area at Distance R:**
The intensity (or radiant energy per unit area) received at the distance R from the star's center is the total power emitted divided by this sphere’s surface area:

I = \frac{P_{\text{total}}}{4\pi R^2} = \frac{\sigma T^4 \times 4\pi r^2}{4\pi R^2}

Upon simplifying, we get:

I = \frac{\sigma r^2 T^4}{R^2}

5. **Conclusion:**

The correct answer is \frac{\sigma r^2 T^4}{R^2} which represents the intensity of radiation received per unit area normal to the direction of incidence.

This matches the provided correct answer option: \frac{\sigma r^2 T^4}{R^2}