The surface temperature of the sun is `T_(0)` and it is at average distance `d` from a planet. The radius of the sun is `R`. The temperature at which planet radiates the energy is

Options

A. `T_(0)sqrt((R)/(2d))`
B. `T_(0)sqrt((2R)/(d))`
C. `T_(0)sqrt((R)/(d))`
D. `T_(0)((R)/(d))^(1//4)`

Answer

To solve the problem, we need to find the temperature at which the planet radiates energy, given the surface temperature of the sun \( T_0 \), the radius of the sun \( R \), and the average distance from the sun to the planet \( d \).

### Step-by-Step Solution:

1. **Understanding the Concept of Thermal Equilibrium**:
- The planet is in thermal equilibrium with the sun, meaning the power radiated by the sun is equal to the power absorbed by the planet.

2. **Power Radiated by the Sun**:
- According to the Stefan-Boltzmann law, the power \( P \) radiated by a black body is given by:
\[ P = \sigma A T^4 \]
- For the sun, the area \( A \) is the surface area of a sphere, which is \( 4\pi R^2 \). Therefore, the power radiated by the sun is:
\[ P_{\text{sun}} = \sigma (4\pi R^2) T_0^4 \]

3. **Power Received by the Planet**:
- The planet is at a distance \( d \) from the sun, so the power received by the planet can be calculated using the same Stefan-Boltzmann law. The effective area from which the planet receives the solar radiation is \( 4\pi d^2 \):
\[ P_{\text{planet}} = \sigma (4\pi d^2) T^4 \]

4. **Setting the Powers Equal**:
- Since the planet is in thermal equilibrium, we set the power radiated by the sun equal to the power received by the planet:
\[ \sigma (4\pi R^2) T_0^4 = \sigma (4\pi d^2) T^4 \]

5. **Canceling Common Terms**:
- We can cancel \( \sigma \) and \( 4\pi \) from both sides:
\[ R^2 T_0^4 = d^2 T^4 \]

6. **Rearranging the Equation**:
- Rearranging gives us:
\[ T^4 = \frac{R^2}{d^2} T_0^4 \]

7. **Taking the Fourth Root**:
- To find \( T \), we take the fourth root of both sides:
\[ T = \left( \frac{R^2}{d^2} \right)^{1/4} T_0 = \frac{R^{1/2}}{d^{1/2}} T_0 \]

### Final Answer:
The temperature at which the planet radiates energy is:
\[ T = \frac{R^{1/2}}{d^{1/2}} T_0 \]