If the radius of a star is `R` and it acts as a black body, what would b the temperature of the star, in which the rate of energy production is `Q`?

Options

A. `Q/4piR^(2)sigma`
B. `(Q/4piR^(2)sigma)^(-1/2)`
C. `(4piR^(2)Q//sigma)^(1/4`
D. `(Q//4piR^(2)sigma)^(1/4)`

Explanation

To find the temperature of a star that acts as a black body, given its radius \( R \) and the rate of energy production \( Q \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Black Body Radiation:**
A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. According to Stefan-Boltzmann law, the power radiated by a black body per unit area is proportional to the fourth power of its absolute temperature \( T \):
\[ P = \sigma T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant.

2. **Calculate the Surface Area of the Star:**
The star can be approximated as a sphere. The surface area \( A \) of a sphere with radius \( R \) is given by:
\[ A = 4\pi R^2 \]

3. **Relate the Rate of Energy Production to Temperature:**
The total power (or rate of energy production) radiated by the star can be expressed as the product of the power per unit area and the surface area:
\[ Q = P \times A \]
Substituting the expressions for \( P \) and \( A \):
\[ Q = \sigma T^4 \times 4\pi R^2 \]

4. **Rearranging the Equation to Solve for Temperature:**
To find the temperature \( T \), we can rearrange the equation:
\[ T^4 = \frac{Q}{4\pi R^2 \sigma} \]
Taking the fourth root of both sides gives:
\[ T = \left(\frac{Q}{4\pi R^2 \sigma}\right)^{\frac{1}{4}} \]

5. **Final Expression for Temperature:**
Thus, the temperature of the star is given by:
\[ T = \left(\frac{Q}{4\pi R^2 \sigma}\right)^{\frac{1}{4}} \]

### Final Answer:
The temperature of the star is:
\[ T = \left(\frac{Q}{4\pi R^2 \sigma}\right)^{\frac{1}{4}} \]