Question

If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is $ Q ? $ ($ \sigma $ stands for Stefan's constant)

• A. $\frac{Q}{ 4 \pi R^2 \sigma}$
• B. $ \bigg(\frac{Q}{ 4 \pi R^2 \sigma} \bigg) ^{-1/2}$
• C. $ \bigg(\frac{4 \pi R^2 Q}{ \sigma} \bigg)^{1/4}$
• D. $ \bigg(\frac{Q}{ 4 \pi R^2 \sigma} \bigg)^{1/4}$

Answer 1

The Correct Option is D

To find the temperature of a star acting as a black body, we can use the Stefan-Boltzmann law, which states that the power emitted per unit area by a black body is proportional to the fourth power of its temperature. Mathematically, it is expressed as:

E = \sigma T^4

where E is the energy radiated per unit area, T is the temperature of the black body, and \sigma is Stefan's constant.

If the star is a sphere with radius R, the surface area A is given by:

A = 4\pi R^2

The total energy emitted by the star per unit time (rate of energy production) is Q. Thus, we have:

Q = A \cdot E = 4\pi R^2 \cdot \sigma T^4

Rearranging this equation to solve for T, we get:

T^4 = \frac{Q}{4\pi R^2 \sigma}

Taking the fourth root of both sides, we find:

T = \left(\frac{Q}{4\pi R^2 \sigma}\right)^{1/4}

Therefore, the correct answer is:

\left(\frac{Q}{4\pi R^2 \sigma}\right)^{1/4}