**The size of the star**– The larger a star is, the more energy it puts out and the more luminous it is.

**The temperature of the star**– If two stars are of the same size but have different temperatures, then the star with higher temperature will be more luminous than the star with lower temperature.

**The Brightness of a Star:**

The brightness of a star which appears to our naked eyes mainly depends upon two factors:

- Luminosity of the star
- Distance of the star from the Earth

**\(B\propto \frac{1}{d^{2}}\)**

**Luminosity Theory:**

- Luminosity depends on the surface area of the star. If radius of a star is R then,

**Surface area of the star Â = 4PR****2**

- Two stars having the same temperature, one with radius 2R will have 4 times greater luminosity than a star with radius R.

- The luminosity of a star also depends upon its temperature. When considering a star to be a completely black body, the radiation emitted per second will be according to the Stefan- Boltzmann law.

**Energy emitted per second (E) = sAT**

**4**

Where,

s= Stefanâ€™s constant with a value of **\(5.7\times10^{-8}Wm^{-2}K^{-4}\)**

A= Surface Area of the Star

T = absolute temperature of the star

- Calculating the energy output for a star which is of the same size as the sun.

R = 6.96x108m and T = 6000K

So, surface area of sun = surface area of star = A = **\(4\pi r^{2}\)**

**\(= 4\times \pi \times ( 6.96\times 10^{8})\\ \\ =6.08\times 10^{18}\)**

Therefore, **\(E=5.7\times10^{-8}\times6.08\times10^{18}\times6000^{4}=4.5\times10^{26}J\)**

A star having twice the temperature of Sun would have 16 times greater energy output.

**Example** – The Sun having a surface temperature of 6000K produces radiation with lmax = 420 nm. Find out the temperature of the Sirius if l max of Sirius is 72nm.

**Answer**–

**Â \(l_{max1}T_{1}=l_{max2}T_{2}\)**

Therefore,

**\(420\times10^{-9}\times6000=72\times10^{-9}\times T_{2}\)**

**\(T_{2}=35000K\)**