Question

A main sequence star has a a peak wavelength of 600 nm. it's apparent brightness is 12 x 10^-12
what is a star's temperature ?
what is the special class and Luminosity of the star?
how far away is the star?

Answer 1

Q: What is the star's temperature?

A: Use Wein's law to answer this question: 

$$T=\frac{b}{\lambda_\text{peak}}=\frac{2.898\cdot10^{-3}}{600\cdot10^{-9}}=4830\text{ K}.$$

Q: What is the special class and Luminosity of the star? 

A: According to the Hertzsprung–Russell diagram, 4830 K corresponds to the spectral class G.

Since this is a main sequence star, its luminosity is 0.1 that of the Sun according to the same diamgram. We know that the brightness of the Sun is 1370 watts/meter². Therefore, the we can calculate how many times the Sun is brighter than our star:

$$\frac{b_1}{b_2}=\frac{1370}{12\cdot10^{-12}}=1.14\cdot10^{14}.$$

Also, we know that the difference in apparent magnitudes $m_2-m_1=1$ corresponds to a ratio of brightness $b_1/b_2=2.512$. The equation for this is

$$\frac{b_1}{b_2}=\sqrt[5]{100^{(m_2-m_1)}}.$$

Therefore, since we know the apparent magnitude of the Sun, we can calculate the apparent magnitude of our star. Just express $m_2$ from the previous equation:

$$m_2-m_1=\frac{5\text{ln}(b_1/b_2)}{\text{ln}100}=2.5\text{ lg}(b_1/b_2),\\\space\\ m_2=−26.7+2.5\text{ lg}(1.14\cdot10^{14})=8.4.$$

Quite a dim star. An unequipped human eye will not be able to observe this star on the night sky.

Now, since we know the luminosity of the Sun, find the luminosity of our star:

$$L_2=0.1L_1=0.1\cdot3.828\cdot10^{26}=3.828\cdot10^{25}\text{ W}.$$

Q: How far away is the star?

A: The distance $d$ can be found since we know the luminosity $L$ and apparent brightness $b$ of the star:

$$L=b\cdot4\pi d^2,\\ d=\frac{1}{2}\sqrt{\frac{L}{\pi b}},\\\space\\ d=\frac{1}{2}\sqrt{\frac{3.828\cdot10^{25}}{3.142\cdot10\cdot10^{-12}}}=5.038\cdot10^{17}\text{ m},$$

or about 53 light years.