Answer:
[tex]\lambda=5.46\times 10^{-8}\ m[/tex]
Explanation:
The hottest ordinary star in our galaxy has a surface temperature of 53,000 K.
We need to find the peak wavelength of its thermal radiation.
Using Wein's law,
[tex]\lambda T=2.898\times 10^{-3}\\\\\lambda=\dfrac{2.898\times 10^{-3}}{53000}\\\\=5.46\times 10^{-8}\ m[/tex]
So, the peak wavelength of its thermal radiation is equal to [tex]5.46\times 10^{-8}\ m[/tex].
