Question

The hottest ordinary star in our galaxy has a surface temperature of 53,000 K. Part A What is the peak wavelength of its thermal radiation

Answer 1

`\lambda=5.46* 10^(-8)\ m`

Step-by-step 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,

`\lambda T=2.898* 10^(-3)\\\\\lambda=(2.898* 10^(-3))/(53000)\\\\=5.46* 10^(-8)\ m`

So, the peak wavelength of its thermal radiation is equal to
`5.46* 10^(-8)\ m`.