# Hot objects with temperatures of 10 million Kelvin give off most of their radiation in which part of the electromagnetic spectrum?

Jun 14, 2017

Well, this can be estimated with Wien's Displacement Law:

${\lambda}_{\max} = \frac{2.8977729 \times {10}^{- 3} \text{m"cdot"K}}{T}$,

assuming your hot object of interest is a blackbody (not at all related to a woman!), i.e. an opaque body that does not reflect any light.

${\lambda}_{\max}$ gives the peak wavelength of a blackbody's radiation curve as a function of surface temperature.

So, if we assume the surface temperature of the object is ${10}^{7}$ $\text{K}$, then its maximum wavelength is approximately:

$\textcolor{b l u e}{{\lambda}_{\max}} \approx \left(2.8977729 \times {10}^{- 3} \text{m"cdot"K")/(10^7 "K}\right)$

$= 2.90 \times {10}^{- 10} \text{m}$

$=$ $\textcolor{b l u e}{\text{0.290 nm}}$,

which is around the X-ray region, with wavelength range $0.01 - \text{10 nm}$.