# Light scattering: Why is the natural frequency in air higher than the light frequencies?

On the other hand, if we take the case of light in the air, we remember that for air the natural frequencies of the oscillators are higher than the frequency of the light that we use.

Why is that?

I could not find the answer elsewhere in the book.

However, we shall discuss the formula we have obtained, in various possible circumstances. First of all, for most ordinary gases (for instance, for air, most colorless gases, hydrogen, helium, and so on) the natural frequencies of the electron oscillators correspond to ultraviolet light. These frequencies are higher than the frequencies of visible light, that is, $ω_0$ is much larger than $ω$ of visible light, and to a first approximation, we can disregard $ω^2$ in comparison with $ω^2_0$. ...
One reason why you would suspect this to be the case is that the orbital radius of an electron in a nitrogen or oxygen atom, which is on the order of $10^{-10}$ meters, is much smaller than the wavelength of visible light, which is on the order of $10^{-7}$ to $10^{-6}$ meters. Since the length scale of visible light is smaller, and frequency has an inverse relationship to wavelength, the frequency must be bigger.