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In the Feynman Lectures on Physics Vol. I Ch. 32 (Radiation Damping. Light Scattering) it says:

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.

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He is referring to the second paragraph of 31-3:

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.

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  • $\begingroup$ How can one deduce a natural frequency (which depends on oscillator reset force, as far as I know, but I am no physicist) from the radius of the atom containing the electron? $\endgroup$ – mak Jul 9 '16 at 14:13
  • $\begingroup$ It's basically just dimensional analysis. For the hydrogen atom, you have three dimensionful quantities: planck's constant, the charge of the electron, and the reduced mass of the electron proton system (which is basically the mass of the electron), since the hydrogen atom can be treated non-relativistically. there is only one way to get a length and this length is the bohr radius. There is only one way to convert the length to a frequency and this gives you the oscillation frequency. $\endgroup$ – Brian Moths Jul 9 '16 at 22:43
  • $\begingroup$ Now the conversion of this frequency to a wavelength involves the speed of light, which is a different speed from the speed you can form from planck's constant, electron mass and electron charge. However, the difference is only a factor of about 137, so the wavelength of light is not a half an angstrom but more like 10-100 nanometers, but this is still smaller than visible light. $\endgroup$ – Brian Moths Jul 9 '16 at 22:47

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