Feynman -- Atomic dipole

  • Thread starter Seidhee
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Hello,

I am reading the volume 2 of the Feynman's Lectures on Physics, and something is bothering me when he calculates the dipole moment of a single atom induced by an extern field :

https://books.google.co.uk/books?id=uaQfAQAAQBAJ&pg=SA11-PA2&lpg=SA11-PA2&dq=feynman+dipole+single+atom&source=bl&ots=6nmZCqHDMk&sig=cd6wOGCKh9E9227ZX-a8KKhTqlM&hl=fr&sa=X&ei=NuI8VbHdI8v9UOSvgOAD&ved=0CDEQ6AEwAQ#v=onepage&q=feynman dipole single atom&f=false

Indeed, he states that : " p = qex "

But why ? I would use in general : " p = Zqex " where Z is the number of electrons in the atom.

x is the displacement of the center of charges of the electrons, and thus x is also the displacement of each electron.

Could you explain his reasoning ? It is not the first time he uses a single electron charge instead of Z in his calculations, and I do not understand.

Thanks.



PS : First, I thought that was because the square of the natural pulsation ω0 depended on Z, which means that ω²0(Z) = Zω²0 (Z=1), which would simplify the Z replacing ω²0(Z) by Zω²0 (Z=1) ; but Feynman seems to use ω0 = ω0(Z) and not ω0 (Z=1) everywhere, so it does not matter.
 

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  • #2
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And qe is obviously equal to the charge of an electron and note equal to Z*(charge), because he uses then e² = q²e/4piε0.

Moreover, when he the does his calculations for the Helium atom, he keeps using qe = charge of one electron and does not use a new value for Z..
 

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