# Solution to Exercise 3.2-3

How large will be the distance d between the (center of gravity) of the positive and negative charges for reasonable field strengths and atomic numbers, e.g. the combinations of
• 1 kV/cm
• 100 kV/cm
• 10 MV/cm
• , the last one being about the ultimate limit for the best dielectrics there are,
• z = 1 (H, Hydrogen)
• z = 50 (Sn (= tin), ...)
• z = 100 (?)

From the backbone we have a relation for d as a function of z,m the radius R of the atom, and the field strength E:
dE   =   4 pe0 · R 3 · E
ze
We need to look up some number for the radius of the three atoms given (try this link), then the calculation is straight forward - let's make a table:
 Atom R d(1 kV/cm) d(100 kV/cm) d(10 MV/cm) z = 1 z = 50 z = 100
Compared to the radius of the atoms, the separation distance is tiny. No wonder, electronic polarization is a small effect with spherical atoms!

Calculate the "spring constant" and from that the resonance frequency of the "electron cloud" (assume the nucleus to be fixed in space).

If you don't know off-hand the resonance frequency of a simple harmonic oscillator - that's fine. If you don't know exactly what that is, and where you can look it up - you are in deep trouble.
Anyway, in this link you get all you need. In particular the resonance (circle) frequency w0 of a harmonic oscillator with the mass m and the spring constant kS is given by
w0  = æ ç è ö ÷ ø
How large are the spring constants? That is question already answered in the backbone, so we import the equation
 kS =   (ze) 2 4 pe0 · R 3 æ ç è ö ÷ ø
Again, let's make a table for the answers:
 Atom Spring constant w0 z = 1 z = 50 z = 100

Exercise 3.2-3

Harmonic Oscillator

© H. Föll (Electronic Materials - Script)