
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: 


d_{E}^{ } = 
4 pe_{0} · 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 offhand 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
w_{0} of a harmonic oscillator with
the mass m and the spring constant k_{S} is
given by 


w_{0} 
= 
æ
ç
è 
k_{S}
m_{ } 
ö
÷
ø 
1/2 




How large are the spring constants? That is question already
answered in the backbone, so we import the equation 


k_{S} = 
æ
ç
è 
(ze) ^{2}
4 pe_{0} ·
R ^{3}

ö
÷
ø 



Again, let's make a table for the
answers: 


Atom 
Spring
constant 
w_{0} 
z = 1 


z = 50 


z = 100 


