2.4.4 Schwingungsfrequenz der Atome in Kristallen

Note: Youngs Modulus (= Elastizitätsmodul) is abbreviated with an "E" in this text; not with a "Y" as is customary in the English literature.
 
We have already employed the picture of an atom or particle oscillating (or vibrating) in its potential well. Now we shall compute the vibration frequency w = 2pn from the binding potential.
As long as the potential increases quadratically with the distance from the equilibrium position ro, the restoring force will be proportional to the deviation x = r - ro from ro; and we have a simple harmonic oscillator.
The harmonic approximation is good enough for getting an order of magnitude estimate of the vibration frequency; i.e. we simply replace the proper potential by its Taylor expansion around ro and stop after the quadratic term. We already did that; we had
U  =  U0 + 1/2U0''  · x2  
       
and  
       
U''(r0)  =  U0 · (nm/r02)  
The basic equation for oscillations in this potential that we have to solve is
ma ·  d2x
dt2
 +  ks · x  =  0
with ma = mass of the vibrating particle (we use the symbol ma instead of m to avoid confusion with the exponent m in the potential equation). In this formulation we also used a "spring constant" ks in order to be able to compare the solutions with standard formulations of classical mechanics.
The resonance frequency w of the system is known from standard mechanics; it is
w  =  æ
ç
è
ks
ma
ö
÷
ø
1/2
(Try it; all you have to do is to see of the solution x = x0cos wt is a solution for the differential equation above).
While for a real oscillator there will always be some friction (or better energy dispersion); i.e. a term kf · dx/dt, we do not have to worry about that because friction does not change the resonance frequency. If you want to know more about this, use the link.
We know the or restoring force Fres of our system, it is simply
Fres  =  –  dU
dx
=  – U0''  · x = U0 · (nm/r02) · x
The spring constant thus is simply ks = U0 · (nm/r02), and the resonance frequency is
w = æ
ç
è
U0 · (nm/r02)
ma
ö
÷
ø
½ = 1
r0
æ
ç
è
U0 · n · m
ma
ö
÷
ø
½
While this is good enough, we remember that we had the second derivative of the potential at some other occasion: When we found a formula for Youngs modulus E.
What we had was
E  =   1
· d2U
  =  n · m ·U0
r0 dr2 r03
It is easy enough to use E instead of the spring constant, we have
ks  =  U0 · n · m
r02 
 =   E · r0
Which gives
w  = æ
ç
è
E · r0
ma
ö
÷
ø
1/2
The vibration frequency of an atom in a lattice thus will be determined - approximately - by the easily obtainable quantities Youngs modulus, lattice constant and mass of the atom. Lets see what we get for some examples
Lets take Silicon. We have
E = 150 GPa = 1,5 · 1011 N/m2   w = 8,4 · 1013 Hz

n = 1,34 · 1013 Hz
ma = 31 · 1,67 · 10–27 kg     Þ
r0 = 0,31 nm = 3,1 · 10–10 m  
That is very satisfactory because it gives us the common result, always just claimed without justification, that the vibration frequency of atoms in a lattice is in the order of 1013 Hz.
That the vibration frequency of atoms in a solid is in the order of n » 1013 Hz is a number we will commit to memory now, and which we will never forget!
Is a frequency of 1013 Hz large or small? Dumb question, you always have to add "In relation to what"?
In electrical engineering, the highest frequencies "commonly" employed are in the (1 - 100) GHz = 109 Hz - 1011 Hz "Microwave" range. However, there is a lot of excitement about novel devices in the "Terahertz" (= THz = 1012 Hz) region. Our atoms, however, vibrate still faster - but not much.
What is the frequency of visible light? Easy. We know its energy E = hn, and we must know that the energy of visible light is in the 1 eV region. It's actually a bit higher, 1 eV is still infrared, but it is good enough for our purpose. With h = 4.13 · 10–15 eV·s (look it up!), we get nlight » 2 · 1014 Hz. So our atoms are a bit slower, but 1013 Hz is a rather large frequency, indeed.

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