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 r_o, the restoring force will be proportional to the deviation x = r - r_o from r_o; 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 r_o 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

m_a ·	d²x dt²	+ k_s · x = 0

with m_a = mass of the vibrating particle (we use the symbol m_a instead of m to avoid confusion with the exponent m in the potential equation). In this formulation we also used a "spring constant" k_s 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	k_s m_a	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 k_f · 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 F_res of our system, it is simply

F_res = –	dU dx	= – U₀'' · x = U₀ · (nm/r₀²) · x

The spring constant thus is simply k_s = U₀ · (nm/r₀²), and the resonance frequency is

w =	æ ç è	U₀ · (nm/r₀²) m_a	ö ÷ ø	½	=	1 r₀	æ ç è	U₀ · n · m m_a	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
E =	r₀	·	dr²	=	r₀³

It is easy enough to use E instead of the spring constant, we have

k_s	=	U₀ ·	n · m r₀²	=	E · r₀

Which gives

w	=	æ ç è	E · r₀ m_a	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
m_a = 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 n_light » 2 · 10¹⁴ Hz. So our atoms are a bit slower, but 10¹³ Hz is a rather large frequency, indeed.