as a simplified model of a crystal, consider three atoms lying along a line, with a distance b between the adjacent atoms . the atoms each have mass m and a charge q. suppose the middle charge is displaced a very small distance x from its equilibrium position and then released. show that the net electrical force on the displaced charge is given approximately F=(q² /b³ piEfslon₀ )x when x<B) the frequency of vibration of the displaced charge in terms of its crystal parameters (q,b, and m)
C) if the atoms r singly ionized carbon and are separated by an equilibrium distance of 4X10^-10 m, what is the numerical value of their vibration frequency?

For simplicity we consider, first, a one-dimensional crystal lattice and assume that the forces between the atoms in this lattice are proportional to relative displacements from the equilibrium positions. Fig.1 This is known as the harmonic approximation, which holds well provided that the displacements are small. One might think about the atoms in the lattice as interconnected by elastic springs. Therefore, the force exerted on n-the atom in the lattice is given by 1 1 ( ) ( ) Fn C n n n n u u C u u = + − + − − , (5.1) where C is the interatomic force (elastic) constant. Applying Newton’s second law to the motion of the n-th atom we obtain 2 2 1 1 1 1 ( ) ( ) (2 ) n n n n n n n n n d u M F C u u C u u C u u u dt = = + − + − − = − − + − − , (5.2) where M is the mass of the atom. Note that we neglected here by the interaction of the n-th atom with all but its nearest neighbors. A similar equation should be written for each atom in the lattice, resulting in N coupled differential equations, which should be solved simultaneously (N is the total number of atoms in the lattice). In addition the boundary conditions applied to the end atom in the lattice should be taken into account. Now let us attempt a solution of the form ( ) n i qx t n u Ae −ω = (5.3) where xn is the equilibrium position of the n-th atom so that xn=na. This equation represents a traveling wave, in which all the atoms oscillate with the same frequency ω and the same amplitude A and have wavevector q. Note that a solution of the form (5.3) is only possible because of the transnational symmetry of the lattice. Now substituting Eq.(5.3) into Eq.(5.2) and canceling the common quantities (the amplitude and the time-dependent factor) we obtain 2 ( 1) ( 1) ( ) 2 iqna iqna iq n a iq n a M ω e C e e e + − − = −
− −  (5.4) This equation can be further simplified by canceling the common factor iqna e , which leads to n−1 n n+1 a un−1 un un+1 Physics 927 E.Y.Tsymbal 2 ( ) ( ) 2 2 2 2 1 cos 4 sin 2 iqa iqa qa Mω C e e C qa C − = − − = − = . (5.5) We find therefore the dispersion relation for the frequency 4 sin 2 C qa M ω = , (5.6) which is the relationship between the frequency of vibrations and the wavevector q. This dispersion relation have a number of important properties