An atom is situated in a simple cubic crystal lattice. If the potential energy o

Question

An atom is situated in a simple cubic crystal lattice. If the potential energy of interaction between any two atoms is of the form . See the approximation formulas in appendix D.] , with similar expressions for where where A and B are constants. (Note: Assume that the six neighboring atoms are fixed and are located at the points (d,0,0), (0,d,0), (0,0,d), and that the displacement (x,y,z) of the given atom from the equilibrium position (0,0,0) is small compared to d. Then , where c and a are constants and r is the distance between the two atoms, show that the total energy of interaction of a given atom with its six nearest neighbors is aproximately that of the three-dimensional harmonic oscillator potential

Explanation / Answer

Potential energy between two atoms is

    V(r) = Cr^(-)

If d is the distance of equilibrium between two atoms, then V(r) is minimum when V(r)= 0.

We need to show thta dV/ dx , dV/dy, dV/ dz are each zeros.

V(r) = c { [ (d -x)^2 + y^2 + z^2 ]^(-/2) ] + [ (d +x)^2 + y^2 + z^2 ]^(-/2) ] + [ x^2 + (d-y)^2 + z^2 ]^(-/2) ] +[ x^2 + (d +y)^2 + z^2 ]^(-/2) ] + [ x^2 + y^2 + (d-z)^2 ]^(-/2) ]+[ x^2 + y^2 + (d +z)^2 ]^(-/2) ] }

So dV/ dx = - ( c /2) { [ (d -x)^2 + y^2 + z^2 ]^(-/2)-1 ][-2(d+x)] - [ (d +x)^2 + y^2 + z^2 ]^(-/2)-1 ] [ 2(d +x) ] - [ x^2 + (d-y)^2 + z^2 ]^(-/2)-1 ][2x] -[ x^2 + (d+y)^2 + z^2 ]^(-/2) -1][2x] - [ x^2 + y^2 + (d-z)^2 ]^(-/2)-1 ][2x] + [ x^2 + y^2 + (d+z)^2 ]^(-/2)-1 ][2x]

   dV /dx = 0

Similarlay we getr dV/dy and dv /dz = 0 , this proves equilibrium.

Second derivates at x= y = z = 0

d^2V/dx^2 = - c [ ( /2) +1 ](d^(- -4)4^2 - 2cd^(--2) - 4cd^(--2)

                  = 2c (-1)d^(--2)

Similarlaly

d^2V/dy^2 = 2c (-1)d^(--2)

d^2V/dz^2 = 2c (-1)d^(--2)

So the potential can be approxmated by V(r) = A + B(x^2+y^2+z^2)

here A = 6cd^-   and B = c ( -1)d^(--2)

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