Homework5_Solutions - Homework#5 Solutions Spring 2014 1.a fcc lattice The distance between nearest neighbors in the fcc lattice is just spheres that

# Homework5_Solutions - Homework#5 Solutions Spring 2014 1.a...

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Homework #5 Solutions Spring 2014 1.a fcc lattice: The distance between nearest neighbors in the fcc lattice is just a 2 2 . Thus the radius of spheres that just touch will be half this value: a 2 4 . The volume of such a sphere is 4 3 π r 3 = 4 3 π a 2 4 3 . There are 4 such spheres per fcc unit cell so that the ratio of the volume occupied by the spheres to the volume of total fcc cubic cell ( a 3 ) is: 4 4 3 π r 3 a 3 = 4 4 3 π a 2 4 3 a 3 = π 2 6 b. bcc lattice: The distance between nearest neighbors in the bcc lattice is just 3 a 2 . Thus the radius of spheres that just touch will be half this value: a 3 4 . The volume of such a sphere is 4 3 π r 3 = 4 3 π a 3 4 3 . There are 2 such spheres per bcc unit cell so that the ratio of the volume occupied by the spheres to the volume of total bcc cubic cell ( a 3 ) is: 2 4 3 π r 3 a 3 = 2 4 3 π a 3 4 3 a 3 = π 3 8 c. Simple Cubic Cell:
The distance between nearest neighbors in the simple cubic lattice is just a . Thus the radius of spheres that just touch will be half this value: a 2 . The volume of such a sphere is 4 3 π r 3 = 4 3 π a 2 3 . There is one sphere per simple cubic lattice unit cell so that the ratio of the volume occupied by the spheres to the volume of total simple cubic cell ( a 3 ) is: 4 3 π r 3 a 3 = 4 3 π a 2 3 a 3 = π 6 d. diamond lattice: The diamond lattice is an fcc lattice with a basis of two atoms located at (0,0,0) and a/4(1,1,1). The basis atoms are nearest neighbors separated by a distance a 3 4 . Thus the radius of spheres that just touch will be half this value: a 3 8 . The volume of such a sphere is 4 3 π r 3 = 4 3 π a 3 8 3 . There are 8 such spheres per fcc unit cell so that the ratio of the volume occupied by the spheres to the volume of total diamond (fcc) cubic cell ( a 3 ) is: 8 4 3 π r 3 a 3 = 8 4 3 π a 3 8 3 a 3 = π 3 16
2. We know that for a lattice with a density of lattice points, n, nv = 1 where v is the volume of the primitive cell. If a 2 and a 3 are primitive lattice vectors, they form a plane with a planar density of lattice points: σ = 1 a 2 × a 3 since there is only one lattice point per primitive cell. The volume of the lattice primitive cell is a 1 i ( a 2 × a 3 ) = a 2 × a 3 a 1 cos θ = a 2 × a 3 ( a 1 i ˆ g ) = v , where ˆ g is a unit vector normal to the plane formed by a 2 and a 3 . However, a 1 i ˆ g is just the spacing between lattice planes, d. So we find that: σ = 1 a 2 × a 3 = a 1 i ˆ g v = d v b. Given the expression above for the density of lattice points in a plane, to maximize this density, we must maximize the spacing between lattice planes. If we think of the spacing between lattice planes as being determined by the magnitude of a reciprocal lattice vector, d = 2 π G , to maximize d we have to find the shortest reciprocal lattice vector of the lattice.

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