Homework #5 Solutions Spring 2014 1.a fcc lattice: The distance between nearest neighbors in the fcc lattice is just a22. Thus the radius of spheres that just touch will be half this value: a24. The volume of such a sphere is 43πr3=43πa24⎛⎝⎜⎞⎠⎟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 (a3) is: 443πr3a3=443πa24⎛⎝⎜⎞⎠⎟3a3=π26b. bcc lattice: The distance between nearest neighbors in the bcc lattice is just 3a2. Thus the radius of spheres that just touch will be half this value: a34. The volume of such a sphere is 43πr3=43πa34⎛⎝⎜⎞⎠⎟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 (a3) is: 243πr3a3=243πa34⎛⎝⎜⎞⎠⎟3a3=π38c. 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: a2. The volume of such a sphere is 43πr3=43πa2⎛⎝⎜⎞⎠⎟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 (a3) is: 43πr3a3=43πa2⎛⎝⎜⎞⎠⎟3a3=π6d. 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 a34. Thus the radius of spheres that just touch will be half this value: a38. The volume of such a sphere is 43πr3=43πa38⎛⎝⎜⎞⎠⎟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 (a3) is: 843πr3a3=843πa38⎛⎝⎜⎞⎠⎟3a3=π316
2. We know that for a lattice with a density of lattice points, n, nv=1where v is the volume of the primitive cell. If a2and a3are primitive lattice vectors, they form a plane with a planar density of lattice points: σ=1a2×a3since there is only one lattice point per primitive cell. The volume of the lattice primitive cell is a1i(a2×a3)=a2×a3a1cosθ=a2×a3(a1iˆg)=v, where ˆgis a unit vector normal to the plane formed by a2and a3. However, a1iˆgis just the spacing between lattice planes, d. So we find that: σ=1a2×a3=a1iˆgv=dvb. 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=2πG, to maximize d we have to find the shortest reciprocal lattice vector of the lattice.