Question on a BCC lattice?

Question

If the lattice parameter in a BCC lattice is 1.9x10E-10m, calculate the atomic positions in the {110} plane taking the lower left atom as the origin.

How would you draw the reciprocal lattice and indicate the primitive lattice vectors chosen and justify the magnitude and direction of the reciprocal vector.

Answer 1

This one takes me back a bit!

The BCC or body centred-cubic lattice is a Bravais lattice such that if the original cubic lattice is generated by primitive vectors: -

ax, ay, az

where x, y, and z are three orthogonal unit vectors, then a set of primitive vectors for the bcc lattice may be written down as: -

a(1) = ax, a(2) = ay, a(3) = (a/2).(x + y + z)

Where the brackets '()' indicate subscripts.

These primitive vectors may be written in a more symmetric form: -

a(1) = (a/2).(y + z - x)

a(2) = (a/2).(z + x - y)

a(3) = (a/2).(x + y - z)

For a set of points 'R' representing a Bravais lattice and a plane wave e^(ik.r). The set of all wave vectors 'K' that yield plane waves with the periodicity of a given Bravais lattice is known as its reciprocal lattice. The reciprocal lattice is also a Bravais lattice.

From the primitive vectors a(1), a(2), a(3) the reciprocal lattice can be generated as follows: -

b(1) = 2π((a(2).a(3)/(a(1).(a(2).a(3))))

b(2) = 2π((a(3).a(1)/(a(1).(a(2).a(3))))

b(3) = 2π((a(1).a(2)/(a(1).(a(2).a(3))))

The b(i) satisfy the condition: -

b(i).a(j) =  2πδ(ij)

Where 'δ(ij)' is the Kronecker delta symbol.

Any vector k may be written as a linear combination of the b(i), such that: -

k = k(1).b(1) + k(2).b(2) + k(3).b(3)

If R is the direct lattice vector then: -

R = n(1).a(1) + n(2).a(2) + n(3).a(3)

Whence n(i) are integers, thus: -

k.R =  2π.(n(1).k(1) + n(2).k(2) + n(3).k(3))

This, finally brings us to the Miller indicies h, k, l (here <110>). Thus, a plane with Miller indices h, k, l is normal to the reciprocal lattice vector

h.b(1) + k.b(2) + l.b(3)

If K.r = A  or the lattice parameter (constant = 1.9 x 10^-10 m) then: -

x(1) = A/2πh

x(2) = A/2πk

x(3) = A/2πl

Hence, when drawn in 3d the plane is a rectangle spaning the x, y axis, with one intercept corner on each axis (l=0 <110>).

I hope this long winded explaination is of some assistance!

P.S I should have added that the atomic spacing is the sum (Pythagoras): -

spacing = √(x(1)² + x(2)²)