Question

A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, and 14.49 Å along the three cartesian coordi- nates. What are the Miller indices of the plane?

Answer 1

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = $4.83 \AA$

Intercepts along the three axes:

$\bar{x} = 9.66 \AA$

$\bar{x} = 19.32 \AA$

$\bar{x} = 14.49 \AA$

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = $\frac{1}{9.66}$

b = $\frac{1}{19.32}$

c = $\frac{1}{14.49}$

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = $\frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}$

b' = $\frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}$

c' = $\frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}$