This example will analyze a simple aspect of X-ray diffraction. So we have X-ray diffraction
of copper and this copper has a face-centered cubic structure. And we'll carry out his analysis
with X-rays that have a wavelength of 0.154 nm. So in the X-ray diffraction pattern we
see at an angle 2 theta of 43.2 degrees. We see a peak. And the question is can we determine
what the Miller Indices are for that peak? So what we're going to do is we're going to
use Bragg's Law which is relating the wavelength. So this is the wavelength of the X-rays. Relates
that to the spacing between the planes of the planes who Miller Indices we're trying
to determine and the angle theta. So this is the inter-plane spacing. So I've just rearranged
this equation here to sole for d. Just a wavelength over 2 sin theta. We're given the X-ray wavelength.
We're given 2 theta so theta is 43.2 divided by 2. And we calculate that the spacing, the
d spacing between these crystal planes that resolve in this X-ray peak is 0.209 nm. So
then we need to use the relationship that basically comes from geometry. And that says
that the spacing between planes who has Miller Indices h k and l is equal to for a face centered
cubic structure unit cell dimension over h squared plus k squared plus l squared. So
h k and l are the Miller Indices. This is the unit cell dimension. And for a face centered
cubic the unit cell dimension is equal to 2 times the atomic radius times the square
root of 2. And this is also just from geometry if we look at the crystal structure. So r
is the atomic radius. That's something we can look up, for example, in a periodic table.
And its value is 0.128 nm. So we know the atomic radius, that means we can calculate
a. We know d is equal to 0.209 nm. So we can solve for h squared plus k squared plus l
squared. So here substituting the atomic radius we now have the unit cell dimension. We substitute
that in here and so we can calculate the square root of h squared plus k squared plus l squared.
And that's a over d h k l. So substitute in the values that we calculated. The unit cell
dimension, inter-plane spacing. Get the square root of the values. Square both sides. H squared
plus k squared plus l squared is equal to 3. Now we need to use some additional information
about face centered cubic structure, namely that the principle diffraction planes, so
the main ones that would show up. The principle diffraction planes are those whose indices
are all even, all odd. So that means (1 1 1), (2 0 0), (2 2 0). All odd, all even, those
are the only ones that are going to show up as the main peaks. What value of h k and l
that are even or odd give us 3.0 when they're summed of squares? Well clearly if one of
these is 1 couldn't have a higher number. It would be much larger than 3. If it were
2 any of them were even it would be much larger than 3 even if the others are 0. So the only
possibilities are that h = k = l = 1. The sum of squares is 3. And so this diffraction
peak then is the (1 1 1) plane. This is the close packed plane, the hexagonal symmetry
plane in the face centered cubic structure. So this is face centered cubic structure,
the (1 1 1) plane that we obtained location. The X-ray diffraction pattern and we can back
up by Bragg's Law and geometry.