braggs-lawcondensed-matterx-ray-crystallography
I want to compute the distance between two (111) planes in a cubic crystalline structure, in order to do some computations involving Bragg reflection.
I have a sketch of which the (111) planes are, however I don't know what these Miller-indices really tell me. Could you explain this to me?
Furthermore, is there an easy way to compute the distance between such planes in general, given the distance of two adjacent atoms?
I could only find this poor-quality picture. It should give you an idea, anyway. For example, consider the first picture in the first row: $(l,m,n)=(1,0,0)$ in that case, and it is easy to verify that the distance between the grey planes is
$$d=a$$
In the second case, $(l,m,n)=(1,1,0)$, and you can see that
$$d=\frac a {\sqrt 2}$$
etc.
Here is my approach - which doesn't quite give the same answer you were looking for...
From Bragg's Law, we know
$$n\lambda = 2d\sin\theta$$
This tells us that the smaller the angle $\theta$, the shorter the wavelength - and the shorter the wavelength, the greater the energy, since wavelength and energy are related through the relationship:
$$E = \frac{hc}{\lambda}$$
The maximum allowed wavelength (lowest energy) is the one for which $n=1$ and $\theta = 90°$. Then $\lambda = 2d$.
The rest is substitution of numbers in the above equations. A handy shortcut is given on wikipedia:
$$E = \frac{1.2398}{\lambda (µm)} ~\rm{eV}$$
with the wavelength given in µm. Given $\lambda = 2d = 2\times10^{-10}$ m, the above evaluates to
$$E = \frac{1.24}{2\times 10^{-4}~µm} = 6.2 ~\rm{keV}$$
Not sure why that is not exactly the same as the answer you were given. There seems to be about a $\sqrt{2}$ difference...
Best Answer
The miller indice is the inverse of the distance between the planes, in the direction of that indice. The higher the index, the more planes in the unit volume.
In the case of $(1,1,1)$, $n=1$,$k=1$ and $l=1$.