- #1
ChrisJ
- 70
- 3
Homework Statement
Q3a [/B]
Copper (Cu) crystallises in a face-‐centre cubic (fcc) structure (with a basis of one copper
atom). Draw a diagram of the unit cell of Cu, on your diagram indicate the (1, 1, 0) lattice plane and the [2, 2, 1] lattice direction.
Q3b: Calculate the angle between the [2,2,1] and [0,0,1] lattice directions
Q3c:
A sample of magnesium oxide, MgO, is investigated using x-‐rays of wavelength 0.154 nm. The (2,2,0) Bragg peak is observed at a diffraction angle of 2θ = 59.4°. Calculate the cubic unit cell parameter a for MgO.
Homework Equations
##2d \sin{\theta} = n \lambda ##
and maybe ## d = \frac{a}{\sqrt{h^2+l^2+k^2}} ## ?
The Attempt at a Solution
This is not for coursework or homework, but for preparation for an exam and this was a question on a past exam paper (which we don't have solutions for). But posted here as thought its where it still needs to go, but was not sure if to post here or in the introductory physics thread, this is part of a 2nd (of 3) years BSc degree. Albeit this is one of the easy questions on the paper.
Q3a:
I am familiar with sketching in the miller planes and lattice directions for a simple cubic structure (aka primitive cubic), so I guess my main question is whether it is the same for a FCC? I have done a lot of searching online on this and have not found anything concrete, but have seen or two people saying it is the same, but my initial instinct was that its not.
If It is the same, then the (1,1,0) plane is just as below
And for the direction of [2,2,1] of, I am not the best at drawing and its easier to just explain it than draw it and take a photo and upload etc. But it would start at the origin, and go through the "roof" of the cube dead centre (the middle of the square roof), and if there were two cubes stacked upon each other would point towards the very top right corner at the "front" (the bit "closest" to the screen).
Q3b: Again if its the same process as for simple cubic, then surely it is just using the definition of the dot/scalar product, and would be 70.5°.
Q3c:
The main reason I posted this question was for this part as I had never come across the second equation I put in the relevant equations section; plus was not sure about just using 1 for n, or is n quite literally the |(2,2,0)| I have used (that makes more sense to me)?
But I just combined the two equations to get
[itex]
a=\frac{\lambda (\sqrt{h^2+l^2+k^2)}}{2 \sin{\theta}} \\
a=\frac{(1.54 \times 10^{-10}) (\sqrt{2^2+2^2)}}{2 \sin{(59.4/2)}} \\
a = 4.38 \times 10^{-10} = 4.38 Å
[/itex]Any help is much appreciated!