Solid State: Diamond lattice and scattering

dimanche 2 mars 2014

I have the following homework question I am working on.



I am given three scattering angles: 42.8, 73.2, 89. (in degrees) without the wavelength of the light used. I am to show that these are consistent with a diamond lattice.



I started with Laue's Law: delta(k) = G and according to the professor in this instance, I am only worried about magnitudes.



By conservation of energy I know |k| = |k'|



This led me to |G| = 2 |k| Sin[theta / 2] where |k| = 2 pi / lambda



Now, if I take the ratios of |G_1|, |G_2|, |G_3| I get:



|G_2| / |G_1| = 1.63; |G_3| / |G_2| = 1.68



To get G, I started with the lattice vectors of the primitive cell of diamond which I believe are the same lattice vectors of the primitive cell of an FCC lattice.



So a_1 = (1/2) a (yhat + zhat); a_2 = (1/2) a (xhat + zhat); a_3 = (1/2) a (xhat + yhat)



I form the reciprocal lattice basis vectors from these.



b_1 = (2 pi / a) (-xhat + yhat + zhat); b_1 = (2 pi / a) (xhat + yhat - zhat); b_1 = (2 pi / a) (xhat - yhat + zhat)



Now one problem is, that I don't know how to construct the G's from this since I don't know how to find the coefficients for the diamond lattice. I know that G = v_1 * b_1 + v_2 * b_2 + v_2 * b_2 but in the end I know that |G| should equal (2 pi / a) Sqrt( v_1^2 + v_2^2 + v_3^2)



Any help would be appreciated, once I get this I can answer the next part of the question where he gives me the wavelength of the x-rays and show that "a" is that for carbon diamond lattice.





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