Question

X-ray Diffraction of Cu (FCC) using X-rays of wavelength =1.54Å gives XRD pattern where one of the peaks occurs at 2=43.2º What are the Miller indices for this peak? Given atomic radius for Cu is 0.128nm (note a=22 r)

Answer 1

The Miller indices for this peak is ( 111 ).

This problem can be solved by applying Bragg's law.

The detail calculation is shown below.

According to Bragg's have Law mx = 2 danky sino where neopher of reflection (it must be treintes = Wave length of the incident light. duki = Interp plannap distance. o = Bragg's angle. dinki) 2 Sino. danki = [let n=1; ie 1st ordere reflection. duki - 1,54x10m. 2= 1.548 = 1.57810 m 2 sim (43.2) 120= 43.2" 7 duki = 2.09 410 tom. 2 Again for enbic system, duki - r where a Lattice parameters. 7 (aki) = Millen in dices of the plane. 0.128 mm Atomic rading for an is 8.129 w Again, system). a = 2N2ir. (For Fce a (220.128) mm. a= 0.362 mm a. 0.3624159 m. - -

a Thus wherekedt = duke (0.362 x 10 9 m (2.09 1010 m. 1:43 = nekteer = 3. when, halkal and el. then wekholt=3 is possible . Thus the correspon millen in dius for this peak is (11).