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X-ray Diffraction of Cu (FCC) using X-rays of wavelength =1.54Å gives XRD pattern where one of...

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)

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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 Braggs have Law mx = 2 danky sino where neopher of reflection (it must be treintes = Wave length of the incidena 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 . T

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