The copper with an atomic radius of 0.1278 nm has an fcc unit cell structure. the wavelength under the x-ray diffraction is 0.1542 nm. (15%) (a) determine the interplanar spacing dnktfor the crystallographic plane (111) (b) calculate the diffraction angle 20 for the crystallographic plane (111)

a) d = 0.2086 nm

b) ]

Explanation:

given data:

atomic radius = 0.1278 nm

\lamda = 0.1542 nm

a) interplanar spacing (d)is given as

a is the size of unit cell

a\sqrt{2} = 4*r

a = 0.3614 nm

miller indices for crystallofgraphic plane [111] is reciprocal i.e.

[h,k,t] = [1,1,1,]

d = 0 .2086 nm

b) diffraction angle

by bragg's law and for first order reflection (n=1)

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explanation: