What is the distance between the adjacent Miller planes if the first order reflection from X-rays of wavelength 2.29
occurs at 27o 8' ?
Solution :
n = 2d sin
n = 1, = 2.29 , = 27o 8'
d = 0.5 * 1 * 2.29 * 10 -8 / sin ( 27o 8' )
= 2.51
18.2)
Gold crystallizes into an FCC structure. The edgelength of the FCC unit cell is 4.07Calculate a) the closest distance between two gold atoms and b) the density of gold if its atomic weight is 197
Solution :
Let b = closest distance between gold atoms; a = 4.07
b 2 = a 2 / 4 + a 2 / 4 = 0.25 * 4.07 2
b = 2.878
Density =
= mass / volume
volume of the unit cell = (4.07)3
mass of the unit cell = 4 * 197 g / 6.02 * 1023
as there are four atoms per unit cell in an FCC Lattice
= 19.419 g/ cm3
18.3)
X rays of = 0.1537 nm from a Cu target are diffracted from the (111) planes of an FCC metal. The Bragg angle is 19.2 o. Calculate the Avogadro number if the density of the crystal is 2698 kg / m 3 and the atomic weight 26.98
Solution :
The distance between the 111planes
d = 0.5 * 1.537 * 10 -10 m / sin ( 19.2o )
= 2.51 * 10 -10 m
We need the edge length a which is related to d by
a = d (h 2 + k 2 + l 2) 1/ 2 where hkl = (111) are Miller indices
a = 2.51 * 10 -10 = 4.0475 * 10 -10 m
= mass / volume = 2698 kg / m3
mass of unit cell = 4 * 26.98 * 10 -3 kg / [NA * (4.0475 * 10 -10)3]
occurs at 27o 8' ? 
= 2d sin 
= mass / volume 
= 4.0475 * 10 -10 m 
