Question

An X-ray diffractoin pattern of an unknown elemental metal shows the crystal structure to be FCC. The (311) peak on this pattern is at a 2(theta) value of 91.59(degrees). If the radiation is Copper K-(alpha) with a wavelength of 0.154051nm, calculate the value of the atomic radius of the elemental metal.

The atom touches______________?

The d-value for the (311) planes is _______ nm?

The edge Length of the unit cell is ________ nm?

Answer 1

1 )The atom touches 6 other atoms as the coordination no for FCC is 6.

2) d₃₁₁= / 2sin = 0.154051nm,/ sin 91.59^{o =} 0.15405/ 0.9996 =  0.1540 nm (2 =  91.59^o )

3) d₃₁₁= a / ( h² +k² + l²)^1/2= (a / 3² + 1² +1² )^1/2= (a/ 11)^1/2

a (egde length )= d₃₁₁ *(11)^1/2 =0.1540 * 3.3166 = 0.51076 nm

4) as it form a fcc we know face diagonal for fcc=

4r = (2)^1/2 a

radius =  (2)^1/2 0.51076 nm /4 = 1.4142 * 0.51076 / 4= 0.1805 nm