Answer
$R = 0.1429 nm$
Work Step by Step
Required:
The metal niobium (Nb) has a BCC crystal structure. If the angle of diffraction for the (211) set of planes occurs at 75.99° (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1659 nm is used, compute the atomic radius of Nb atom. Value of $d_{211} = 0.1347 nm$
Solution:
Using Equation 3.22 to calculate the value of $a$,
$a = d_{211} \sqrt (2^{2} + 1^{2} + 1^{2}) = (0.1347)\sqrt 6 = 0.3299 nm$
Using Equation 3.4 to calculate value of $R$,
$R = \frac{a\sqrt 3}{4} = \frac{(0.3299 nm)\sqrt 3}{4} = 0.1429 nm$
