Question

mot MSE 260 Materials Science and Engineering - Homework 4 1. (3.70) 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 following: (a) the interplanar spacing for this set of planes and (b) the atomic radius for the Nb atom. 2. (3.71) For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 44.53° for FCC nickel (Ni) when monochromatic radiation having a wavelength of 0.1542 nm is used? Radius of nickel atom is 0.1246 nm.

Answer 1

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4 Angle of diffraction, 20 = 75.99° 0 = 37.995 = 0.1659 hm, n=1 (a) We know that a drino = na & d = nx = 10.1659 = 0.13478 nm 2 sino 2x & in (37.995) (b) We know that, Here, h= 2, K=1, l=1 d = - a 7h²+k²+1² a = d xy h²+k²tl? = 0.13478 x122+*+12 = 0.330142 nm For Bcc a= 4r 9 ra ang = 0-330142x13 M = 0.142956 mm ③ Diffraction angle, 20 = 44.53° = 0 = 22.265 We know that adlino = nx a d = n = 0.1542 = 0.20 34 89 m 2 sino 2 sin 22.265" Here, de = 0.12 46 nm For, FCC : a = 42 72 7 a = 4x0.1246 = 0.352 4 22 nm V We know that a = - a Th²+k²+1² + K² + 12 = a. 0.352422 - 1.73 18 97 0. 203489 - À 0982422 - 1731597

7 h ²+k² + 1² = 2.999468 - (1 for Fcc, hkl values should be either odd number 60 I even numbel. i h=K= d = 1 Satisfies the eg. De Crystallographic plane will be (111)