Question

3-77 Figure 3-43 shows the results of an X-ray diffrac- tion experiment in the form of the intensity of the diffracted peak versus the 20 diffraction angle. If X-rays with a wavelength of 0.15418 nm are used, determine: Intensity ULLA mmmmmmmmmmmmm 0 10 20 30 40 50 28 Figure 3-43 XRD pattern for Problem 3-77. (a) the crystal structure of the metal; (b) the indices of the planes that produce each of the peaks; and (c) the lattice parameter of the metal.

Answer 1

a)$\Theta$ = 17 20.5 28.5 33.5 36 41.5 45 47 (in degrees from the picture analysis)

$\ \Theta$ = 8.5 10.25 14.25 16.75 18 20.75 22.5 23.5 (in degrees)

sin$\Theta$ = 0.1475 0.1779 0.2462 0.2882 0.3090 0.3543 0.3827 0.3987

sin²$\Theta$= 0.02184 0.03165 0.0606 0.0831 0.0955 0.1255 0.1465 0.1590

Three types of cubic  lattice are there. PRIMITIVE BODY CENTERED FACE CENTERED

Different planes from which diffraction occur are

100 110 111 200 210 211 220 221 or 300 310 311 222 320 321 400 and( K=$\Lambda$²$\div$4a^{2 SINCE a is not given     we cannot calculate K by this eqn)}

primitive 1k 2k 3k 4k 5k 6k 8k 9k 10k 11k 12k 13k 14k   16k

fcc    2k 4k 6k 8k 10k 12k 14k 16k

bcc 3k   4k 8k 11k 12k 16k (THAT IS PEAKS IN THE DIFFRACTION PATTERNS CORRESPONDS TO THIS K VALUES)

BY TRIAL AND ERROR METHOD WE CAN CHOOSE K VALUE AS 0.00728 BY DIVIDING THE FIRST SIN²$\Theta$=0.02184 BY 3 SO THAT IT MATCHES WITH ALL OTHER VALUES. SO IT IS BCC PATTERN.

b) SO DIFFERENT PLANES ARE 111,200,220,311,222 ,400

c) LATTICE PARAMETER

n$\Lambda$= 2d sin$\Theta$

SO d=n$\Lambda$/2sin$\Theta$

HERE FOR FCC n=4

THEREFORE d=4*0.15418nm/0.1478*2

=2.086nm