Example 4 X-rays of wavelength 1:54 ˚
A are used to calculate the spacing between
(200) planes in aluminum. The first-order Bragg’s angle corresponding to this
reflection is 22.4°. Determine the lattice parameter of aluminum crystal.
Solution: Given: k ¼ 1:54 ˚
A; n ¼ 1; hkl
ð Þ 200
ð
Þ; crystal is Al, h ¼ 22:4
; a ¼ ?
From the first-order Bragg’s equation, we can write
d ¼
k
2 sin h
¼
1:54
2 Â sin 22:4
Further, from the relationship between interplanar spacing and lattice parameter,
we can write
a ¼ d h
2
þ k
2
þ l
2
À
Á 1=2
¼ d 2
2
þ 0
2
þ 0
2
À
Á 1=2
or a ¼ 2d ¼
2 Â 1:54
2 Â sin 22:4
¼ 4:041 ffi 4:05 for Al
Example 5 A bcc molybdenum sample was studied with the help of X-rays of
wavelength 1.543 Å. Diffraction from {200} planes was obtained at 2h ¼ 58:618
:
Assuming this to be of first order, determine the lattice parameter.
Solution: Given: The crystal structure of molybdenum is bcc,
k ¼ 1:543 ˚
A; hkl
f g 200
f
g; 2h ¼ 58:618
; so that h ¼ 29:309
, a = ?
For {200} planes in a cubic crystal, the interplanar spacing is given by
d 200 =
a
h
2 + k
2 + l
2
À
Á 1=2 =
a
2
2 + 0
2 + 0
2
À
Á 1=2 =
a
2
Further, from Bragg’s equation, we have
a ¼
k
sin h
¼
1:543
sin 29:309
¼ 3:15 ˚
A
Example 6 X-rays of unknown wavelength are diffracted from an iron sample.
First peak was observed for (110) planes at 2h ¼ 44:70
: If the lattice parameter of
bcc iron is 2.87 Å, determine the wavelength of the X-ray used.
Solution: Given: The crystal structure of iron is bcc hkl
ð Þ 110
ð
Þ;
2h ¼ 44:70
; so that h ¼ 22:35
a ¼ 2:87 ˚
A; k ¼ ?
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7 Diffraction of Waves and Particles by Crystal