h ¼ sin
À1 k
2d
= sin
À1
1:43 Â 10
À10
2 Â 3:14 Â 10 À10
¼ sin
À1
1.43
2 Â 3:14
= sin
À1 0.2277
ð
Þ
¼ 13:16
¼ 13
; 10
0
Example 3 Calculate the energy (in eV) associated with an electron of wavelength
3 Â 10
À2 m:
Solution: Given: k = 3 Â 10
À2 m; E = ?
We know that the wavelength associated with a moving electron is given by
k¼
h
ffiffiffiffiffiffiffiffiffi ffi
2mE
p
or
E ¼
h
2
2mk
2
=
6:626 Â 10
À34
ð
Þ
2
2 Â 9:1 Â 10 À31 Â 9 Â 10 À4 J
¼
6:626 Â 10
À34
ð
Þ
2
2 Â 9:1 Â 10 À31 Â 9 Â 10 À4 Â 1:6 Â 10 À19 eV
¼ 1:68 Â 10
À15 eV
Example 4 Calculate the de Broglie wavelength of electrons and their velocity
when the first Bragg’s maximum of electron diffraction in a nickel crystal
(d = 0.4086 Å) is found to occur at a glancing angle of 65°.
Solution: Given: n = 1, d = 0:4086 ˚
A ¼ 0.4086 Â 10
À10 m, h¼65
; k = ?, v = ?
The Bragg’s equation for n = 1 is
k ¼ 2dsinh
¼ 2 Â 0:4086 Â 10
À10
 sin 65
¼ 2 Â 0:4086 Â 10
À10
 0:9063
¼ 0:74 Â 10
À10 m
Now, for the de Broglie equation, the velocity of the electron is given by
v ¼
h
mk
¼
6:626 Â 10
À34
2 Â 9:1 Â 10 À31 Â 0:74 Â 10 À10
¼ 9.84 Â 10
6 m/s
286
7 Diffraction of Waves and Particles by Crystal