Example 5 Through what potential should a beam of electrons be accelerated so
that their wavelength becomes 1.54 Å. Assuming that an electron beam can
undergo diffraction by crystal.
Solution: Given: n = 1, k ¼ 1:54 ˚
A ¼ 1:54 Â 10
À10 m; V ¼ ?
We know that the wavelength associated with the electrons is given by
k ¼
h
mv
or v =
h
mk
Also; eV =
1
2
mv
2
¼
1
2
m
h
mk
2
or
V ¼
h
2
2m k
2 e
¼
6:626 Â 10
À34
ð
Þ
2
2 Â 9:1 Â 10 À31 Â 1:54 Â 10 À10
ð
Þ
2 Â1:6 Â 10 À19
¼ 63.6 volts
Example 6 Electrons are accelerated to 344 V and are reflected from a crystal. The
first reflection occurs when the glancing angle is 60°. Determine the interplanar
spacing of the crystal.
Solution: Given: V = 344 V, h ¼ 60
, interpanar spacing d = ?
We know that the wavelength associated with the electrons is given by
k ¼
h
ffiffiffiffiffiffiffiffiffi ffi
2mE
p
¼
h
ffiffiffiffiffiffiffiffiffiffiffiffi
2meV
p
¼
6:626 Â 10
À34
2 Â 9:1 Â 10 À31 Â 1:6 Â 10 À19 Â 344
ð
Þ
1=2
¼ 0:662 Â 10
À10
¼ 0:662 ˚
A
Now, the Bragg’s equation for n = 1 is
2d sin h ¼ k
or d ¼
k
2 sin h
¼
0:662
2 sin 60
¼ 0:38 ˚
A
7.4 Other Diffraction Methods
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