ELECTRON DIFFRACTION
Introduction
In this lab it will be studied to observe diffraction of the beam of electrons on a graphitized carbon target, and to calculate the intra-atomic spacings in the graphite.
Theory
Consider planes of atoms in a crystal separated by distance d. Electron ”waves” reflect
from each of these planes. Since the electron is wave-like, the combination of the reflections
from each interface will lead to an interference pattern. This is completely analogous to light
interference, arising, for example, from different path lengths in the Fabry-Perot or Michelson interferometers. The de Broglie wavelength for the electron is given by: λ = h/p, where
p can be calculated by knowing the energy of the electrons when they leave the “electron
gun”:
𝑝2
2𝑚
= e UA
(1)
where UA is the accelerating potential. The condition for constructive interference is that the
path length difference for the two waves is a multiple of a wavelength. This leads to Bragg’s
Law:
n λ = 2d sinθ
(2)
which for small angles and first order diffraction becomes
r=
2R
d
n𝜆
(3)
where n = 1, 2, . . . is integer, R is the distance from the graphite target(65mm) and r is the
diameter of the diffraction ring. In this experiment, only the first order diffraction n = 1 is
observed. Therefore, the intra-atomic distance in a crystal can be calculated by measuring
the angle of electron diffraction and their wavelength (i.e. their momentum):
d=
𝜆
1
=
2sinθ 2sinθ
h
√2𝑒m𝐔A
(4)
where h is Planck’s constant, e is the electronic charge, m is the electron’s mass, and UA is
the accelerating voltage.
Results
As it is seen from the experimental result table below, seven measurements of r1 and r2
have been performed along with changing UA values. The wavelength 𝝀 is calculated by
varying UA as below formula;
𝜆=
h
√2𝑒m𝐔A
Where
e = 1.602 x 10-19 C
m = 9.109 x 10-31 kg
h = 6.625 x 10-34 J.s
UA (kV) λ (pm) r1in(mm) r1out(mm) r1ave(mm) r (mm) r2out(mm) r2ave(mm)
2in
4.00
19.4
10.5
13.1
11.8
19.2
23.1
21.2
4.50
18.3
10.4
12.4
11.4
18.6
21.8
20.2
5.00
17.3
10.0
11.7
10.9
17.4
21.4
19.4
5.50
16.5
9.2
11.2
10.2
16.9
19.1
18.0
6.50
15.2
8.6
10.1
9.4
16.1
18.4
17.3
7.00
14.7
8.3
9.4
8.9
15.2
17.7
16.5
7.40
14.3
8.1
9.1
8.6
14.8
17.2
16.0
Plotting the graph r vs 𝝀 for two interference rings and getting its slope by using linear fit
can be used to find d in accordance with Equation (3)
Slope
=
2R
𝐝
λ (pm) vs r1ave (pm)
1,3E+10
y = 5,8E+08x + 7,8E+09
1,2E+10
r1ave (pm)
1,1E+10
1,0E+10
9,0E+09
8,0E+09
7,0E+09
6,0E+09
14,3
14,7
15,2
16,5
λ (pm)
As seen on the graph of r1 vs 𝝀 the slope is calculated as ;
Slope1 = 5.8 x 10+8
d1
=
2R
Slope
= 224 pm
17,3
18,3
19,4
λ (pm) vs r2ave (pm)
2,2E+10
y = 9,0E+08x + 1,5E+10
2,1E+10
r2ave (pm)
2,0E+10
1,9E+10
1,8E+10
1,7E+10
1,6E+10
1,5E+10
1,4E+10
14,3
14,7
15,2
16,5
λ (pm)
As seen on the graph of r2 vs 𝝀 the slope is calculated as ;
Slope2 = 9.0 x 10+8
d2
=
2R
Slope
= 144 pm
17,3
18,3
19,4