Physical Principles of Electron Microscopy: An Introduction to TEM ...

TEM Specimens and Images 95
v 0
m
M
z
(a) before (b) after
Figure 4-2. Kinematics of electron-nucleus scattering, showing the speed and direction of the
motion of the electron (mass m) and nucleus (mass M ) before and after the collision.
In the TEM we detect scattered electrons, not recoiling atomic nuclei, so
we need to eliminate the nuclear parameters V and � from Eqs. (4.1) – (4.3).
The angle � can removed by introducing the formula (equivalent to the
Pythagoras rule): sin 2 � + cos 2 � = 1. Taking sin � from Eq. (4.3) and cos �
from
Eq. (4.2) gives:
M 2 V 2 = M 2 V 2 cos 2 �+ M 2 V 2 sin 2 �
= m 2 v0 2 + m 2 v1 2 � 2m 2 v0 v1 cos� (4.4)
where we have used sin 2 � + cos 2 � = 1 to further simplify the right-hand side
of Eq. (4.4). By using Eq. (4.4) to substitute for MV 2 /2 in Eq. (4.1), we can
eliminate
V and obtain:
mv0 2 /2 � mv1 2 /2 = MV 2 /2 = (1/M)[ m 2 v0 2 + m 2 v1 2 - 2m 2 v0 v1 cos� ] (4.5)
The left-hand side of Eq. (4.5) represents kinetic energy lost by the electron
(and gained by the nucleus) as a result of the Coulomb interaction. Denoting
this energy loss as E, we can evaluate the fractional loss of kinetic energy
( E/E0) as:
E/E0 = 2E/(mv0 2 ) = (m/M) [ 1 + v1 2 /v0 2 � 2 (v1/v0) cos� ] (4.6)
The largest value of E/E0 occurs when cos� ��1, giving:
E/E0 � (m/M) [ 1 + v1 2 /v0 2 + 2 (v1/v0) ] = (m/M) [1+v1/v0] 2
(4.7)
This situation corresponds to an almost head-on collision, where the electron
makes a 180-degree turn behind the nucleus, as illustrated by the dashed
trajectory in Fig. 4-3a. (Remember that a direct collision, in which the
electron actually collides with the nucleus, is very rare).
Kinematics cannot tell us the value of v1 but we know from Eq. (4.1) that
v1 < v0, so from Eq. (4.7) we can write:
E/E0 < (m/M) [1+1] 2 = 4m/M = 4m /(Au) (4.8)
where u is the atomic mass unit and A is the atomic mass number (number of
x
V
M
�
m
�
v 1