Download PWE 17-6: Transmission Electron Microscope

Example 17-6 Transmission Electron Microscope
A transmission electron microscope forms an image by sending a beam of fast-moving electrons rather than a beam of light
through a thin sample. As we will see in Chapter 26, such fast-moving electrons behave very much like a light wave. If the
electrons have sufficiently high energy, the image that they form can show much finer detail than even the best ­optical microscope. The electrons are emitted from a heated metal filament and are then accelerated toward a second piece of metal
called the anode that is at a potential 2.50 kV (1 kV = 1 kilovolt = 103 V) higher than that of the filament. If the electrons
leave the filament initially at rest, how fast are the electrons traveling when they pass the anode?
Set Up
As the electrons move through the potential
difference between the filament and the anode,
the electric potential energy will change in
accordance with Equation 17-6. Each electron
has a negative charge q0 = 2e, so an increase
in electric potential (V 7 0) means a decrease
in electric potential energy (Uelectric 6 0). The
total mechanical energy (kinetic energy plus
potential energy) is conserved as the electron
moves because there are no forces acting on it
other than the conservative electric force, so
the electron kinetic energy will increase as the
potential energy decreases.
Solve
Solve Equation 17-6 for the change in electric
potential energy as the electron moves from the
filament to the anode, which is at a potential
2.50 kV higher than the filament.
Electric potential difference
related to electric potential
energy difference:
V =
Uelectric
q0
filament
(17-6)
electrons
Mechanical energy is conserved:
E = K + Uelectric = constant
anode
∆V = Vanode – Vfilament = 2.50 kV
From Equation 17-6,
Uelectric = q0 V
The electron has change q0 = 2e, so
Uelectric = 2e V
= 2(1.60 * 10219 C)(+2.50 * 103 V)
= 24.00 * 10216 J
(Recall that 1 V = 1 J>C, so 1 J = 1 C # V.)
The conservation of mechanical energy tells
us that the change in the kinetic energy of the
electron is equal to the negative of the change
in the electric potential energy.
Mechanical energy is conserved:
E = K + Uelectric = constant
So the change in mechanical energy is zero:
E = K + Uelectric = 0
The change in the kinetic energy of the electron is
K = 2Uelectric = 2(24.00 * 10216 J)
= +4.00 * 1016 J
Each electron begins with zero kinetic energy,
so the change in its kinetic energy is equal to its
final kinetic energy as it reaches the anode.
Since the electron begins with zero kinetic energy at the filament, the
change in its kinetic energy is
K = +4.00 * 10216 J = Kanode 2 Kfilament = Kanode
Use this to find the speed of the electron at the anode:
1
K anode = melectron v 2anode so
2
vanode =
214.00 * 10-16 J2
2K anode
=
B melectron
C 9.11 * 10-31 kg
= 2.96 * 107 m>s
Reflect
The electrons are accelerated to nearly one-tenth of the speed of light (c = 3.00 * 108 m>s).