Download B A C a b c

A DETEMINATION OF THE CHARGE-TO-MASS RATIO, e/m,
FOR ELECTRONS
1. Introduction
Consider a thermionic diode (Figure 1) consisting of a cylindrical anode, A, with a fine wire cathode,
C, along its axis. If the anode is maintained at a positive potential with respect to the cathode and the
latter is heated to a sufficiently high temperature, electrons from the cathode will travel to the anode
(see trajectory a, figure 1) and a current can be observed.
A
b
If the diode is placed in a uniform magnetic field B, parallel to its
axis, the electron paths will be curved. At a certain critical value of
a
c
B, Bc say, the electrons will only just be able to reach the anode (see
C
trajectory b, figure 1). When B is increased beyond this value,
B
electrons cannot reach the anode and the current will be cut off (see
trajectory C, figure 1). The trajectories shown in figure 1 are
schematic, but in figure 4 (page 5) you will find some typical electron
Figure 1
paths calculated by using a computer to perform a numerical
integration of the equations of motion.
In this experiment you will study the anode current as a function of the applied magnetic field, and
from measurements of the critical field (together with other quantities), you will be able to determine
the charge-to-mass ratio, e/m, for electrons.
2. Summary of Theory
A fuller discussion is given in section 5. The main result is quoted here so that you can proceed with
the experiment immediately and read the more detailed treatment in your own time.
The critical value of the magnetic field, Bc, is given by:8V
Bc2 = 2
b (e/m)
(1)
where V is the potential difference between anode and cathode and b is the internal radius of the
anode.
In your experiment, the axial magnetic field is provided by a solenoid. The magnetic field, B, on the
axis of a long solenoid is given by:B = µoNI
(2)
where I is the current in the solenoid and N is the number of turns per unit length. µo is the magnetic
constant, its value in SI units being 4π x 10-7 tesla metre amp-1.
From equations 1 and 2 it follows that if the current producing the critical magnetic field is Ic,
8V
Ic2 = (e/m)
1
(µoNb)2
(3)
1
Thus if Ic is measured for several different values of V, a graph of Ic2 against V should be a straight
line, from the slope of which e/m can be deduced, provided that N and b are known.
3. Experimental Procedure
You will notice that the diode has three cylindrical electrodes. The middle one is the anode; it is the
current flowing from the cathode to this electrode that you want to measure, and it is the internal radius
of this electrode that is denoted by "b" in equations 1 and 3. The two outer cylinders are guard rings,
to be maintained at the same potential as the anode, thus helping to ensure that the electric field
between cathode and anode is radial, and is not distorted by end effects.
The electrical circuit to be wired-up, and the observations to be made with it are described in section
3a, and later (section 3b) some suggestions are made about the method of determining "b"; but you
can do the two parts of the experiment in whichever order you prefer. If the travelling microscope
(shared between four experimenters) is available, you might start by determining "b".
3a
Electrical measurements
The circuit is shown in figure 2; wire it up carefully, and have it checked by a demonstrator before
switching on.
A
M
+
GR
Anode
D is the diode with its two guard rings both
D
Supply
connected to the terminal on the base labelled
V
10R
10R
GR. The anode is connected to the terminal
labelled A, and the two ends of the filament to
F
C
F
those labelled F. The two 10 ohm resistors are
already built into the base of the diode. Voltages
Voltstat
R
Transformer
are measured on the volt meter, V, connected
Figure 2
between the anode-guard-ring assembly and the
AC Mains
centre tap, C, of the resistor chain. Notice that
the microammeter, M, measures the current to the anode only, and not that going to the guard rings.
The microammeter contains an amplifier to increase its sensitivity. Remember to switch it off when
you have finished work.
The "Volstat" transformer provides a reasonably stable AC. voltage, 12 volts RMS, and the rheostat,
R, is used to control the current through the diode filament. It is most important that the filament
should not be overloaded; so always make sure that the rheostat is set to its maximum value before
switching on.
The anode voltage and a dc. current supply for the solenoid are obtained from the e/m power supply
unit on your bench which has two outputs; they can be varied independently. The solenoid (not
shown in figure 2) should be connected to its supply in series with an ammeter. This solenoid has
marked on it the total number of turns, so, by making obvious measurements, you can calculate N, the
number of turns per unit length. How good is the claim that it is a "long" solenoid in the sense implied
by equation 2?
In this experiment (and in most others, too) it is well worthwhile to pay attention to the lay-out of your
equipment. Put the components that you need to adjust frequently in positions where you can reach
them easily; meters that have to be read carefully should be straight in front of you. Pieces of
equipment which do not need much adjusting can be placed towards the back of your bench.
2
When your circuit has been checked, follow the procedure set-out below. The diode should not be
inside the solenoid at this stage, so that you can see the bright glow of the filament.
1. Keep the rheostat at its maximum value, set the anode potential to about 15 volts and switch on the
microammeter. The microammeter should not yet indicate any current.
2. Carefully reduce the value of the rheostat until, as the filament glows brightly, anode current is
recorded.
3. Adjust the rheostat so that the anode current is between 20 and 30µa and insert the diode in the
solenoid. The experiment is now ready to proceed.
Measure the anode current as a function of the current in the solenoid for two different values of the
anode voltage (15 volts and 5 volts would be suitable). It would be wise to check the reproducibility
of the observations for at least one of these curves. You need not aim for identical current values - just
do a second run-through and then plot your results. At some solenoid current, the anode current will
begin to decrease sharply; but the fall-off is not infinitely rapid (why not?) so you will have to
consider what you mean by the "cut-off current" Ic (eqn. 3). Think about this problem and discuss
your conclusions with a demonstrator.
Determine Ic for several different values of V, the anode potential, between 15 volts and 5 volts, and
plot Ic2 vs V (see equation 3).
3b Determination of the Radius of the Anode
The external diameter of the anode can be measured with a travelling microscope, looking through the
side of the glass envelope. Make several measurements at different points along the length and for
different orientations of the diode. Strictly, you should allow for refraction in the glass; but the effect
of this is small enough to neglect. To obtain a value for the internal radius of the anode you need to
know the thickness of the material, and this you cannot measure directly. Do the best that you can by
looking through the end of the envelope and measuring the thickness of the material of the end guardring.
It is important to get as reliable a value as you can for the internal radius, b, since the final value for
e/m is proportional to 1/b2.
4. Discussion of Results
Deduce as good a value as possible for the statistical error on your final result, and then compare your
value for e/m with published data. A significant difference between the two values indicates the
presence of important systematic errors, and you should try to suggest what these might be. The
theoretical treatment in section 5 makes a number of assumptions in order to derive the final, simple
result; try to consider which of these may not be met adequately by your equipment. Remember,
among other things, that the diode is a piece of commercial equipment, and has not been carefully
constructed for research purposes.
It would be interesting to know something about the magnitude of the critical magnetic field Bc
(Why?). Calculate one value of Bc, corresponding to some average value of anode voltage.
If you had more time to spend on checking the performance of your equipment, what tests would it be
useful to carry out?
3
5. Calculation of the Critical Magnetic Field
The following method, which uses only simple vector angles leads to the condition for an electron,
starting from the cathode, radius a, at C, just to reach the anode, radius b, at P. (See figure 3).
P
Potential V
Consider a particle of mass m, charge q, travelling in a
^
θplane perpendicular to the axis of the diode. Take an q=charge (-ve)
-v
F=q(vxB)
origin 0 at the point where this plane cuts the axis and m=mass
^
Anode
-r
describe the position of the particle by the usual planeq,m
^ . Take
polar co-ordinates r and θ; unit vectors ^r and θ
b
B, z^
the diode axis as the z axis, unit vector ^z.
− Let the velocity of the particle at some arbitrary point on
Figure 3
2a
^ ). The force exerted on the
its path be v = (vr^r + v θ θ
particle by the magnetic field B is F where F = q(v + B); so, if B is directed along the z-axis, (B = Bz ,
say):^ + vθ ^r )
F = qB(vr^r x ^z + vθ θ,^ x ^z) = qB( − vrθ
The torque about O is G where:G = r x F = − rqBv r^z
If the electric field, E, is radial, it produces zero torque about O.
The angular momentum, J, of the particle about O is related to G by dJ/dt = G. Thus:dJ = − (rqBvredt)z^ = − (Bqrdr)z^
The negative sign in the last equation simply means that a particle travelling outward from C will
begin to twist clockwise if q is positive, and anticlockwise, as shown in figure 3, if q is negative, as is
the case for an electron. Since we need to consider only the magnitude of the angular momentum we
can write:dJ = B|q|rdr ; Jf − Ji = «B|q|(b2 − a2)
where Jf and Ji are the final and initial values, respectively of the angular momentum. If the electron
starts off radially, Ji - O; also a<<b, so Jf = «B|q|b2.
If the electron just reaches the anode, travelling tangentially to it a t P, Jf = bpf where pf is the final
linear momentum, related to the final kinetic energy, T, by pf = 2mT. If the electron started from the
cathode with zero velocity and was accelerated to the anode through a potential difference V, T = |q|V.
Thus if Bc is the value of the magnetic field at which the electrons are just capable of reaching the
anode, we have:Jf = «Bc|q|b2 = bpf = b 2m|q|V
which gives:8V
Bc2 = b2(|q|/m)
4
To calculate the actual details of the electron path it would be necessary to write down, and solve, the
equations of motion. An exact analytical solution of these differential equations cannot be found.
Figure 4 shows the results of numerical solutions obtained using the Radial Ridge Cyclotron Group's
IBM 1130 computer. The scale is about 20 times life size. The curves a, b and c represent the electron
trajectories for magnetic fields equal to 0.7, 1.0 and 1.5 times the critical field for an anode potential of
15.0 V.
Figure 4
5