# Ratio of Charge to Mass (e/m) for the Electron I. Theory ```Ratio of Charge to Mass (e/m) for the Electron
In this experiment we observe the motion of free electrons in a vacuum tube.
From their response to electric and magnetic fields the ratio of charge to mass for
the electron can be determined.
At the turn of the century several crucial experiments were performed to demonstrate the
atomic structure of matter. In 1897 J.J. Thompson was able to observe the motion of single
electrons in electric and magnetic fields, and so determine the ratio of the electron’s charge, e, to
its mass, m. This demonstrated some interesting things about the atom. First of all, this ratio is
exactly the same for all electrons, suggesting that they are fundamental particles rather than just
fragments of matter. Secondly, since the typical charge on atomic particles was known
approximately at the time, the mass of the electron could be estimated, and it turns out to be very
tiny compared to the mass of an atom. This was the first suggestion that the electron in an atom
might be a small particle orbiting a larger one, like the Earth in the solar system.
I. Theory
A. The apparatus.
This experiment is carried out in a special vacuum tube, which contains a small amount of
mercury vapor. Electrons emitted by a heated cathode are accelerated by the voltage applied
between the cathode and anode. Some of the electrons come out in a narrow beam through a
circular hole in the center of the cylinder. This emission is then focused into a narrow beam by
the grid of the tube. When electrons of sufficiently high kinetic energy leaving the cathode
collide with mercury atoms a fraction of the atoms become ionized. Upon recombination of the
ions with stray electrons, a characteristic blue color is observed. This makes the path of the
beam of electrons visible as the electrons travel through
the mercury vapor.
B (out of
A magnetic field is applied using a pair of coils of
paper)
wire. Two coils, separated by a distance equal to their
of this configuration is that it gives a rather uniform field
over a rather large area around the centerpoint. The
magnetic field at the center of the tube is calculated from
a relation given in its manual,
F
B = 7.7 x10−4 I
where I is the current in amps, and B is in Teslas.
B. Electrons in a magnetic field.
A charged particle moving in a magnetic field
experiences a force to the side (perpendicular to the
particle’s motion) and perpendicular to the magnetic field.
If the particle’s initial velocity is perpendicular to a
uniform magnetic field, it will move in a circle. This
e/m - 1
v
Figure 1. Force on an electron in a
uniform magnetic field.
situation is shown in figure 1.
The magnetic force, equal to evB, is the only force on the electron; so, Newton’s second law
(F=ma) gives
mv 2
evB =
r
The direction of the force on the electron is given by the right-hand rule. Walker gives this rule
as follows: &quot;To find the direction of the magnetic force on a positive charge, start by pointing
the fingers of your right hand in the direction of the velocity, v. Now, curl your fingers toward
the direction of B. Your thumb points in the direction of F. If the charge is negative, the force
points opposite to the direction or your thumb.&quot; You should be able to use this rule to confirm
that the force on the electron shown in figure 1 is towards the center of the circle.
The velocity, v of the electron can be related to the accelerating voltage, using energy
conservation:
Drop in potential energy = gain in kinetic energy
1
eV = mv 2
2
Solving this expression for v, and substituting into eB = mv/r, gives us the relationship:
e
2V
= 2 2 (1)
m B r
II. Experimental Procedure
Turn the apparatus on, and wait for warm-up (about one minute). Turn the voltage up to
about 200 volts. A beam will now be observed on the glass envelope opposite the anode. Adjust
the current through the coils and observe how the beam curves up into the sphere of the tube and
eventually bends over to complete a full circle.
A. Measuring the sign of the electron’s charge:
First method: Using a compass, determine the direction of the magnetic field applied by the
coils. Then, from the direction the electrons bend, determine whether their charge is positive or
Second method: Turn off the current to the coils. Use a permanent magnet with known
north and south poles to apply a known magnetic field to the electron beam. From the direction
of deflection of the electrons, determine whether their sign is positive or negative. Explain your
reasoning.
Now turn the magnetic field back on, and bring one end of a bar magnet as close as you can
to the electrons' path and observe the spiral path which the electrons now follow. Record what
you see.
Q1. Can you explain why the presence of the extra field distorts the electron's path?
e/m - 2
Q2. How will the Earth's field affect the motion of the electrons?
B. Radius as a function of voltage.
Before you begin taking data, calculate a predicted value for e/m using known values: e = 1.6
x 10-19 C, m = 9.11 x 10-31 kg. If your measured results differ greatly from this number, e.g. you
your mistake; it might be simply that you converted something incorrectly.
Here we plan to measure the radius r of the electron’s circular path as a function of the
accelerating voltage V. Note that equation (1) can be solved for V, giving
⎡ B 2e ⎤ 2
V =⎢
⎥r
⎣ 2m ⎦
or
y = ax
where y = V, x = r2, and the slope a of the line is given by
B 2e
a=
.
2m
Thus if we plot V as a function of r2, the curve should be a straight line, whose slope is related to
e/m. Solving for e/m gives
e 2a
=
.
m B2
Set the current to 1.5 A. This determines the value of the magnetic field produced by the
coils; the relationship, from the equipment manual, is
B = k I, where k = 7.7 x 10-4 T/A .
Calculate the value of B corresponding to the current you have set.
Now, vary the accelerating voltage V and measure r, the radius of the orbit, as a function of
V, over the widest range of voltages for which a measurement is feasible. Note that it might be
easier to measure the diameter first, and then to get the radius from the diameter. The most
convenient way to analyze this data is to open an EXCEL spreadsheet, and that’s what we’ll do
here. See the first experiment, &quot;Data Analysis and use of Excel on the IBM PC,&quot; for instructions
on plotting and linear fits. Plot V versus r2 and find the slope of the graph (linest or trendline).
Then use your value of the slope to determine e/m, using the relation given above. Note: to get
the slope in the right units, r should be expressed in meters.
Q3. Describe the sources of error in this part of the experiment.
C. Radius as a function of current.
Here we plan to measure the radius r of the electron’s circular path as a function of the
magnet current I, at constant V. Note that if we substitute B = kI into equation (1), it can be
solved for I, giving
e/m - 3
I = (2Vm/(k2e))1/2 (1/r)
Thus if we plot I as a function of 1/r, the curve should be a straight line, whose slope is related to
e/m. In this case the slope a is given by
a = (2Vm/(k2e))1/2,
and the charge-to-mass ratio is calculated from it using the relation
e/m = 2V/(k2a2).
Now set V to a fixed value (200 V). Then determine r for various settings of I, the current
through the coils. Plot I against 1/r and find the slope of the graph. Then use your value of the
slope to determine e/m, using the relation given above. Note: to get the slope in the right units,
r should be expressed in meters.
Q4. Describe the sources of error in this part of the experiment.
D. Error analysis
Now you have two experimentally determined values for e/m. Use them to calculate a best
value (the average) and the error on the average value. [See the first experiment, &quot;Data Analysis
on the IBM PC,&quot; for instructions on calculating averages and standard deviations.] State your
final result, complete with error.
Compare the expected value of e/m with your measurement in the standard way
(discrepancy, number of sigmas, quality of agreement
Q5. Is your result credible? In other words, does the uncertainty (or experimental error) which
you calculated account for your discrepancy?
Q6. The difference between the two values of e/m measured by different methods has been used
to determine the uncertainty in your final result. Explain how this procedure can allow for
unknown sources of error.
III. Equipment
Daedalon e/m apparatus
Bar magnet
Compass
e/m - 4
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