The Determination of
The Charge to Mass Ratio of the Electron
A magnetic field produced by Helmholtz coils is used to deflect electrons into circular
paths whose radii are known. By knowing the energy of the electrons and the magnetic field
strength, the ratio of the charge to mass of the electron (e/m) is determined.
Theory
A tube designed specifically to determine the charge to mass ratio of the electron is used
in this experiment. This e/m tube consists of an electron gun composed of a straight filament
surrounded by a coaxial cylindrical anode containing a single axial slit. The beam of electrons is
produced by this gun. (Refer to Figure 1.)The various parts in Figure are labeled as follows:
A)
B)
C)
D)
E)
F)
G)
five crossbars attached to a stiff
wire
a typical path of the beam of
electrons
the cylindrical anode
the distance from the filament to
the far side of the crossbar
the lead wire and support for
anode
the filament
the lead wires and support forthe
filament
Figure I (a) A top view of the e/m tube.
(b) A crossectional view of
the filament assembly.
Electrons emitted from theheated filament F are accelerated by the potential difference
applied between F and the anode C. Some of the electrons emerge from the slits as a narrow
beam. When electrons of a sufficiently high kinetic energy (10.4 eV or more) collide with the
mercury atoms in the tube, a fraction of the atoms will be ionized. On recombination of these
ions with stray electrons, the mercury arc spectrum is enitted with its charactefistic blue color.
This makes the path of the beam of electrons visible as the electrons travel through the mercury
vapor.
The e/m tube is cradled at the center of Helmholtz coils producing a magnetic field that deflects
the electrons into a circular path.
When a charged I particle such as an electron moves in a magnetic field in a direction at
right angles to the field, it is acted on by a force, the value of which is given by
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F = qvB,
(1)
where B is the magnetic field strength, q is the charge on the particle (in this case q = e because
the particle is an electron), and v is the velocity of the particle. This force causes the electron to
move in a circular path in a plane perpendicular to the magnetic field. The radius of this circle is
such that the required centfipetal force is furnished by the force exerted on the electron by the
magnetic field. Therefore,
mv 2
(2)
evB 
,
r
where m is the mass of the electron and r is the radius of its circular path.
If the velocity of the electron is due to its being accelerated through a potential difference
V, then it has a kine.tic energy of
1
mv 2  eV .
(3)
2
Substituting the value of v from (3) into (2)
e
2V
 2 2.
m B r
(4)
Thus, when the accelerating voltage, the field strength, and the radius of the circular path
described by the electron are known, then the value of elm can be computed.
The magnetic field B that causes the electron to move in a circular path is produced by
Helmholtz coils and has a value of
8 NI
B o ,
(5)
125a
where N is the number of tums of wire on each coil, I is the current through the coils, a is the
mean radius of the coils, and m. is the permeability of free space. Substituting (5) into (4) gives
e  3.91a 2  V 


.
m   o2 N 2  I 2 r 2 
(6)
Equation (6) is the working equation for the apparatus used. The quantity within the first set of
parentheses is constant for any given pair of Helmholtz coils. (For the Helmholtz coils with
which we are working, the number of turns of wire N = 130 and the mean radius of the coils a is
about 0.15 meters, but you should measure it.) The value of r, the radius of the circle in which
the electron beam travels, can be varied by changing either the accelerating voltage or the
Helmholtz coil field current “I”, which is used to deflect the electrons.
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Apparatus
o
e/m vacuum tube, with the outside edge of the crossbars located at the following
distances from the filament, which correspond to the diameters of the paths:
crossbar
o
1
2
3
4
5
6
7
0.050 ± 0.001 meters
0.060 ± 0.001
0.070 ± 0.001
0.080 ± 0.001
0.090 ± 0.001
0.100 ± 0.001
0.110 ± 0.001
Helmholtz coils apparatus, 130 tums of wire on each coil and a mean radius of
about 0.15 ±0.01 m. Apparatus contains needed voltage and current meters.
Procedure
1) To minimize the effect of the Earth’s magnetic field, orient the Helmholtz coils parallel to
the north-south direction.
2) Turn on the power switch. The unit will perform a 30-sec self-test. When the self-test is
complete, the display will stabilize and show “000”. A 5-10 minute warm-up time is
recommended before taking measurements.
3) During the warm-up period, measure the diameter of the Helmholtz coils. Since they are
not perfectly circular, measure both inside and outside diameters on several axes to get a
good average. Also measure the separation between the coils which is supposed to equal
the radius of the coils. The slight differences in the measured coil radii will give you a
good idea of the number of significant digits of this measurement.
4) Turn the Voltage Adjust control up to 200 V and observe the bottom of the electron gun.
The bluish beam will be traveling straight down.
5) Turn the Current Adjust control up and observe the circular deflection of the beam.
Increase the current to the Helmholtz coils until the electron beam describes a circle. The
diameter of the beam can be measured using the internal fluorescent scale inside the tube.
Make fine adjustments to the field current until the outside edge of the electron beam is
tangent to the outside edge of one of the crossbars.
6) For the 200 V accelerating voltage, measure the coil current values required to bend the
beam into as many of the eight available radii as possible. Because the beam spreads out,
take separate readings for the inside and the outside of the beam at each radius. This will
give a rough idea of the uncertainty of the measurements.
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7) Repeat the experiment for accelerating voltages of 300 V and 400 V. Arrange all this in a
neat data table. If time permits repeat for voltages 250 V and 350 V.
Analysis
Calculate the charge-to-mass ratio for all the data values that you have. It will save time if
you create a data table in an EXCELL spreadsheet and program other columns to calculate
values. In this experiment we will use statistical methods to determine uncertainties.
Calculate values of e/m for each crossbar intersection with the beam (at the inner and
outer edges of the beam). For each separate voltage, calcualte the average (or mean) and the
standard deviation of the e/m ratios, keeping inner and outer edge data separate. In a results
table, list these averages and standard deviations, compare your average ratios to the theoretical
value.
Draw one large error-bar graph displaying all your ratio averages and standard deviations
for each voltage. Include the theoretical value in the graph.
Conclusions
Answer the basic question: Did the calculated mean values agree with the known value of
the charge-to-mass ratio to within a standard deviation of the mean? Explain what factor(s)
caused the greatest error in your determination of the charge to mass ratio of the electron and
why.
Questions
1) Start with the magnetic field produced by a circular loop, and derive the expression for
the magnetic field produced by Helmholtz coils (5). Also derive the final working
equation (6).
2) Explain how electrons being ejected in slightly different directions by the accelerating
apparatus can cause the beam to spread as observed. Use your explanation to argue that
the outer edge of the beam is closer to the correct diameter of the electron-beam path. Do
your results favor this argument?
3) Explain the meaning of a standard deviation.
4) Do you notice any systematic differences in the results for the different accelerating
voltages? Can you think of any reason for these differences? A plot of the e/m ratios vs.
voltage could help with this question. EXCELL can make this plot easy to do.
5) Calculate average e/m ratios for each different radius, regardless of voltage. Do you
notice any systematic differences in the results for different radii? Can you think of any
reason for these differences? A plot of the e/m ratios vs. radii could help with this
question. You can use EXCELL.
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