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Physics Department, University of Windsor
64-311 Laboratory
Experiment 1
Determination of e/m for an Electron (A)
Introduction
The ratio of charge to mass of the electron was first determined by J.J.
Thomson in 1897. For many years afterwards, only the ratio was known; the
charge and the mass being so small they eluded measurement. This
remained the case until Millikan measured e by using charged oil drops
(1911). In Thomson's classical experiment a stream of electrons was
deflected initially by a magnetic field; an electric field was then applied to
bring the deflected beam back to its original position. This current
experiment uses a different apparatus which was later developed by
Bainbridge (1938). Here, electrons of a well defined energy are directed into
a circular path by a uniform magnetic field. A measurement of the energy
together with the radius of curvature and the magnitude of the magnetic
J J Thomson
field yields e/m. This experiment will also serve as a useful introduction to electron beams.
Description
The e/m vacuum tube is a spherical bulb of about 3 litres volume. It is placed in the plane
midway between a Helmholtz coil pair (two parallel coils of radius R ≈ 33 cm, set apart by a distance
equal to R). Each coil has 72 turns of copper wire with a resistance of ~1 Ω. The mean radius of each
coil is 33.0 cm. The coils are supported in a rigid frame which can be tilted so that their magnetic field
will be parallel to the earth's magnetic field. A graduated scale indicates the angle of tilt.
Inside the e/m tube (see Figure 1) is a simple electron gun consisting of a tungsten filament (F)
surrounded by a coaxial anode (C) which has a slit (S) parallel to the filament. When the filament is
hot, the emitted electrons are accelerated by the anode. Those emerging from the slit form an electron
beam. The beam is made visible by the presence of mercury vapour in the tube. This is shown as a
dashed line in Figure 1 (a). When electrons of sufficiently high energy collide with the mercury atoms,
the resulting ions recombine with stray electrons to give the characteristic blue colour of mercury
vapour discharge lamps.
There are five posts inside the tube placed at a distance D (6.5, 7.8, 9.0, 10.3, and 11.5 cm,
respectively) from the filament. D is measured from the far side of each post. With a suitable
adjustment of the magnetic field B and accelerating voltage V, the electron beam emerging from the slit
can be made to travel in a circular path and strike the posts.
The magnetic field B at the centre of the Helmholtz coils is given by:
⎛ 4⎞
B=⎜ ⎟
⎝5⎠
3/ 2
μ 0 NI
R
=
32πNI
5 5R
x10 −7 T
(1)
where N is the number of turns in one coil. This equation, together with: FB = evB = mv 2 / r and
eV = mv 2 / 2 , will enable you to determine e/m from the experimental variables.
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Figure 1 (a) Sectional view of the tube and filament assembly. (b) Filament assembly at right angles to Figure 1
(a). The parts are: C - cylindrical anode, S - slit, P - post, F - filament, I - wires.
+
A
0-5 A, 0-5 V
(a)
V
0-50 V
mA
+
+
(b)
to filament
to filament
0-5 A, 0-5 V
to anode
Helmholtz
Coils
A
Figure 2 Wiring diagram for the electron gun (a) and Helmholtz coils (b).
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Method
You should place the entire apparatus in a dark room to easily observe the electron beam.
During operation, the electron beam will be deflected by the field B from the Helmholtz coils and
the external field BLocal from the Earth and the steel structure of the Essex Hall. Normally, one would
orient the coils with a compass and dip needle so that the axial field of the coil is parallel to BLocal.
However, the magnetic field inside the physics building is not uniform, and a proper orientation of the
coils is difficult (to say the least). Fortunately the magnitude of BLocal is small compared to the field
produced by the coils (provided you are in a good location) and introduces only a small error in the
result.
Make electrical connections to the tube and coil as shown in Figure 2. Note that the two coils are
connected in series. If it is not clear that the polarity is correct, it will become apparent later, and is
easily remedied.
The filament of the e/m tube has the electron emission when carrying 2.5 to 4.5 amperes.
B
B
B
Please ensure you do not exceed 4.5 A, as that will seriously shorten the life of the filament.
Good results can be obtained using an accelerating voltage of 20 to 45 V. An electron emission
current of 5 to 10 mA should produce a visible beam. For maximum filament life operate the tube with
the minimum filament current necessary to produce a visible electron beam. It is helpful and good
practice to apply 25 V of accelerating potential to the anode before heating the filament. Always start
with a low filament current and carefully increase it until the proper electron emission is obtained. The
electron beam will appear suddenly so do this slowly to avoid burning out the filament!
At this point you should see an electron beam bouncing off of the walls of the tube. Adjust the
filament current to keep the emission current between 5 and 10 mA. The emerging electron beam may
have a slight curvature to it due to BLocal.
Now send a small current through the Helmholtz coils. If the magnetic field from the coils bends
the beam away from the five posts, the coil connections must be reversed. If there is no effect at all,
then the field from one coil is opposing the other and the connections to one of them must be reversed.
Make the necessary changes and adjust the magnetic field until the beam describes a circle which does
not strike the inside walls of the tube .
Slowly rotate the tube in its cradle. What happens? Shut off the filament supply and reverse the
connections to the filament. What happens to the beam? Why does this happen?
Before you begin measurements allow a few minutes for the equipment and filament to warm up
and stabilize. Rotate the tube in its cradle so that the beam describes a circle.
Adjust the Helmholtz coil current until the sharp outside edge of the electron beam strikes the
outside edge of a post. The outside edge is used because those electrons have the greatest velocity.
Electrons leaving the negative side of the filament are accelerated through the greatest potential
difference between the filament and the anode. Thus electrons on the outside edge will have a kinetic
energy closest to eV where V is the voltage on the voltmeter.
Determine and record the current through the coils required to cause the outside edge of the
electron beam to strike the outside edge of each post. Calculate the magnetic field generated by each
value of current and record B on the table. Now repeat the procedure for at least two other values of
accelerating voltage.
For each accelerating voltage, plot the magnetic field versus 1/r. The points should fall on a
B
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straight line given by:
⎡ 2V ⎤
B=⎢
⎥
⎣ (e / m) ⎦
1/ 2
⎛1⎞
⎜ ⎟ + B Local
⎝r⎠
(2)
which you should derive. The last term is the component of BLocal parallel to B. Returning to
Fig 2, your results were taken with the negative terminal of the anode power supply connected to the
negative terminal of the filament supply. There is, however, a potential difference across the filament,
VF. Assuming that the electrons are emitted from the center of the filament, where it is hottest, then the
actual (average) energy that the electrons have corresponds to V = V A − VF / 2 , where VA is the
measured anode potential difference. Substitute this more accurate expression for V in equation (2)
above. Calculate the slope and the error in the slope for each plot by using the least squares method of
best fit. Using propagation of errors, now find your value of (e / m ) ± Δ(e / m) for each accelerating
voltage. Average, appropriately, all your (e / m ) ± Δ(e / m) values to give a final ‘best’ determination of
e/m.
How does your value for e/m compare to the accepted value of 1.75882 x 1011 Ckg-1? In addition,
you should be able to obtain a value for the magnitude of the component of BLocal parallel to B from
each plot, and these should be the same provided there were no changes in the apparatus.
B
B
Questions
List the main sources of random error and estimate the error in e/m introduced by them compare with the values of Δ(e/m) calculated from each plot. Discuss any potential systematic errors.
One obvious source of error lies in assuming that the magnetic field calculated at the center of
the Helmholtz coil pair is uniform over the path of the electron beam. While this is proably not quite
true, the magnetic field is quite uniform in a fairly large region around the center. It can been shown
that the magnitude of the field deviates by less than 1% over a radius of about 10 cm for this apparatus
(Cacak and Craig, 1969). The key to this remarkable result lies in the spatial dependence of the field
near the center. As an optional exercise you may want to show that the first three derivatives of the
field with respect to the axial variable vanish at the center.
References
Ainslie D.S. Am. J. Phys. 26, 496 (1958)
Bainbridge K.T. American Physics Teacher 6, 35 (1938)
Cacak R.K. and Craig J.R. Rev. Sci. Instrum. 40, 1468 (1969)
Christy R.W. and Davis W.P. Jr. Am. J. Phys. 28, 815 (1960)
Ficken G.W. Jr. Am. J. Phys. 35, 968 (1967)
lona M., Westdal H.C. and Williamson P.R. Am. J. Phys. 35, 157 (1967)
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