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Demonstration Experiments: The charge to mass ratio of the electron
27 September 2012
The charge to mass ratio of the electron
In the standard model of particle physics, the fundamental properties of the electron are its charge
and mass, which are fundamental constants, and its spin and lepton number, which are quantum
numbers. It is not easy to measure the electron’s mass directly. It is usually easiest to apply an
electric or magnetic force to the electron to measure its charge to mass ratio and deduce its mass
from an independent measurement of its charge. In this experiment you will determine this
fundamental ratio.
The apparatus that we have to measure the charge to mass ratio of the electron consists of an
electron source in an evacuated tube and a set of Helmholtz coils* that produce a uniform
magnetic field. The tube contains a trace amount of hydrogen gas that is ionized through electronmolecule collisions and the resulting emission allows you to see the trajectory of the electron
beam. By varying electrical potentials applied to the source you can vary the energy and focus of
the electron beam.
Background


A particle of mass m with a charge q moving with velocity v in a magnetic field B experiences a
force given by the Lorentz for law,

 
(1)
F = q v×B .
Because of the cross product, the force is always at right angles to the motion, thus the magnitude
of the velocity never changes.† For some radius r, the magnitude of the Lorentz force will just
equal the centripetal force needed to keep the particle in circular motion,
v2
(2)
FC = m .
r
Equating the magnitudes of these forces, we can solve for the charge to mass ratio,
q
v
.
(3)
=
m rB
Our apparatus produces a beam of electrons, so we make the substitutions q → −e , m → me . The
electron beam is accelerated by an electrical potential U, so an individual electron has kinetic
energy eU. This allows us to determine the electron’s velocity through the equality
(4)
eU = 12 me v 2 .
Substituting this into eq. (3), it is easy to show (you should verify this!)
e
2U
.
(5)
=
me (rB )2
The experimental strategy should be clear: you can control U through the voltage applied to the
electron source and B through the current in the Helmholtz coils. By measuring the radius r for
different values of these parameters you can determine the charge to mass ratio of the electron.
(
)
Experimental procedure
Safety warning: The electron tube operates at high voltage, up to 500V. All
connections should be made using the special shielded banana cables, not ordinary
cables. The tube itself is made of very thin glass. Take care not to bump it or drop
things on it, as it will implode if broken.
*
†
The magnetic field produced by the Helmholtz coils is discussed in the appendix.
See Young and Freedman, §27.4.
Blackett Laboratory, Imperial College London
89
Demonstration Experiments: The charge to mass ratio of the electron
27 September 2012
The electrical connections to the electron tube are shown in figure 1. It should already be
connected to its power supply, but it is worth understanding how the tube works. The cathode is
heated by the current from a 6.3VAC supply. The hot cathode emits electrons via thermionic
emission. The cathode is held at ground potential, 0V. It is surrounded by a metal cylinder with a
small hole in its end, called a Wehnelt cylinder. The Wehnelt cylinder is held at a negative
voltage with respect to the cathode and helps to focus the electron beam. You’ll probably find a
bias voltage of about -20V to be optimal. The anode is the conical metal structure you can see in
the tube. It is held at a large positive voltage, typically about 250V. The electrons born at the
cathode are accelerated towards the anode and emerge from the hole in its tip. There are two
small deflector plates near the anode; they should be electrically tied to the anode using the
terminals on the tube base. One meter should be set up as a voltmeter to monitor the anode
voltage.
The Helmholtz coils are powered by the separate DC supply, you can use its built in ammeter to
monitor the coil current. The B field at the center of the coils is proportional to the coil current I.
It has the theoretical value
3/ 2
⎛4⎞ n
(6)
I ≡ β I,
B = µ0 ⎜ ⎟
⎝5⎠ R
where µ 0 = 4π × 10 −7 N A −2 , n = 130 is the number of turns in each coil and R is the coil radius,
nominally 150mm. For convenience we
have defined β as the proportionality
between the B field and the current.
We can rearrange equations (5) and (6) to
get
e r 2β 2 2
(7)
U=
I ,
me 2
in other words, a plot of the anode voltage
U vs. I 2 for an electron beam with a given
radius r should give a straight line whose
slope is proportional to the charge to mass
ratio.
Starting with U = 250V, adjust I to given
an electron orbit with diameter about 8cm.
Use the slides and the mirror reflector to
measure the edges of the electron beam.
Make 5 or so additional measurements
with U between 200V and 300V, always
adjusting the coil current to keep the beam
diameter constant.
Figure 1: Electrical connections to the electron beam tube.
The electron beam loses energy due to collisions with the background hydrogen gas.
According to eq. (5), lower energy electrons follow orbits with a smaller radius. This
implies that you should measure the outside diameter of the beam to measure the highest
energy electrons. On the other hand, the electrons in the beam repel each other, causing the
beam to expand. At the same time the ionized background gas tends to cancel this space
charge and therefore focus the beam. You might think this would mean you should measure
from the center of the beam. The manufacturer of the equipment does suggest using the
outside edge, but you should at least consider this as a possible source of systematic error.
Blackett Laboratory, Imperial College London
90
Demonstration Experiments: The charge to mass ratio of the electron
27 September 2012
Analyse your (U,I) data to extract a value of e/me. Remember the diameter of the orbit is 2r! You
should also calculate the uncertainty in your measurement. There are a few different approaches
to this, so you might want to discuss your analysis strategy with your demonstrator.
Details
The expression for the B field given by eq. (6) is only exact for the single point in the center of
the coils. You should take additional sets of measurements for smaller beam diameters to see if
your e/me ratio varies systematically with the orbital diameter.
The earth’s magnetic field adds an offset to the field generated by your coils. If you rotate the
apparatus by 180º and repeat your measurement for a particular diameter, you can subtract off
this background. In fact, you can also use your results to get the value of the component of the
earth’s field parallel to the coil axis.
Assuming the spreading in the beam is due entirely to scattering energy loss, can you determine
how much energy the beam loses? Hint: how much voltage is required to move the beam by its
width?
It probably makes sense to use the center of the windings on the coils to define R in eq. (6). How
much difference would it make to your results if you took the inner or outer edge instead? Can
you think of any other factors which could systematically alter your results?
The accepted value of e/me is 1.7588×1011 C kg-1.
Blackett Laboratory, Imperial College London
91
Demonstration Experiments: The charge to mass ratio of the electron
27 September 2012
Appendix: Helmholtz coils
It is well known‡ that the magnetic field at a distance z along the axis of a current loop of radius R
lying in the x-y plane is
R2
.
(a1)
Bz = µ 0 I
3/ 2
2 R2 + z2
We can simplify the notation by measuring all distances in units of R. The field from two coils
separated by their radius, i.e. located at ±½, is then
⎞
µ I⎛
1
1
⎟.
(a2)
+
Bz = 0 ⎜
3/ 2
3/ 2 ⎟
~
1 2
1 2
2 ⎜ 1 + (~
)
(
)
+
+
−
z
1
z
2
2
⎝
⎠
~
At z = 0, this reduces to
(
(
)
)
(
)
3/ 2
⎛4⎞
(a3)
Bz = µ 0 I ⎜ ⎟ ,
⎝5⎠
which is exactly eq. (6) when we take account of the number of turns and put back in the radius
R. The magic of Helmholtz coils is that both dBz / dz and d 2 Bz / dz 2 vanish at the center.§ This
means the magnetic field the coils produce is very uniform. In fact, it is possible** to derive a
general expression for all of the field components. It turns out the field is also very uniform as
you move in the x and y directions around the origin, though the mathematics becomes a bit
messier.
B. E. Sauer, Sept. 2008
‡
See Young and Freedman, §28.5. See also problem 28.67.
The first non-constant term in a series expansion of eq. (a2) around z = 0 goes as z4.
**
See §5.5 in J. D. Jackson, Classical Electrodynamics, 2nd Ed., (Wiley, New York, 1975).
§
Blackett Laboratory, Imperial College London
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