Lab 6: The Charge-to-mass ratio of the electron

In a series of experiments performed just prior to the turn of the century, J. J. Thompson
was able to show that electrons behaved as particles of mass **m** that carried a fixed
amount of (negative) electrical charge **e**. These particles moved in trajectories
governed by the laws of electricity, magnetism and classical mechanics. Thompson received
the Nobel Prize for this work in 1906. (J. J. Thompson's son, G. P. Thompson, received the
Nobel Prize in 1937 for experiments that showed that electrons could exhibit wave-like
properties as well.) Thompson was not able to measure the charge or mass separately but
only the ratio of these two quantities, **e/m**.

In this exercise you will repeat some of Thompson's experiments using apparatus similar to his. This apparatus consists of an evacuated clear glass bulb containing an electron "gun" similar to that found in a modern TV picture tube. This gun is a heated filament that emits electrons that are then accelerated toward a metal plate kept at a positive potential relative to the filament. A narrow slit in the plate allows the electrons to pass through, and they eventually strike a flat mica sheet on which there is printed a centimeter scale. Metal electrodes above and below this mica sheet allow us also to apply electric fields to the region.

The glass bulb is supported between two large coils of wire. The arrangement of these
two coils (suggested by Helmholtz) is such that the separation between them is equal to
one-half of their average diameter. A current through these coils will create a magnetic
field B which is substantially uniform over the central region between the coils. The
strength of this magnetic field is given by

where

N = number of turns of wire (320)

r = mean radius of coils (0.068m)

I = current in Amperes

There are two power supplies on the table in front of you, one for the Helmholtz coils and the other for the electron tube. For simplicity we will refer to these power supplies by brand name:

Lambda and TEL-Atomic.

Notice that each Helmholtz coil has a terminal labeled A and one labeled Z. Connect the Z terminals to one another and the A terminals to the Lambda power supply. The knob labeled CURRENT should be rotated to its extreme clockwise position and left there. The current through the Helmholtz coils is controlled by rotating the knob labeled VOLTAGE and may be read from the meter on the front panel of the Lambda power supply. The current required is so small that you will want to include the multimeter in the circuit to give you a better value for the current.

The filament in the electron gun should be connected to the two yellow terminals labeled 0V and 6V4A on the TEL-Atomic power supply. Also, connect the 0V terminal to the terminal that carries the symbol for electrical ground. The left-most black terminal on the TEL-Atomic supply (which is the negative terminal of the high voltage power supply) should be connected to this same ground. Finally, connect the electrode that sticks out of the neck of the glass tube to the terminal labeled +5kV on the TEL-Atomic power supply. A double female banana plug connector placed over the exposed end of this high voltage wire will protect against electrical shock.

**When you have completed wiring the circuit, ask the laboratory instructor to check out
your circuit before turning on either of the power supplies.**

Part 1: MEASURING e/m

Once your circuit has been approved, turn on the TEL-Atomic power supply and rotate the 5kV VOLTAGE CONTROL in a clockwise direction. When the voltage exceeds about 1500 Volts, you will begin to see the blue glow of the electron beam superimposed upon the light beam from the filament. (The electron beam is easier to see in a darkened room.) Once you are able to see the beam, turn on the Lambda power supply and observe the effect of the magnetic field on the moving electrons. Vary the accelerating voltage as well as the magnetic field. Observe the effect of reversing the direction of the magnetic field by reversing the direction of the current through the Helmholtz coils.

Check your knowledge of the direction of the magnetic field and the direction of the force it exerts on a moving negatively charged particle.

**What is the direction of the magnetic field between the Helmholtz coils for a given direction of the current?****What is the direction of the force that this field exerts on a moving electron?**

An electron moving at speed v through a uniform magnetic field B will trace out a
circular path of radius R given by

The speed v is obtained by recognizing that an electron which starts from rest and
accelerates through a potential difference of V volts will acquire a kinetic energy given
by

We can eliminate v from these equations to obtain

The voltage V in equation (4) is read directly from the TEL-Atomic power supply. For a
given current, the magnetic field B may be computed from equation (1). All we need is a
value for R in order to be able to compute the ratio of electronic charge to electronic
mass. A little analysis will show that, for circles passing through the origin (which is
at the exit aperture of the electron beam) and points (x,y) or (x,-y), the radius R is
given by

(See below for a suggested spreadsheet to organize your data.)

For a given high voltage V, adjust the coil current I so that the electron beam passes
through the point: x = 10, y = + 2.6.

Keeping this same voltage, reverse the magnetic field and alter its strength until the
beam passes through the point: x = 10, y = - 2.6.

Use the average value of these two currents when computing the magnetic field in equation
(1). Repeat this process for five different values of the accelerating voltage and compute
five different values for e/m. The accepted value is 1.7588 x 10^{+11} Coulomb/kg.
Because of various misalignments during the manufacture of the tube, your value will
probably differ from this.

As a consistency check (on the apparatus as well as on yourself), use the accepted
values for the charge (1.60219 x10 ^{-19} Coulomb) and mass (9.10953 x 10 ^{-31}
kg) to compute values for the velocity in equation (2) and equation (3).

Part 2: Crossed Electric and Magnetic Fields

Notice that the metal plates above and below the mica sheet are connected to electrodes on the top and bottom of the tube. If the upper electrode is connected to the electrode at the neck of the tube while the lower electrode is connected to that filament supply lead which is grounded, an electric field will be created which will deflect the electron beam upward along a parabolic path. Turn off the power to the Helmholtz coils to observe this effect.

A combination of electric and magnetic fields applied at right angles forms what is known as a velocity selector and is a common occurrence in mass spectrometers and particle accelerators. The physical principle is quite simple: an electric field E exerts a force qE on a charged particle while a magnetic field exerts a force qvB on the same particle. When these two forces are equal in magnitude but oppositely directed, the particle will move in a straight line. Its velocity will be given by v = E/B.

(See below for a suggested spreadsheet to organize your data.) You can approximate this condition over a narrow portion of the electron beam. Adjust the potential difference applied to the electrodes above and below the beam and the magnetic field until you see the beam move in a straight line across the middle portion of the tube (where the magnetic field is most nearly uniform). Since you have an approximate value for the velocity of the particles (see above), use this equation to compute the strength of the applied electric field E. For parallel plates of infinite extent, the electric field is given by V/d where d is the plate separation. Does this computation give a very accurate value for the electric field between the upper and lower electrodes? If not, why not?

Spreadsheet Setup

Setup the following spreadsheets using Excel:

**MAGNETIC FIELD ONLY**

Accepted value is: e/m = 1.7588 x 10^{+11} C/kg