Specific Charge of an Electron

Specific Charge of an Electron

(last edited October 15, 2018)

Dr. Larry Bortner

Learning Objectives

Explain what happens when a charged particle moves in a uniform magnetic field.

Explain how the mass of an electron can be found by knowing the radius of its orbit in a magnetic field.

Purpose

To observe how a magnetic field affects a moving charge. To find the charge to mass ratio of an electron.

Background

Weighing an Electron?

We can’t build a balance scale small enough and even if we could, the electron wouldn’t stand still long enough to make a measurement. But we can infer what the mass m of an electron is by making other measurements. In particular we can find the specific charge (the mass-normalized charge) of this fundamental particle. Since we can find the charge e on an electron from other experiments (like Millikan’s oil drop experiment), an experimental e/m ratio can also give the electron’s weight (that is, its mass).
The charge-to-mass ratio of an electron has been determined using many methods requiring different combinations of applied electric fields and magnetic fields. In 1897, J. J. Thomson determined the ratio of charge to mass of particles in cathode rays, using different residual gases in the discharge tubes. He discovered that e/m for each particle was the same regardless of which gas he used. Thompson concluded that these different atoms all contained the same particle. Today, this particle is known as the electron.
We will determine the charge-to-mass ratio by Lenard’s method of measuring a beam of electrons (the cathode ray) bent into a circular path by a known magnetic field. The value of e/mcan be determined from the radius of curvature of the path of the electrons.

Force on Moving Charge in Magnetic Field

Suppose a charged particle like an electron is moving in a magnetic field and that the direction of the motion is at right angles to that field. It is acted upon by an electromagnetic force F which is perpendicular both to the direction of the field and to the direction of the velocity of the particle (Fig. 1).

Figure 1The magnetic force on a negatively-charged electron moving in a magnetic field.

The magnitude of the force depends on the charge of the electron e, its velocity v, and the magnetic field B:

(1)

Circular Motion of Electron

Since the resulting force is perpendicular to the velocity, it causes the electron to move in a circular path of radius r (Fig. 2).

Figure 2Circular path of an electron in a uniform magnetic field.

Hence F is a centripetal force:

(2)

Equating Eqs.(1)and (2), we have

(3)

This gives us the expression

(4)

“Falling” Electrons

A heated filament in a helium-filled tube provides the electrons. As the metal of the filament is heated, some of the “free” electrons of the metal gain enough kinetic energy to break away from the surface of the metal. A constant electric field is set up between the filament and a metal plate, giving the electrons that reach the plate a specific velocity. From conservation of energy, the kinetic energy which the electrons have gained going from the negative filament (cathode) to the positive plate (anode) is equal to the potential energy which they lose by falling through the potential differenceV between the anode and cathode. This potential energy is the work done on the electron by the electric field, or Ve.
Therefore:

(5)

The speed that the electron acquires is given by

(6)

Substituting Eq. (6) into Eq. (4) leads to

(7)

Eq. (7) gives the ratio of charge to mass of an electron, knowing the accelerating potential V, the flux density of the magnetic field B, and the radius r of the circular path of the beam of electrons. To calculate e/m, use the SI system, where ris in meters, V in volts, and B is in tesla.

Current Dependence of Magnetic Field

The experimental apparatus consists of a large vacuum tube that produces a collimated beam of electrons (a cathode ray). The tube sits at the center of a pair of Helmholtz coils that provides a nearly uniform magnetic field over a considerable volume between them. The two identical coils are separated by a distance equal to their radius. The magnetic field at the center is parallel to the axis of the coils and is proportional to the current:

(8)

whereγis a geometrical/physical conversion factor given by

(9)

For Eqs.(8)and (9),

1.B is the magnetic field in tesla.

2.Iis the current in amperes going through each coil.

3.N is the number of turns of wire on each coil, which is 130 for these coils.

4.dcoil is the mean diameter of each coil.

5.is the permeability of empty space.

Experimental Relation to Verify

We can now relate the specific charge to directly measurable quantities and constants, specifically the electron accelerating voltage V, the current I that produces the magnetic field, and the radius r of the electron orbit. Substituting our expression for B from Eq. (8) into Eq. (7) and rearranging, we get

(10)

This means that if we plot V vs. ½(γIr)2, we should get a straight line with a slope that is the fundamental value of the specific charge of the electron, e/m.
As mentioned, the tube contains low pressure helium. When an electron with sufficient kinetic energy collides with a helium atom, the atom ionizes. As these ions recombine with stray electrons, the characteristic bluish violet color of the helium spectrum is emitted and the path of the electron beam is seem.

Figure 3Diagram of experimental setup, turned on its side.

Fig. 3 represents the experimental layout. There are three different circuits: that of the filament, of the anode, and of the Helmholtz coils.

• The filament is tungsten. When the proper amount of current runs through it (controlled by internal logic), it heats up enough that some electrons escape the metal.
• The anode voltage is the potential difference between the metal plate and the filament. There is a small hole in the center of the plate that allows a small proportion of the electrons with a speed given by Eq. (6)to continue traveling in the uniform magnetic field.
• The magnetic field is produced by a current running through two identical coils in series.

Earth’s Magnetic Field Contribution

In this experiment, the earth’s magnetic field is almost large enough to visibly affect the electron’s path. The magnitude is less than one per cent of the central field produced by the coils, so we can ignore the geomagnetic field for the most part. To minimize the effect, we'll align the axis of the coils perpendicular to the north-south axis.

Procedure

You need the following items:

• Daedalon EP-20 e/m of the Electron Apparatus
• blackout cloth
• dip needle
• tape measure
• magnifying glass

1.Turn on the instrument. It will go through a warm up and internal check lasting thirty seconds before you can use it. You should see a small yellow light that is the glow of the filament.
2.Double check with the dip needle to make sure that the coils are properly aligned. Is the tube axis perpendicular to the N-S direction? (Is the long side of the apparatus parallel to the N-S direction?)

Figure 4Diameter of the coil.

3.Use the tape measure to find the diameter dcoilof one of the Helmholtz coils. (See Fig. 4.) Measure from the center of the bundle of wires and assume an uncertainty of 0.5 cm.
4.Choose two coil currents between 1.50 A and 2.00 A that are at least 0.25 A apart.
a.Use thecurrent adjust knob to set thecoil currentat one of these values.
5.Be sure the blackout cloth is draped over the coils.
a.Twist the voltage adjust knob clockwise. At an accelerating voltage of about 100 V, a thin bluish-greenish beam appears. These are the electrons recombining with the helium. Since there is an external magnetic field, the pathof these charged particles is a small circle.
6.Continue increasing the voltage until the electron beam hits the 5 cm mark on the glass rod going through the center of the bulb. (The magnifying glass may help.) Record the current I, the path diameter dorbitand the voltage V.
7.Repeat Step 6 for path diameters of each half centimeter up to and including 11 cm. (dorbit(cm)=5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5, 10.0, 10.5, 11.0)
a.Return the voltage to zero.
8.Repeat steps 5-7 for the second coil current (and a third if you're so inclined).
a.You should have about 26 data points to analyze. For the higher magnetic fields, you may not achieve the larger orbits, since the voltage cannot go higher than 500 V.
9.Estimate and record uncertainties in I, dorbit, and V.
10.Turn off the apparatus.

Analysis

1.Find the Specific Charge template and fire it up.
2.Enter in values of μ0, N, dcoil, u{dcoil }, u{ dorbit }, u{I}, and u{V}.
3.Enter the coil currents, path diameters, and voltages into the appropriate blue-shaded columns.
4.Calculate the radii in meters for each orbit.
5.Let y=½(γIr)2. Calculate y in Vkg/GC for each point.
a.Use SI units and multiply by 109.
6.Propagate the uncertainty of y.
a.Calculate u{y} for each point.
7.Calculate values of y−u{y}and y+u{y}in their appropriate columns.
8.The anode voltage V is automatically plotted vs. ½(γIr)2. You need to add the error bars.
a.Click on the plot. A Chart Tools tab appears on the Ribbon.
b.Click onLayout> Error Bars in the Analysis group> More Error Bars Options…> Custom> Specify Value.
c.Click on the little spreadsheet to the right of the data entry window for the Positive Error Valuethen select all of the values of u{y}.
d.Repeat this last step for the Negative Error Value
e.Click OK and Close the Format Error Bars box.
Note that the error bars are relatively small at the lower end of the graph and get larger as the line rises and that most of the points are fairly close to the Trendline. This means that the measurement errors are greater than the inherent random error and that the standard deviation of the slope generated by LINEST does not give the best error estimate.
9.To complete the analysis of this data:
a.Find the least squares slopes of V vs y−u{y}and V vs. y+u{y}. You need just the slope, nothing else. This is one of the times where you can use LINESTwithout adding the 1 and 1 at the end and hitting Shift/Control/Enter.
b.Find the average of these two slopes for the experimental value of e/m.
c.The standard error of these values is u{e/m}.
10.Compare your experimental value to the accepted value of

Questions

1.Derive Eq. (7) using Eqs. (4)and (6).
2.Suppose that the filament in the tube emitted muons instead of electrons and that they interacted with helium in the same way (their path in the tube glowed). For one of your coil currents, how would the accelerating voltage change to give the same orbital diameters?