Charge to mass ratio for an electron
OBJECTIVES
To measure the charge to mass ratio of the electron and thereby gain familiarity with some of the basic principles utilized in studying charged particles.
APPARATUS
e/m tube and Helmholtz coils (Figure 1Figure 1)
Power supplies for Helmholtz coils, filament and accelerating potential
Voltmeters, ammeters and switches as required
Dip needle
Meter stick
Figure 1: e/m tube and Helmholtz coil
To measure the properties of the electron, we first need a source of electrons. A hot filament (like that of a light bulb) gives off electrons as well as light. We will use a filament in a tube that has all of the air removed, and then a very low pressure of gas (mercury) is added back (see Figure 3Figure 3). The electrons that come fromemitted by the filament are accelerated by a voltage applied between the filament (the cathode, labeled as F in Figure 3Figure 3) and a metal cover surrounding the filament (the anode labeled as C in Figure 3Figure 3). After they have escaped through the slit, some of the electrons that leave the filament excite the gas molecules and make the electron trajectory visible. The ionized gas emits a faint blue light which can easily be seen. The low pressure of the gas in the tube will keep too much ionization from occurring, which could result in a “space charge” problem.
This lab also requires a source of uniform magnetic field. A uniform magnetic field can be produced by use of Helmholtz coils. Helmholtz coils consist of two coils of wire that are placed parallel to each other and separated by a distance equal to their radius. The magnetic field intensity produced by ideal Helmholtz coils is quite uniform in the center of the coils.
INTRODUCTION
The work that led to the concept of quantization of charge was done by the English physicist, Michael Faraday, in his famous work on electrolysis (1831 - 1837), in which he established the Faraday laws of electrolysis and the quantization of charge. In 1897, J.J. Thomson measured the charge-to-mass ratio of cathode rays (electrons). Then, in 1909, R.A. Millikan measured the charge on the electron by means of the Millikan oil-drop method. This confirmed the quantization of charge, and permitted the calculation of the mass of the electron.
When a charged 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 that force is given by
F = eBv (Equation 1)
where F is the force in Newtons, B is the magnetic flux densityfield in tesla (Webers/meter2T), e is the charge of the electron in coulombs, and v is the velocity of the electrons in meters/second.
Figure 2: Path of an electron in a magnetic field
This force causes the particle to move in a circle in a plane perpendicular to the magnetic field (Figure 2Figure 2). The radius of this circle is such that the required centripetal force is furnished by the force exerted on the particle by the magnetic field. The centripetal force necessary to move in a circle or radius r, you may recall, is given by . Therefore:
(Equation 2)
where m is the mass of the electron in kilograms, and r is the radius of the circle in meters.
Finally, to use Equation 2 in this lab, we will need some way of determining the velocity of the electron. Recall that an electron accelerated through a difference of potential V will have a kinetic energy, KE, of eV. The kinetic energy, KE, can also be expressed in terms of its mass and velocity:
or
(Equation 3)
where V is the accelerating potential in volts.
Equations 2 and 3 can be solved for the ratio of the electron’s charge to mass, e/m, in terms of measurable or calculable values: the accelerating potential, V; the radius of curvature of the electron beam, r; and B, the magnetic field intensity. You will do this calculation for your writeup. See the CALCULATIONS section at the end of the lab.
The uniform magnetic field we will need will be produced by use of Helmholtz coils. Helmholtz coils consist of two coils, with N turns ofr wire each and radii a, that are placed parallel to each other and separated by a distance equal to their radius. The magnetic field intensity produced by the Helmholtz coils can be calculated from the equation:
(Equation 4)
where N is the number of turns of wire on each coil (72 for this coil), I is the current in Amperes, a is the mean radius of the coils in meters, µ0 is the permeability of empty space (4π´10-7 TWb•Am-1•Am-1). Using the units above, B will be in webers/meter2 (tesla).tesla.
Figure 33: A top view of the e/m tube (left) and a side view closeup of the filament assembly (right)
A Five cross bars attached to staff wire
B Typical path of beam of electrons
C Cylindrical anode
D Distance from filament to cross bar
E Lead wire and support for filament
F Filament
G Lead wires and support for filament
L Insulating plugs
S Slit in cylindrical anode
Using a uniform magnetic field to bend the path of charged particles as we are doing with the e/m tube and the Helmholtz coils has applications beyond verifying the well-known constants for e and m. It is also the principle behind a mass spectrometer. Particles carrying the same charge but differing in mass move in orbits of different radii in a given magnetic field. It is, therefore, possible to separate ions of different atomic mass. Indeed, the first evidence that all chlorine atoms were not identical, but consisted of 75% 35Cl and 25% 37Cl was obtained in this way.
PROCEDURE
1. Measure and record the radius of the Helmholtz coils.
2. The magnetic field produced by the coils will be perpendicular to the plane of the coils. Therefore, the e/m tube should be oriented in the coils such that the electrons from the filament, F, (see Figure 3Figure 3) when they exit from the slit, S, will be traveling at right angles to the magnetic field, B.
3. Since the magnetic field of the earth will also exert a force on the moving electrons, the coils should be positioned such that the magnetic field produced by the coils will be in the opposite direction to that of the earth’s magnetic field. A dip needle should be used to determine the dip of the earth’s magnetic field. The Helmholtz coils then should be elevated to the complement of this angle. The magnetic field of the coils will now either add to or will neutralize the earth’s magnetic field depending on the direction of the current through the field coils.
4. Set the accelerating potential to 25 volts.
Figure 4: Path of electrons with no applied field from the Helmholtz coils.
5. Adjust the filament current to 4 amperes. Do not exceed a filament current of 4 amperes. You should see some white/yellow light coming from the slit, as well as a bluish beam. That beam is due to the electrons that are emitted interacting with the mercury gas in the tube. Note that the light comes straight out, while the beam is probably curved a bit (Figure 4Figure 4). The curvature of the beam is produced by the earth’s magnetic field. Since light is not a charged particle, the white light is not bent like the blue electron beam.
6. Adjust the current in the field coils until the electron beam is undeviated. You can tell when it is undeviated by having the blue beam aimed directly into the center of the light when it hits the side of the tube. The current necessary to straighten the beam produces a magnetic field from the coils which will exactly cancel the earth’s magnetic field. You may need to be tall or sit on a high stool to get a good view of the path of the beam. Record this current as I0.
7. Increase the current in the field coils so that the outside of the electron beam just touches crossbar number 1 (letter B in Figure 3Figure 3 shows the desired touching electron beam position for crossbar 2). Record this current as I1.
8. Repeat step 7 for crossbars 2, 3, 4, and 5. The diameters of the circles, D (see Figure 3Figure 3) scribed by the electron beam when it strikes the various posts are shown in Table 1Table 1 below.
9. Repeat steps 6 to 87 for accelerating potentials of 40, and 60 volts.
Calculations and Questions FOR LAB WRITEUP
CrossbarNumber / Distance to Filament
(D) in meters
1 / 0.115
2 / 0.103
3 / 0.090
4 / 0.078
5 / 0.065
Table 1: Diameters of the electron path for each crossbar
1. Derive an expression for e/m in terms of the measurable quantities: the accelerating potential V, the magnetic field intensity (B), and the radius (r) of the circle of the electron beam (=D/2). Start with eqns 2 and 3 and solve for e/m. Do not plug in known values of e and m.
2. Calculate the current (I) in the Helmholtz coils necessary to produce the magnetic field that deflects the electrons I = I1 – I0. See steps 6 and 7. You will have a current I for crossbar number 1, 2, 3, 4, and 5 for an accelerating potential of 25 volts, and a similar set of currents for accelerating potentials of 40 and 60 volts.
3. Calculate, using I0, the earth’s magnetic field. How does this compare with published values?
4. For each circular path described by the electron beam, compute the values of e/m. This will consist of a set of 15 values, one for each crossbar at each accelerating potential.
5. Calculate the experimental value of e/m and explain your approach, rationale, or criteria for the procedure used (compare with accepted value).
6. From your calculated value of e/m and the known value of e, calculate the mass of the electron.
7. Name two sources of error which are most likely to affect the accuracy of your results.
8. If a television tube has an acceleratingon potential of 30,000 volts, what will be the velocity of the electrons as they strike the screen? Express the results in m/s.
9. Calculate the magnetic field intensity, B, required to bend electrons that have been accelerated through a potential difference of 100 volts into a circle 20 cm in diameter. Express your answer in tesla.
10. Calculate the time, T, required for an electron to make a complete circle in the e/m tube. Assume the magnetic field is held constant. How is this time related to the accelerating potential?
11. In deriving an expression for e/m, it has been assumed that the final velocity of the electrons is small compared with that of light, i.e., no relativistic corrections are necessary, Is this assumption justified?