Figure 1. Leybold-LaPine apparatus for measuring q/m of electron
Measurement of e/m for the electron, using a Leybold-LaPine apparatus
The purpose of the experiment is (1) to study the motion of charged particles in a magnetic field as a function of field intensity and of particle velocity, and (2) to measure the charge/mass ratio for the electron.
To see how a charged particle really moves through a magnetic field, and how its motion varies with its velocity and the strength of the magnetic field.
Motion of charged particle in magnetic field: Cutnell and Johnson 21.2, 21.3
Magnetic field produced by a current: Cutnell and Johnson 21.7. See Figures
21.33 and 21.37.
The Leybold-LaPine gas-focused cathode ray tube apparatus is shown in the picture below. The apparatus consists of a spherical glass bulb, filled with hydrogen at a fairly low pressure. An electron gun inside the bulb produces a beam of electrons, whose path is made visible by a pale blue glow from hydrogen ions excited by the electron beam. The presence of the hydrogen ions, with their relatively small mobility, also serves to focus the beam by attracting straying electrons back into the beam. The velocity of the electrons in the beam is controlled by controlling the accelerating voltage applied to the electron gun.
A pair of circular coils of wire, called Helmholtz coils, is used to produce a uniform magnetic field within the region occupied by the bulb. This configuration consists of parallel circular coils, of equal turns and equal radius, separated by an axial distance equal to the radius. In this apparatus N, the number of turns of each coil is 130, and R0, the coil radius is 150 mm. They are connected in series to a 6-volt DC source, and can be loaded up to approximately 5 amperes.
Figure 1. Leybold-LaPine apparatus for measuring q/m of electron
A regulated high voltage dc power supply can be used to provide the accelerating voltage, which can be as high as 300 volts. This same power supply also provides the necessary 6.3-volt ac filament voltage for the electron gun.
The apparatus should be set up when you arrive in lab. If so, do not fiddle with the wiring.
In the unlikely event that it is necessary to connect the wiring yourself, follow these instructions.Connecting the high-voltage supply (HVPS) to the Leybold-LaPine (LL) apparatus:
- from "0 to 400 Volts +" on HVPS to "Electron Gun electrodes upper +" on LL
- from "0 to 400 Volts common" on HVPS (on right side of GND) to "Electron Gun electrodes lower -" on LL
- from "25 VA max, 6.3 VAC 4 amps" left terminal on HVPS to "Electron Gun heater AC or DC 6.3V" lower terminal on LL
- from "25 VA max, 6.3 VAC 4 amps" right terminal on HVPS (which is the central terminal in a set of 3) to "Electron Gun heater AC or DC 6.3V" upper terminal on LL
Connecting BK precision DC power supply (DC) to Leybold-LaPine (LL) apparatus
- from red terminal (rightmost in set of 3) on DC to "Helmholtz coils 6~9 V2A +" upper red terminal on LL
- from black terminal (leftmost in set of 3) on DC to "Helmholtz coils 6~9 V2A -" lower black terminal on LL
Initial adjustments to Leybold-LaPine apparatus:
- set Helmholtz coils current adjustment knob to "10"
- set up/down switch to "e/m measure"
Initial adjustments to high-voltage power supply
- set "DC on/standby" toggle switch to "DC on"
- set "Voltmeter" rocker switch to right position ("B+ volts")
In order to measure the diameter of the circular path of the electrons in the electron beam tube, use is made of the virtual image produced by a plane mirror. An illuminated centimeter scale is viewed by reflection from the surface of a plane sheet of transparent plastic.
To measure the diameter of the blue circle, do the following: close one eye, and always use the other to make readings. Move your head to the left, so that the reflection of the left-hand section of the circle lines up with the left-hand section of the circle itself. Read the scale at the point where this section of the circle crosses it, and write down the number. Now, move your head to the right-hand side of the bulb, so that the reflection of the right-hand side of the blue circle lines up with the right-hand side of the circle itself. Once again, read the scale at the point where the blue circle crosses it, and write down the number. The diameter of the circle (in cm) is the sum of the two numbers.
Note that it may be difficult to read the diameter of large circles (larger than 5.5 cm radius) due to the optical distortion produced by curved glass in the bulb. Just do the best you can.
If the initial velocity of the electron beam is not at right angles to the magnetic field, the component of velocity parallel to the field will be unaffected by the magnetic force, and the path of the electrons will be a helix. This may make it difficult to measure the diameter of the electron's path accurately.
The velocity of the electrons in the beam depends on the accelerating voltage V applied to the electron gun by the relationship:
1 2
--- m v = qV Equation 1
2
where m is the mass of an electron, v is its velocity,
and q is
its charge. The velocity is thus proportional to the square root
of the voltage.
The magnitude of the force on the electrons due to the magnetic
field B is given by the expression:
F = q v B sin(theta) Equation 2
where theta is the angle between the magnetic field lines
and the direction of the electon's velocity v.
If the velocity and the field are at right angles to each another,
the path is a circle, and the centripetal force is simply
F = qvB.
Equating the force to the centripetal acceleration times
the mass, one obtains the expression:
m v
R = ----- Equation 3
q B
for the radius R of the circular path.
The student should eliminate the velocity v by combining equations (1) and (3), the rearranging terms to obtain an expression for the ratio of the charge on an electron to its mass, q/m, in terms of the magnetic field B, the accelerating voltage V, and the radius of the electrons' path R.
The magnetic field at the center of a Helmholtz coil configuration can be shown to be given by:
3/2 N I
B = mu (4/5) -----
r
where
mu is the permeability of free space, N is the number
of turns of wire in the Helmholtz coils, I is the
current flowing through the coils (in amps),
and r is the radius of the coils.
For the coils you are using, this turns into
-4
B = 7.793 x 10 I Tesla
where I is the current passing through each coil in amps.
For each value of the current I to the Helmholtz coils, calculate the corresponding strengh of the magnetic field B.
You should have derived above an equation for the ratio q/m as a function of V, B, and R. In order to find q/m graphically, we need to rearrange the equation so that it looks like this:
y = (q/m) x
where x and y are some combinations
of V, B and R.
Then, plot points calculated from your all your data on a single
graph, and fit a straight line to the points.
The slope of that line is equal to the ratio q/m.
Make sure that the units you use are correct: meters, volts,
and Tesla.
Compare your value of q/m with the accepted value (look up the charge of an electron and its mass in the front of your textbook). Calculate the uncertainty in each direction for each plotted point, based upon estimated uncertainty in your measurements of the current, voltage and radius. Place error bars on each point in your graph (unless they are too small to see). Estimate the uncertainty in the slope of your best-fitting line, which translates into an uncertainty in your value of q/m. Does your value agree with the book value, within its uncertainty? If not, why?
A derivation of the formula for q/m.
Step 1. Solve equation (1) for v^2:
2 2 q V
v = ------ Equation 4
m
Step 2. Solve equation (3) for v, then square it
2 2 2
q B R 2 q B R
v = ------ -> v = --------- Equation 5
m 2
m
Step 3. Combine Equations (4) and (5)
2 2 2
2 2 q V q B R
v = ------ = ---------- Equation 6
m 2
m
Step 4. Solve for q/m
q 2 V
--- = -------- Equation 7
m 2 2
B R
DATA SHEET FOR CHARGE/MASS OF ELECTRON
Write in the radius of the circle (R), in centimeters, made by electrons
for each combination of voltage to the electron gun (V), in volts,
and current flowing through the Helmholtz coils (I), in amps.
For each value of I, calculate the magnetic field strength in the
center of the coils (B), in Tesla.
When you use the values for radius (R) in your calculations, remember
to translate from centimeters into meters!
Current Mag Voltage to electron gun (Volts)
to coils field
(amps) (Tesla) 75 100 125 150 175 200 225 250 275 300 350
-----------------------------------------------------------------------------
|
1.0 |
|
_____________________________________________________________________________
|
1.5 |
|
_____________________________________________________________________________
|
2.0 |
|
_____________________________________________________________________________
|
2.5 |
|
_____________________________________________________________________________
|
3.0 |
|
_____________________________________________________________________________
Last modified 1/23/2002 by MWR