Physics307L:People/Le/Notebook/071008
e/m Ratio
Purpose
We are looking at the ratio of the charge of an electron, to its mass.
Equipment
SJK 01:11, 13 November 2007 (CST)
It's great that you put model number and manufacturer on the main apparatus... you should do the same for the other equipment! Also, somewhere you will want to mention what software you are using for data analysis.
Uchida e/m Experimental Apparatus, Model TG-13
3 DC power suppies
2 Multimeters
Procedure
SJK 01:13, 13 November 2007 (CST)
Good description. You will also want to link to the lab manual and say something like, "following the procedure here, with these changes and / or specific details."
Our apparatus is an electron gun, inside a bulb filled with helium, inside a Helmholtz Coil. The electron gun has a metal element that is heated to release electrons, which are then accelerated through an electric field into the bulb. The bulb, filled with the gas, then interacts with these electrons, causing the gas to emit a green light. We can then turn on the Helmholtz Coil to create a magnetic field that is perpendicular to the electron beam, causing a deflection.
We measure the radius of curvature of the beam with a ruler in the background of the bulb. Lining up the right side of the beam with its image in the ruler and vice versa with the left, we can try to eliminate the paralax between the beam and the ruler behind it. We then adjust the strength of the accelerating plates in the gun and the strength of the magnetic field and measure the radii accordingly.
Data
Set 1
B-field constant, change accelerating voltage
| Accelerating Voltage | Coil Current | Left Radius (cm) +/-0.1 | Right Radius (cm) +/-0.1 | Avg Radius (cm) +/- |
|---|---|---|---|---|
| 246 | 1.02 | 4.5 | 4.9 | 4.7 |
| 246 | 1.04 | 4.4 | 4.8 | 4.6 |
| 246 | 1.06 | 4.4 | 4.6 | 4.5 |
| 246 | 1.08 | 4.4 | 4.6 | 4.5 |
| 246 | 1.10 | 4.4 | 4.6 | 4.5 |
| 246 | 1.12 | 4.4 | 4.5 | 4.45 |
| 246 | 1.14 | 4.4 | 4.5 | 4.45 |
| 246 | 1.16 | 4.4 | 4.5 | 4.45 |
| 246 | 1.18 | 4.4 | 4.4 | 4.4 |
| 246 | 1.20 | 4.4 | 4.3 | 4.35 |
| 246 | 1.22 | 4.3 | 4.2 | 4.25 |
| 246 | 1.24 | 4.2 | 4.2 | 4.2 |
| 246 | 1.26 | 4.2 | 4.0 | 4.1 |
set 2
B-field constant, change accelerating voltage
| Accelerating Voltage (V) | Coil Current (A) | Left Radius (cm) +/-0.1 | Right Radius (cm) +/-0.1 | Avg. Radius (cm) +/- 0. |
|---|---|---|---|---|
| 246 | 1.02 | 4.5 | 4.6 | 4.55 |
| 246 | 1.04 | 4.5 | 4.6 | 4.55 |
| 246 | 1.06 | 4.5 | 4.5 | 4.5 |
| 246 | 1.08 | 4.5 | 4.5 | 4.5 |
| 246 | 1.10 | 4.5 | 4.5 | 4.5 |
| 246 | 1.12 | 4.5 | 4.5 | 4.5 |
| 246 | 1.14 | 4.5 | 4.4 | 4.45 |
| 246 | 1.16 | 4.4 | 4.4 | 4.4 |
| 246 | 1.18 | 4.4 | 4.3 | 4.35 |
| 246 | 1.20 | 4.4 | 4.3 | 4.35 |
| 246 | 1.22 | 4.3 | 4.3 | 4.3 |
| 246 | 1.24 | 4.3 | 4.2 | 4.25 |
| 246 | 1.26 | 4.2 | 4.1 | 4.15 |
Set 3
B-field constant, change accelerating voltage Changed heater voltage to 6.0 from 6.2
| Accelerating Voltage (V) | Coil Current (A) | Left Radius (cm) +/-0.1 | Right Radius (cm) +/-0.1 | Avg. Radius (cm) +/- |
|---|---|---|---|---|
| 246 | 1.02 | 3.9 | 4.0 | 3.95 |
| 246 | 1.04 | 3.9 | 4.0 | 3.95 |
| 246 | 1.06 | 3.9 | 3.9 | 3.9 |
| 246 | 1.08 | 3.9 | 3.9 | 3.9 |
| 246 | 1.10 | 3.9 | 3.9 | 3.9 |
| 246 | 1.12 | 3.9 | 3.9 | 3.9 |
| 246 | 1.14 | 3.9 | 3.9 | 3.9 |
Set 4
B-field changed, constant accelerating voltage
| Accelerating Voltage (V) | Wire Current (A) | Left Radius (cm) +/- .1 | Right Radius (cm) +/- .1 | Avg Radius | Radius SqauredAvg |
|---|---|---|---|---|---|
| 250 | 1.31 | 4.3 | 4.0 | 4.15 | 17.2225 |
| 255 | 1.31 | 4.3 | 4.1 | 4.2 | 17.64 |
| 260 | 1.31 | 4.4 | 4.1 | 4.26 | 18.0625 |
| 265 | 1.31 | 4.5 | 4.2 | 4.35 | 19.9225 |
| 270 | 1.31 | 4.5 | 4.2 | 4.35 | 19.9225 |
| 275 | 1.31 | 4.5 | 4.3 | 4.4 | 19.36 |
| 280 | 1.31 | 4.5 | 4.3 | 4.4 | 19.36 |
| 285 | 1.31 | 4.5 | 4.4 | 4.45 | 19.8025 |
| 290 | 1.31 | 4.6 | 4.4 | 4.5 | 20.25 |
| 295 | 1.31 | 4.6 | 4.5 | 4.55 | 20.7025 |
| 300 | 1.31 | 4.6 | 4.5 | 4.55 | 20.7025 |
Calculations
To calculate e/m ratio we can use a sum of forces (ignoring gravity) and with Newton's 2nd law and centripetal motion we have
[math]\displaystyle{ F=ma }[/math]
[math]\displaystyle{ qvB=\frac{mv^2}{r} }[/math]
solve for q/m using:
[math]\displaystyle{ qV=\frac{1}{2}mv^2 }[/math]
and solving for v and plugging into above
[math]\displaystyle{ \frac{q}{m} = \frac{2V}{B^2 r^2} }[/math]
where [math]\displaystyle{ B=7.8 x 10^{-4} \frac{weber}{amp-meter^2} x I }[/math]
