# Physics307L:People/Gibson/Notebook/070924

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## Contents |

## E/M Ratio Lab

__DH 1__

__DH 1.2__To begin this set up, we followed the procedure of connecting all the wires ect. as described in the lab manual from the previous semester. Here I will enter the paragraph listed from that. The equipment used is as follows: Two multimeters one used to measure the voltage from the high voltage source which is used to determine the average velocity of the electrons. The 2*n**d* multimeter is used to read the current running through the Helmholtz coil.

The Helmholtz coils require us to warm up the electron gun for 2 minutes before we can view electrons in the helium tube.

## Data

__DH 2__

__DH 3__

^{SJK 23:04, 17 October 2007 (CDT)}The key to measuring the e/m ratio is to start taking the radius of the electron beam circle. This is done with a mirror scale behind the beam itself measured in cm. We continued to take 10 measurements first taking all measurements at constant voltage (making sure to be above 250V for precision of measurements), then took 10 at constant voltage then took 10 at both varying voltages and currents.

**Measurements**

1st 10 measurements are made at constant voltage, with varying current. 2nd 10 measurements are made at constant current, with varying voltage.

This group of data is under constant current of 1.20 ASet 1:Set 2:Voltage: 270V Voltage: 270V Current: 1.04 A Current: 1.01A Radius: SOON cm (5.4 from right, 4.1 from left) Radius: 4.5 cm (5.5 from right, 4.1 from left)Set 3:Set 4:Voltage: 270V Voltage: 270V Current: 0.95 A Current: 1.18 A Radius: 4.6 cm (5.6 from right, 4.0 from left) Radius: 4.6 cm (5.0 from right, 4.0 from left)Set 5:Set 6:Voltage: 270 V Voltage: 270V Current: 1.24 A Current: 0.99 A Radius: 4.0 cm (4.8 from right, 3.9 from left) Radius of circle: 4.55 cm (5.5 from right, 4.1 from left)Set 7:Set 8:Voltage: 270V Voltage: 270 V Current: 1.22 A Current: 1.13 A Radius: 4.35 cm (4.9 from right, 3.9 from left) Radius: 4.20 cm (5.2 from right, 4.0 from left)Set 9:Set 10Voltage: 270V Voltage: 270 V Current: 1.08 A Current:1.30 A Radius: 4.3 cm (5.3 from right, 4.1 from left) Radius: cm (4.6 from right, 3.9 from left)

^{SJK 23:38, 17 October 2007 (CDT)}Set 1:Set 2:Voltage: 270 V Voltage: 289 V Current: 1.35 A Current: 1.35 A Radius of circle: 3.7 cm Radius: 3.9 cmSet 3:Set 4:Voltage: 298 V Voltage: 307 V Current: 1.35 A Current: 1.35 A Radius: 3.95 cm Radius: 4 cmSet 5:Set 6:Voltage: 313V Voltage: 320 V Current: 1.35 A Current: 1.35 A Radius: 4.1 cm Radius: 4.15 cmSet 7:Set 8:Voltage: 330V Voltage: 348V Current: 1.35 A Current: 1.35 A Radius: 4.25cm Radius: 4.35 cmSet 9:Set 10:Voltage: 360 V Voltage: 397V Current: 1.35 A Current: 1.35 A Radius: 4.45 cm (5.0 from right, 4.2 from left) Radius: 4.6 cm (5.1 from right, 4.3 from left)

TOTALLY RANDOM

Set 1:Set 2:Voltage: 180 V Voltage: 189 V Current: 1.21 A Current: 1.02 A Radius: 3.7 cm (3.9 from right, 3.5 from left) Radius: 4.15 cm (4.7 from right, 3.6 from left)Set 3:Set 4:Voltage: 198 V Voltage: 207 V Current: 1.09 A Current: 1.18 A Radius: 4.15 cm (4.6 from right, 3.7 from left) Radius: 4.1 cm (4.4 from right, 3.8 from left)Set 5:Set 6:Voltage: 212V Voltage: 220 V Current: 1.25 A Current: 1.30 A Radius: 3.95 cm (4.2 from right, 3.7 from left) Radius: 3.85 cm (4.1 right, 3.6 from left)Set 7:Set 8:Voltage: 230V Voltage: 248V Current: 1.33 A Current: 1.14 A Radius: 3.85 cm (4.2 from right, 3.7 from left) Radius: 4.5 cm (5.0 from right, 4.0 from left)Set 9:Set 10:Voltage: 260 V Voltage: 288V Current: 1.01 A Current: 1.33 A Radius: 4.8 cm (5.5 from right, 4.1 from left) Radius: 4.3 cm (4.6 from right, 4.0 from left)

## Interpreting the data (Finding *e* / *m*)

**Here to determine e/m we used the following from the lab manual: for constant velocity, plotting r vs. I^-1 with a least squares plot gives a value for e/m. For constant current, the plot goes like r^2 vs V. As shown below:**

__DH 4__

^{SJK 23:43, 17 October 2007 (CDT)}Constant Velocity

Constant Current: Here is the excel file used to determine the various parts of the e/m lab. This also includes the graph of r^2 vs. V

Although I can't get my percent error to show using the formula actual-theoretical/theoretical *100 gives me an error of 92.3%. I realize this must be some experimental error. Other reports of error for data set 1 and the random set is included in the excel file above, each display 200+ % error and ~75% error. I honestly don't know how to compensate for this.

## Conclusions

**In our search for determining the ratio e/m, we used the formula**

__DH 5__**to determine the e/m ratio from the average of our 3 data sets. We then compared this to the actual value of e/m = 1.76E+11 C/Kg we saw quite a discrepancy. To account for this several things were considered and we first looked to another groups data to see if they were reporting answers that were in the ball park of what we were getting. This turned out to be the case that they too showed similar orders of magnitude for their e/m ratio tests (Lorenzo and Tomas).**

^{SJK 23:24, 17 October 2007 (CDT)}The first using data involved a constant voltage, the 2nd constant current and the third were completely random values as can be seen above in the measurement tables.

1st Data set measurement

e/m = 6.28343E+11 +/- .725E+11 C/Kg (Koch says (6.3 +/- 0.7) * 10^11)

2nd Data set measurement

e/m=3.39E+11 +/- .0272E+11 C/Kg so, (3.39 +/- .03) E+11

3rd Data set measurement

e/m = 3.07111E+11 +/- .00840E+11 (3.07 +/- .01)*10^11

Questions: Why do we see a beam at all? -A current is supplied to a metal plate in the hydrogen tube, then when this plate gets hot it burns off electrons which then are accelerated between two plates then arced through in a circle due to the B field caused by the Helmholtz coils.

We ignored the earth's magnetic field in this experiment. How much error does this actually induce?

-http://www.ngdc.noaa.gov/seg/geomag/ has an awesome calculator for figuring this out, however I'll skip it for now and move on but it does pose an interesting error.

^{SJK 23:49, 17 October 2007 (CDT)}Suppose the beam emitted protons instead of electrons... how does this change? -Well if you keep the current in the same direction through the helmholtz coils now, you would see nothing, however if you reversed the current you could possibly see the beam again.

^{SJK 23:49, 17 October 2007 (CDT)}