Physics307L:People/Meyers/e/m ratio lab summary

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Contents

Purpose

SJK 22:08, 13 October 2010 (EDT)
22:08, 13 October 2010 (EDT)future informal summaries can improve from this one.  The main thing is the presentation of the final result, along with uncertainty, and a statistical comparison with accepted value(s).  Now, with Poisson, that will be a challenge, since it's a weird lab.  However, with Millikan and other labs, you will measure a quantity and an uncertainty in that measurement.  It's probably important to talk with me about this.
22:08, 13 October 2010 (EDT)
future informal summaries can improve from this one. The main thing is the presentation of the final result, along with uncertainty, and a statistical comparison with accepted value(s). Now, with Poisson, that will be a challenge, since it's a weird lab. However, with Millikan and other labs, you will measure a quantity and an uncertainty in that measurement. It's probably important to talk with me about this.

The purpose of this lab is:

1)To calculate, experimentally, the ration of charge to mass, e/m, of an electron.

2)Understand better the physics behind this principle.

Steve Koch 21:53, 13 October 2010 (EDT): Remember to include a link to your primary notebook! http://openwetware.org/wiki/User:Richard_T._Meyers/Notebook/Phys307l/E/m_Lab

Procedure

We used the online lab manual, linked [here]. When we got to the lab we realized that the equipment was already set up for us, thank you Emran and Randy.

The Apparatus
The Apparatus

We checked with the manual to be sure that this was the correct setup. After this and the safety quiz we began to take data. After taking points for accelerating Voltage, Constant Voltage and Constant Current we stopped.

Most of the second day of the lab was taken up with analyzing the data.

Data

SJK 22:06, 13 October 2010 (EDT)
22:06, 13 October 2010 (EDT)In the summary you do not need to include the raw data unless it's of unusual interest to the reader.  Instead, you want to focus on the final measurements with uncertainty and discussion of whether results are consistent with accepted values and whether systematic error is apparent, etc.
22:06, 13 October 2010 (EDT)
In the summary you do not need to include the raw data unless it's of unusual interest to the reader. Instead, you want to focus on the final measurements with uncertainty and discussion of whether results are consistent with accepted values and whether systematic error is apparent, etc.
View/Edit Spreadsheet
View/Edit Spreadsheet

Calculations

We used the equations in the manual and the fact that F_B=F_c\,\!:

 B=\frac{\mu R^2NI}{(R^2+x^2)^{3/2}}

 F_c=eV=\frac{mv^2}{2} and

 F_B=evB\,\!

So we got:

\frac{e}{m}=\frac{2}{((7.79x10^{-4})(1.42875))^2 slope}=6.747x10^8\,\!

and

\frac{e}{m}=2.741x10^{11}

Upon averaging the two values, one being very large and the other being very small we get: SJK 22:02, 13 October 2010 (EDT)
22:02, 13 October 2010 (EDT)WHOAH!  Does it make sense to average together these two values???  If you took two measurements of someone's height and got 100 cm with one measurement and 1 mm with the other measurement, you would never average those together.  The same in this case:  clearly there is a big problem with at least one of the measurements.  There's no way they represent the same parent mean, and thus averaging is not appropriate.  When we talk about weighted averages in class soon (next week or the following), we'll discuss this further, but with less severe examples.  In your case, I think your intuition could tell you that averaging is not appropriate, since one value is almost zero compared to the other.
22:02, 13 October 2010 (EDT)
WHOAH! Does it make sense to average together these two values??? If you took two measurements of someone's height and got 100 cm with one measurement and 1 mm with the other measurement, you would never average those together. The same in this case: clearly there is a big problem with at least one of the measurements. There's no way they represent the same parent mean, and thus averaging is not appropriate. When we talk about weighted averages in class soon (next week or the following), we'll discuss this further, but with less severe examples. In your case, I think your intuition could tell you that averaging is not appropriate, since one value is almost zero compared to the other.

\frac{e}{m}=1.374x10^{11}\,\!

Error

We found the accepted value to be:

\frac{e}{m}=1.758x10^{11} \frac{C}{kg}\,\!

So we compared our value and the accepted value to get a percent error:

 error=\frac{1.758x10^{11}-1.374x10^{11}}{1.758x10^{11}}x100=21.84%

21.84% error is not bad for twenty measurements. SJK 22:05, 13 October 2010 (EDT)
22:05, 13 October 2010 (EDT)As noted above, the average is not appropriate.  Taking your answer as is, though, how do you assess "not bad?"  The answer is you need to compare your discrepancy from the accepted value with your range of uncertainty.  If it's close, then there's a chance your measurements aren't bad.  If it's very far away, as would be the case for you, then you know that you have significant systematic error that is dwarfing your random error (uncertainty).  We probably need to discuss this further in person before you complete your next report.
22:05, 13 October 2010 (EDT)
As noted above, the average is not appropriate. Taking your answer as is, though, how do you assess "not bad?" The answer is you need to compare your discrepancy from the accepted value with your range of uncertainty. If it's close, then there's a chance your measurements aren't bad. If it's very far away, as would be the case for you, then you know that you have significant systematic error that is dwarfing your random error (uncertainty). We probably need to discuss this further in person before you complete your next report.

Conclusion

We found the ratio to be:

\frac{e}{m}=1.374x10^{11}\frac{C}{kg}\,\!

and a percent error to be:

%error=21.84%\,\!

I did notice that our values for the Voltage versus radius squared had a smaller correlation than that of the Inverse Current versus the Radius. I assume that the sigmoid shape of the Constant Current graph is because of unsystematic error, parallax error, in reading the radius.

With a few more data points I am confident that we could produced a better estimation of e/m.

Thanks

1)Nathan for being my excellent lab partner and for taking down the equipment list when I had failed to.

2)Emran for setting up the lab previously and for a general outline of my report.

3)Randy also for setting up the lab previously.

Citation

1) I found the accepted value of e/m here

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