Physics307L F07:People/Knockel/Lab2

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Electron's e/m ratio summary

SJK 01:44, 10 October 2007 (CDT)
01:44, 10 October 2007 (CDT)Well, I wouldn't say it was a disaster at all, so much as real physics...but I do understand your frustration.  I guess a careful reading of the previous years' lab manual does say that you won't be able to measure the correct value for various reasons.  You are also spoiled from the Millikan lab.However...you still need to report your uncertainty with a result like this! What if your random error were large enough that you did overlap with the correct value?  And the only reason you know your answer is "wrong" is from the availability of the accepted measurement.  But even then, your answer is only "wrong" if your random uncertainty makes it very unlikely that the "true" value overlaps with your measurements.
01:44, 10 October 2007 (CDT)
Well, I wouldn't say it was a disaster at all, so much as real physics...but I do understand your frustration. I guess a careful reading of the previous years' lab manual does say that you won't be able to measure the correct value for various reasons. You are also spoiled from the Millikan lab.

However...you still need to report your uncertainty with a result like this! What if your random error were large enough that you did overlap with the correct value? And the only reason you know your answer is "wrong" is from the availability of the accepted measurement. But even then, your answer is only "wrong" if your random uncertainty makes it very unlikely that the "true" value overlaps with your measurements.

The e/m ratio (charge to mass ratio) for an electron can, in theory, be measured by seeing how an electron beam behaves in a magnetic field. Using a Helmholtz coil to generate the field and an electron gun in a vacuum tube filled with helium, a beam can be created, formed into a circle with the magnetic field, and measured (since the electrons makes the helium glow).

This experiment was a disaster since there was ridiculous drag from the helium and for other reasons. The only thing I could conclude was that

\frac{e}{m} < 3.13\times10^{11}\frac{C}{kg}.

Taking e/m = 3.13x1011 C/kg, there is a relative error of 78% from the actual value of 1.76x1011 C/kg.

In my lab notebook, I go into great detail explaining the theory, why there is so much systematic error, and alternative methods for finding e/m without so much error. I, of course, also explain the setup, equipment, procedure, etc., and I give my data and calculations.

link to e/m ratio notebook

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