User:Ryan P. Long/Notebook/Physics 307L/2009/11/09: Difference between revisions
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==Objective== | ==Objective== | ||
The objective of this lab is to find the charge to mass ratio for an electron. We used an apparatus with an electron gun that fired electrons into a helium filled bulb placed inside a magnetic field to excite a ring of Helium atoms. We measured the radius of the ring, and knowing about the Lorentz force, we calculated e/m. | |||
==Equipment== | ==Equipment== | ||
Revision as of 16:56, 22 November 2009
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e/m LabObjectiveThe objective of this lab is to find the charge to mass ratio for an electron. We used an apparatus with an electron gun that fired electrons into a helium filled bulb placed inside a magnetic field to excite a ring of Helium atoms. We measured the radius of the ring, and knowing about the Lorentz force, we calculated e/m. Equipment
SetupFirst we connected the HP power supply to the Helmholtz coils on the Uchida apparatus with one of the BK multimeters in series. Next the other BK multimeter was connected to the voltmeter jacks on the Uchida apparatus. The Soar power supply was connected to the heater jacks on the Uchida. Finally, the Gelman was connected to the electrodes on the Uchida. We followed the procedure in Professor Gold's manual, which included letting the heater warm up for 2 minutes before applying voltage to the electrodes on the apparatus. To take data, we let either let the voltage of the electrodes or the current running through the coils be constant, while varying the other parameter. We measured the radius of the rings on the right and the left side using the built in ruler.
DataNote: At a current of -1.35A, maximum accelerating voltage is 320V, the minimum is 165V. At -1.05A, the maximum voltage is 200V and the minimum is 100V. 11/16/09 Voltage applied to heater jacks: 6.24 V Raw Data:{{#widget:Google Spreadsheet |
key=0Ajaf_99vseiXdEIyXzJxWGJhLUQ2SE9qNGo4UURmMUE | width=760 | height=300
}} AnalysisGiven that: (From Alexandra Andrego's Lab Notebook):
We can calculate [math]\displaystyle{ \frac{e}{m}\,\! }[/math] based on the slope of a [math]\displaystyle{ \frac{V}{{r}^{2}}\,\! }[/math] line holding current constant, or the slope of a [math]\displaystyle{ \frac{r}{\frac{1}{I}}\,\! }[/math] line holding voltage constant. Part 1
I calculated the slope and slope uncertainty of the line from my plot using the linest array function in excel. The radius used in my analysis is the mean of the left side and right side radii, per Professor Gold's manual. [math]\displaystyle{ Slope=8.141\cdot 10^{-6}\pm 3.989\cdot10^{-7}\frac{m^{2}}{V} }[/math]
with my slope and constant current, e/m can be calculated: [math]\displaystyle{ \frac{e}{m}=\frac{1}{slope}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}=\frac{1}{.00000814}\times \frac{2}{(7.8\cdot 10^{-4}\times 1.35)^{2}}\cong 2.216\cdot 10^{11}\frac{C}{kg} }[/math] using my slope uncertainty, I calculated a lower and upper bound value for my e/m ratio: [math]\displaystyle{ lower bound: 2.11\cdot 10^{11}\frac{C}{kg} }[/math] [math]\displaystyle{ upper bound: 2.33\cdot 10^{11}\frac{C}{kg} }[/math] The accepted value (from wikipedia) is:
Part 2
[math]\displaystyle{ Slope=15.697\pm 0.68610\frac{1}{A\cdot m} }[/math] [math]\displaystyle{ \frac{e}{m}=\frac{(slope)^2\times 2\times V}{(7.8\times10^{-4})^{2}}=\frac{(15.697)^2\times2\times240}{(7.8\times10^{-4})^{2}}\cong 1.944\cdot 10^{11}\frac {C}{kg} }[/math] again, using my slope uncertainty, I calculated a lower and upper bound value for my e/m ratio: [math]\displaystyle{ lower bound: 1.78\cdot 10^{11}\frac{C}{kg} }[/math] [math]\displaystyle{ upper bound: 2.118\cdot 10^{11}\frac{C}{kg} }[/math]
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