User:Johnny Joe Gonzalez/Notebook/Physics 307L/2009/11/11
Plank's Constant | <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page <html><img src="/images/c/c3/Resultset_previous.png" border="0" /></html>Previous entry<html> </html>Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html> | ||||||
Plank's ConstantSJK Incomplete Feedback NoticeSJK 18:11, 16 December 2009 (EST)SafetyThis is a relatively safe experiment, only one power cord is needed, and the mercury lamp is inside the container (the HG light source). However, the equipment should still be treated with care as utter lack of regard for the light source, h/e apparatus and lens/grating apparatus, can still obviously damage the equipment.
MaterialsMercury Vapor Light Source and Light Block(Model OS-9286) ProcedureFirst we connected the multimeter with the h/e apparatus, after that we waited for the lamp to heat up. Once heated we aligned the h/e apparatus to one of the first diffraction patterns created by the mercury lamp. Afterward we measured the stopping potential of the colors, starting with yellow, we aligned the yellow and green band to the h/e device and then placed the yellow filter on the white reflective mass. We then measured the stopping potential, the yellow filter was then removed and replaced with the green filter. the stopping potential was then measured for the green. The rest of the lines did not require filters we just lined them up so that the line would be centered on the slit of the mask before taking the stopping potential. between each measurement we pushed the zero button to reset the h/e device. DataSJK 18:06, 16 December 2009 (EST){{#widget:Google Spreadsheet |
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}} As the intensity of light is decreased the stopping potential takes longer to reach stabilization and appears to have an exponential relation as can be seen through the plot below. The plot was made using microsoft excel, I the small dark lines are trendlines, while the two different colored lines represent the two colors we chose to measure(red is the violet, blue is the deep violet.) These measurements were made by taking the recharge times when they reached 90% of the stabilized stopping potential. Data AnalysisSince the intensities of the filters used only affected the recharge times and not the actual stopping potential (or at least affected it only minimally), I assume that this has to do with the fact that even though photons are partially blocked, they are still coming through at the same frequencies, therefore, the stopping potential will still be the same, however, since the intensity filters are not allowing as many photons to come through at once it takes the stopping potential longer to reach stabilization. As the frequencies of light increased (as the wavelengths became smaller), from violet([math]\displaystyle{ \nu =7.4e^{14}Hz, \lambda =404.6nm }[/math] ) to ultraviolet([math]\displaystyle{ \nu =8.2e^{14}Hz, \lambda =365.5nm }[/math] ) the time it took for the stopping potential to stabilize increased, this could be due to the fact that the stopping potentials are higher at higher frequencies. This experiment shows strong evidence for the quantum/photon model of light. experiment 2Unfortunately we were unable to get a second order measurements for this experiment, I was a little confused as to what to do at first and by the time I figured out that we still needed to make second order measurements it was too late. However, we did manage to take a few first order measurements I'm hoping that this will still be enough for some sort of accurate result. Also we were able to filter out some light in order to get a reading off of the red band, using 3D glasses. My plot has two trendlines, one includes the red band stopping potential, while the other excludes it. Using the equation [math]\displaystyle{ E=h\nu =KE_{max}+W_{0} }[/math], Since E is proportional to the negative potential needed to stop the flow of electrons, we can rewrite this to be: [math]\displaystyle{ h=\frac{e\Delta V}{\Delta \nu } }[/math] where [math]\displaystyle{ \frac{e\Delta V}{\Delta \nu } }[/math] is the slope of a line, by plotting this and using a least squares fit we can determine the value of h.
my V vs f plot for the stopping potential vs the frequenciesSJK 18:09, 16 December 2009 (EST)
Further more to find the work function's value we can look at our value at the y-intercept. Unfortunately my results were full of random error, I believe the primary reason for this was not aligning the color bands in the same area every time, I believe that this caused some of my measurements to be off slightly, however, slightly is enough to affect my plot for frequency vs stopping potential. As for the red band, although it was included in some of our experimentation, when added in it made our results and plot less accurate. However, the red part of the 3D glasses did have an affect and the numbers produced from it were somewhat agreeable to the rest of our results. Weather or not we were actually testing the red color or if it was ultraviolet interference is another matter since it is likely that the red wavelength would not have enough energy to excite and eject the electrons from the sheet.
Error&Resultswhen we tried using the filter at 60% and down our stopping potentials increased significantly. We set the filter back to 100% and even removed it to see if this error came from random error (accidentally moving the arm), but as soon as the filter was removed or adjusted back to 100% our stopping potential returned to its original value. Our difference in potential seemed to increase by about 0.5V.
More important is I'm not sure how to correctly analyze the data, since I don't think we made enough different measurements therefore actual error analysis is difficult. my results did vary from the accepted value by about 3.29% at best, but by 27.5% when we include the red results, this was also just using the trendline feature on excel; when I used LINEST I came up with [math]\displaystyle{ h=5.4900(7)*10^{-34} }[/math] when including red and [math]\displaystyle{ h=6.190(2)*10^{-34} }[/math] without the red result.
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