User:Garrett E. McMath/Notebook/Junior Lab/2008/10/13
 Electron Diffraction
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Electron Diffraction Lab
Purpose
FormulasIn the early 1900s, deBroglie postulated that matter must have wavelike properties, just like light waves can have particle like nature. He postulated that matter waves have wavelength: [math]\displaystyle{ \lambda = \frac{h}{p} }[/math] Not long after his proposition, experimental evidence for his postulate was provided, when electrons were observed to diffract. The relationship for light diffraction passing through a slit of width d, is: [math]\displaystyle{ d sin \theta = \lambda }[/math] For small angles, this relationship simplifies to: [math]\displaystyle{ \frac{dD}{2L}=\lambda }[/math] where D is the spacing between the maxima on the screen, a distance L away. [math]\displaystyle{ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_{k}}}=\frac{h}{\sqrt{2meV_{a}}} =\frac{2\pi\hbar}{\sqrt{2meV_{a}}} }[/math] Setting the wavelength of the electron equal to the wavelength in diffraction, we get: [math]\displaystyle{ \frac{Dd}{2L}=\frac{2\pi\hbar}{\sqrt{2meV_{a}}} }[/math] [math]\displaystyle{ d=\frac{4\pi\hbar L}{D\sqrt{2meV_{a}}} }[/math] [math]\displaystyle{ r=slope\cdot\frac{1}{\sqrt{V}} }[/math] SJK 17:11, 1 November 2008 (EDT) What is r? And I think you forgot a factor of 2 in your final answer [math]\displaystyle{ slope=\frac{Lh}{\sqrt{2me}\cdot d} }[/math] [math]\displaystyle{ d=\frac{Lh}{\sqrt{2me}\cdot slope} }[/math] Equipment
CautionsGiven the tiny current, electrical shock is probably the least worries.SJK 02:54, 1 November 2008 (EDT)It's microamp current when the electrons are traveling through the vacuum bulb...if you made a direct connection with your body, much more current could flow! Thus you should assess the maximum current of the power supply, not the maximum current draw of the device Also, please take a look at My comments on Paul's notebook, since your notes are the same. Procedure
DataWeek1
SECOND TRIAL
We started taking more data, but the measurements were so similar that we thought it would be pointless to continue. We will fit it before the next period and see if we have any major outliers. If this is the case, we will go back and make some adjustments. Week2After looking at our data from the first week, we decided that it would be a good idea to retake measurements some measurements. An important distinction is that this time we will be measuring the inner diameter of the inner ring, not the outer diameter as we did the first time. We also used a new caliper: Electronic Digital Caliper: Carrera Percision.
CorrectionsSJK 17:36, 1 November 2008 (EDT)None of this is in your excel sheet, and thus I have no idea what you're doing with this information. More generally, it's not clear from your notebook or excel sheet how you converted slopes to lattice spacing. (the formula I see shows a factor of 2 error) L=130.0±2mm
Glass thickness=1.5mm
radius of curvature=66.0mm
Accepted seperation of carbon atoms=.123nm and .213nm
[math]\displaystyle{ {delta}{L} = {R}({1-cos(arcsin({D}/{2R}))}) }[/math]
Sources of Error
Data Analysis
Your excel sheet does not match your answers! I See what you did: you mixed up the "Y" and "X" columns in your linest function. You used the slope from the graphs, but the uncertainty from the LINEST...thus, your uncertainties are incorrect. Inner Diameter
Outer Diameter
Calculated Spacing of Carbon Inner Diameter
Outer Diameter
Summary
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