Physics307L:People/Meyers/Electron Diffraction Lab Summary

Purpose

Low Voltage Correction Device

The purpose of this lab is to become more familiar with the DeBroglie relationship. We calculated the wavelengths of the primary and secondary scattering of a beam of electrons. We showed that the length of the diameters is related to the voltage the electrons are accelerated through. We also compared the wavelengths we calculated with the accepted values.

Procedure

High Voltage and Heater Supply

Along with the Lab Manual the procedure for this experiment is primarily just adjusting the high voltage and measuring the associated diameters of the projected diffraction pattern. We used a micrometer to measure the diameters of the rings of the diffraction pattern. The wiring diagram for the set up is shown Here.

Data

The Diffraction Bulb

This is the Raw data from which we calculated the wavelengths:

{{#widget:Google Spreadsheet |key=0AlAK7TMhL37adFBlSzdqR3pjdGcwWExmVGJHRHBwRXc |width=775 |height=575 }}

After correcting the diameters for going from a sphere to a plane by these equations:

[math]\displaystyle{ y=R-\sqrt{R^2-\frac{D_{observed}^2}{4}}\,\! }[/math]

[math]\displaystyle{ tan(\theta)=\frac\frac{D_{observed}}{2}{L-y}\,\! }[/math]

[math]\displaystyle{ D_{corrected}=2Ltan(\theta)\,\! }[/math]

We get this graph of the inverse square root of the Voltage versus the Diameters.

height 500

Using the experimentally found slopes and h being Plank's constant, e being the elementary charge, m being the mass of an electron and L and R being defined in the manual as 13cm and 66mm respectively. we can use the following equation:

[math]\displaystyle{ d=\frac\frac{2hL}{\sqrt{2me}}{slope}\,\! }[/math]

to calculate d as:

[math]\displaystyle{ d1=2.59(5)*10^{-10}m=0.259(5)nm\,\! }[/math]

[math]\displaystyle{ d2=1.59(3)*10^{-10}m=0.159(3)nm\,\! }[/math]

Error

The error I calculated using STDEV in EXCEL is 2.272 and 3.696 for the inner and outer ring respectively. From there I put these into the above equations to get the reported diameters. To calculate the percent error see below:

[math]\displaystyle{ %=\frac{0.259-0.213}{0.213}*100=21.59%\,\! }[/math]

[math]\displaystyle{ %=\frac{0.159-0.123}{0.123}*100=29.27%\,\! }[/math]

For how unsteady we were at taking the data with the calipers I think to have 30% error is fair.

Thanks

1)To exce2wiki.net for the conversion of an excel doc to wiki code converter. It saved me for hours of menial data input.

2)To Kirstin from who I got the relations for the diameter correction.

3)To Nathan for being a great lab partner.