I Online Experiment: Compton Scattering

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The discussion centers on an online experiment involving Compton Scattering, where the author sought simulations or videos to analyze the relationship between energy changes and scattering angles. They initially plotted the data incorrectly, leading to a linear relationship that did not match the expected gradient based on the electron rest energy. Upon realizing the mistake, the author corrected the axes for the plot, suggesting that the relationship should align with the theoretical gradient when properly represented. The corrected gradient calculation showed agreement with the expected value, indicating the importance of accurate plotting in experimental analysis. This highlights the significance of understanding the underlying physics in data representation for Compton Scattering experiments.
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Plotting a compton scattering Dataset
I recently thought it might be interesting to see if there were any online simulations or videos of Compton Scattering experiments. The search result yielded following:

Given the standard equation: $$1-\cos\theta_d=\frac{E_0 \Delta E}{E_1 E_2} ,$$ we should get a straight line with gradient ##\frac {E_0}{E_1}## if we plot ##1-\cos\theta_d## on the x-axis against ##\frac{\Delta E}{E_2}## on the the y-axis. ##E_2## is the energy (at various angles) of the scattered photon(s) and ##\Delta E## is the change in energy referenced against that recorded at ##0^\circ##. So here is my data set and plot based on the above video. Well we do get a clearly linear relationship but my concern is that the regression obtained linear gradient is nowhere near ##\frac {E_0}{E_1}## where ##E_0## is electron rest energy.

Any ideas on what the problem is ? Have to say I thought we needed much higher energy photons than 17.48 keV as in this video experiment (?).

1636142000019.png
 
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Oops - I see the mistake. Should have plotted ##\frac{\Delta E}{E_2}## on the x-axis and ##1-\cos\theta_d## on the y-axis. Will fix and show ammended plot.
 
1636143442537.png


$$\frac{E_0}{E_1}=\frac{512}{17.44}=29.36$$ so in good agreement with the regression obtained gradient above.
 
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