Physics307L:People/Klimov/Eoverm

=e/m Ratio Lab Summary=

In this lab, the charge to mass ratio of the electron was measured. The experiment was based on the fact that moving electrons feel a force in the presence of a magnetic field, which causes them to spin. So, in this experiment, electrons were accelerated into a magnetic field, and the radius of the resulting circular beam was measured. Using the values of the radius, the accelerating voltage of the electrons and the current flowing through the coils, the e/m ratio was calculated.

This lab consisted of 2 main experiments. First, we held the current in the coils constant while varying the accelerating voltage. This data was fit with a least squares line and its slope was related to the e/m ratio. In the second experiment, the accelerating voltage was kept constant while the current was varied. Once again, the least squares fit was found, and its slope was related to the e/m ratio.

A detailed walkthrough of the experiment can be found below in the Lab Notebook.

Important Links (all link to different parts of the same page):
 * Lab Notebook


 * Data


 * Post Experimental Data Analysis


 * Lab Manual Questions

Summary of Data
With each measurement, there was also some ± uncertainty that resulted from our limited ability to measure the radius of the beam correctly. While I have presented these uncertainties here, I should mention that I had a hard time doing so because of how many other errors there were in the lab. It seems wrong of me to account for only one possible source of error (our inaccuracies in measuring the beam) when there were clearly so many. The uncertainties given below were produced by my linear fit program based on the amount of reasonable tilt the slope could have, beyond the best fit. The uncertainty given for the mean is the mean size of my error bars which are provided in the lab notebook.

Actual e/m Ratio
$$\frac{e}{m}_{act}=1.7563 \cdot 10^{11} \frac{C}{kg} $$

Constant Current Experiment
$$\frac{e}{m}=3.4409 \pm .046 \cdot 10^{11}\frac{C}{kg} $$

$$ %_{error}=95.9%$$

Constant Voltage Experiment
$$\frac{e}{m}=3.3154 \pm .122 \cdot 10^{11}\frac{C}{kg} $$

$$ %_{error}=88.7% $$

Mean e/m ratio from all data points
$$\frac{e}{m}=3.3142 \pm .198 \cdot 10^{11}\frac{C}{kg} $$

$$ Standard Deviation = 0.31407 \cdot 10^{11} \frac{C}{kg} $$

$$ %_{error} = 88.7% $$

Conclusions

 * I am not really happy with the result, given how much effort was put into making accurate measurements. However, after analyzing the possible sources of error, I think its clear that it would be very hard to get a great measurement of the charge to mass ratio. The entire setup was pretty bad, and in my lab notebook I wrote down an extensive list of things that could improve the experiment in the future.
 * The discrepancy from the accepted value is likely due to systematic error. The apparatus has a great number of limitations, and I don't think it is possible to get a great value for the e/m ratio using it. I am not claiming that the experiment was completely rid of random errors, but I think that they are incomparable in magnitude given the large number of systematic errors.
 * If I had more time with the experiment, and I could do only one thing, I would take apart the electron gun and rebuild it to shoot electrons straight up (instead of out of the side). From there, I would have to set up a set of plates that could produce an electric field perpendicular to the magnetic field produced by the coils. This would allow me to make the e/m ratio measurement by balancing the force from the E field with the force from the B field, like Thompson did. Such an experiment would remove many possible sources of error. Once again, in my lab notebook, i provide a great list of things that could be done to improve this lab.
 * Even though I'm not happy with our results, I am glad that we had as many problems as we did. This lab has showed me that experiments don't always go as planned, and one must always consider all possible errors.


 * Lab manual questions answered in lab notebook.