# Physics307L F09:People/Andrego/e/m Ratio

## E/M RATIO FOR ELECTRONS LAB SUMMARY

Please note that Anastasia Ierides was my lab partner for this lab. You can find her lab summary by following this link.

### Brief Overview

The purpose of this lab was to measure the charge to mass ratio (e/m) for electrons passed through an electron gun into a magnetic field created by current running through a series of helmholtz coils. We were able to obtain results by observing the relationships between the radius of the electron beam loop and the voltage across the apparatus as well as the current being passed through the helmholts coils.

### Data Results

We first looked at the relationship between the radius and the voltage (holding the current constant at 1.34 A) where we found...
$\displaystyle{ r^2=\frac{2V}{({7.8\times10^{-4}{I})}^{2}}\times\frac{m}{e}\,\! }$
By graphing the radius squared versus our voltage values we were able to use Excel to plot a linest fit line for our plot and determine the slope, or the proportionality constant between r squared and the voltage. Because this constant of proportionality includes the value e/m we were able to set our slope equal to the slope in our equation and solve for the ratio of e/m.

Our best calculated value was...

$\displaystyle{ \simeq3.666\times10^{11}\frac{C}{kg}\,\! }$

When we included the uncertainty in our linest fit slope we obtained the following range for our calculated value of e/m...

$\displaystyle{ 3.204\times10^{11}\frac{C}{kg}\leq \frac{e}{m}\leq 4.283\times10^{11}\frac{C}{kg}\,\! }$
We then looked at the relationship between the radius and the current (holding the voltage constant at 350 V) where we found...
$\displaystyle{ \frac{1}{r}=\sqrt{\frac{(7.8\times10^{-4})^{2}}{2V}\times\frac{e}{m}}\times I\,\! }$
By graphing one over the radius versus our current values we were able to use Excel to plot a linest fit line for our plot and determine the slope again, or the proportionality constant between 1/r and the current. Because this constant of proportionality includes the value e/m we were able to set our slope equal to the slope in our equation and solve for the ratio of e/m.

Our best calculated value was...

$\displaystyle{ \simeq3.301\times10^{11}\frac{C}{kg}\,\! }$

When we included the uncertainty in our linest fit slope we obtained the following range for our calculated value of e/m...

SJK 15:20, 14 November 2009 (EST) 15:20, 14 November 2009 (EST)
You're including too many digits on these values. Since your uncertainty is more than 10%, you really only need one decimal place -- that would make it easier to read. Also, while I like reporting the range like you did, another way to do it is to say mean +higherro -lowerror, which is a good way to do it when you have asymmetric uncertainty
$\displaystyle{ 2.408\times10^{11}\frac{C}{kg}\leq \frac{e}{m}\leq 3.699\times10^{11}\frac{C}{kg}\,\! }$

### Error

For ALL RECORDED accounts of error in our experiment methods and procedures please see the Notes about Our Uncertainty section in our e/m Ratio Lab Notebook.

$\displaystyle{ \frac{e}{m}accepted=1.76\times10^{11}\frac{C}{kg}\,\! }$ (Steve Koch 15:34, 14 November 2009 (EST):Remember to cite the source of your accepted value!)
We have to note that because our measured ranges of values for the e/m ratio do not include the accepted value, we must have a fairly large source of error. I believe that this is mainly due to the lack of precision in our measurements of the radii of the electron beams. Measuring with the naked eye is not the best method for this. We did ponder over the idea of using a camera to take still shots of the beams, but this would require a lot of photo analysis and would have its own systematic error.

SJK 12:45, 14 November 2009 (EST) 12:45, 14 November 2009 (EST)
Good job comparing accepted value to your 1 sigma range. However, it would be better to be more explicit about saying how far outside your range is the accepted value (it's many sigma away). If it were just slightly outside your range, you wouldn't feel confident about saying it's inconsistent. But in this case, the chance of your discrepancy being due to random error is very very low, and thus the conclusion that there's systematic error. Do you really think it's only you ability to measure the radius is the major problem? I don't think so. Unfortunately, we didn't get to discuss this during the lab time, sorry about that! There are many things about the apparatus, which basically make it impossible to eliminate systematic error.

### Conclusions

Even though we know our error to be very large for this lab, I feel as though we did accomplish a great feat in obtaining values of the 10^11 order. This lab was made significantly difficult by the lack of means for taking specific or at least more specific measurements of the radii. This lab proves to be more about the analysis of our experimentation and good use of our linear fit skills. In the future it might be better to struggle with the digital photo analysis than the great factor of human error.