# Physics307L:People/Young/Young's over m ratio for electron

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## Brief Summary

SJK 02:10, 18 December 2008 (EST)
02:10, 18 December 2008 (EST)
Since you did your formal report on this lab, I have already given you a bunch of feedback over there on your rough draft.

This lab's purpose is to calculate the e/m ratio for an electron using an electron gun, Helmholtz coils, and our knowledge of how magnetic fields effect electrons. Our apparatus contains a electron gun within a air tight bulb. The electron gun consists of a heater that heats up some material until it begins to shoot electrons off. Then a potential difference takes the loose electrons and shoots them straight out to the right. The electron beam can be viewed in a dark room as a light purple/blue beam. Around our bulb there are some Helmholtz coils that we can use to bend the electron beam by creating a B field through the coils to cause a constant angular acceleration to our beam.

## Error Analysis

• Measurements were made by measuring the image of our beam on a mirror ruler in the back of the apparatus. The glowing given off by the Helium gas in the bulb would often make measurements difficult.
• Our apparatus due to it's age was not giving us as intense of an electron beam as we would have liked to have a thin well defined line on the reflecting ruler.
• The magnetic field through our Helmholtz coils are assumed to be constant, however this is not the case for Helmholtz coils with a finite length.
• The beams image on the reflecting ruler was slightly diffracted by the bulb so we got our measurements should be slightly larger.

## Derivation of e/m Ratio

Relation of charge over mass ratio using forces.

$\displaystyle F_e = ma \$

$\displaystyle B_Hq = \frac{mv}{r} \$

Relation of charge over mass ratio using Energy

$\displaystyle KE = eV \$

$\displaystyle \frac{1}{2}mv^2 = qV \$

Where,

$\displaystyle V \$ is the accelerating Voltage in Volts
$\displaystyle v \$ is the velocity of the electron in m/s
$\displaystyle q \$ is the charge which in this case is the charge of an electron
$\displaystyle m \$ is the mass of an electron which is
$\displaystyle B_H \$ is the B field produced by the Helmholtz coils in (V s/m²)
$\displaystyle r \$ is the radius of the electron beam path in meters

The general form of the equation for a Helmholtz coil is $\displaystyle B_H = \frac{\mu_0 I R^2}{2(R^2+r^2)^{3/2}}$

After simplifying all of our equations we have

$\displaystyle \frac{q}{m} = (\frac{5}{4})^3 \frac{2V(R^2)}{{\mu_0}^2 n^2 I^2 r^2} \$

Where,

$\displaystyle R \$ is the Radius of the Helmholtz coils which is .15 meters
$\displaystyle \mu \$ permeability of free space 1.2566371 × 10-6 (N/A^2)
$\displaystyle n \$ is the number of loops in the Helmholtz coils which is 130
$\displaystyle I \$ is the current through the coils

## Analysis

• For the x we are assuming x=R/2 since we are about in the middle of our Helmholtz coils.

Accepted value for from Wikipedia

e/m ratio= 1.76E+11 C/kg

### Constant Voltage Vary Current

Using the relation between the B field and the effect it has on the electron beam I found.

q/m= 2.3090(.27)e+011 C/kg

• σ = 2.7163e10
• sem= 3.3954e9
• % difference=31.2%

This value is about 2 standard deviations away from the accepted value. The percent difference is the closest of all the values that I found. This leads me to believe that this method is one of the most effective forms of calculating the ratio.

### Constant Current Vary Voltage

Looking only at the relation between accelerating Voltage and radius of Curvature I found that

q/m= 2.7373e+011(.29)C/kg

• σ = 2.9998e10
• sem=3.7497e9
• % difference=55.5%

This value is about 3 and a half standard deviations away from the accepted value. Current would probably be the largest contributer to error since the method where only current is varied produced the worst results for our confidence interval.

### Vary Voltage and Current

Now we vary both the current and the Voltage. I found that...

q/m= 3.0618(.8)e+011 C/kg

• σ = 8.0372e10
• sem = 1.0047e10
• % difference=74%

This value is about 1.5 standard deviations away from the accepted value. Which is interesting because the percent difference is the largest of all the methods. This leads me to believe that the varying the voltage and current adds up to give us the worst value ,but with a standard deviation large enough to say we have a much higher confidence interval than all the other methods.

Matlab Code for data Analysis

## Final Thoughts

When taking measurements from the reflecting ruler I found it easier to measure the inside of the loop. The focus dial on the apparatus seemed to change the radius as well as the size of the beam. The fat beam that was measured had a width that could have effected our data. If the radius was in fact larger then our measurements then since q/m goes by 1/r^2 we would have a smaller q/m ratio. This would account for why all of our values for q/m are substantially larger than the accepted value.