User:John Callow/Notebook/Junior Lab 307/2009/10/28

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Electron diffraction
Link to lab manual used

Link to my summary

note
I realized the morning of 11/11/09 that I made a few errors that I'm correcting, but after I sent the email to turn it in. (Steve Koch 21:53, 14 November 2009 (EST): No problem, thanks for correcting and making the note!)

Materials
(1) Teltron 813 kV power unit (high voltage supply)

(1) Tel 2501 universal stand

(1) Tel 2555.02 magnet

(1) Tel 2555 electron diffraction bulb

(1) Hewlett Packard 6216B power supply (regulating voltage source)

(6) connector cables

Safety
For the Experimenter

This experiment uses a high voltage power supply, be careful not to get shocked.

Make sure everything is off when setting up and taking down the circuit.

For Equipment

The graphite in the bulb can be punctured by current overload, be sure not to allow the current to go above .25 mA. (This wasn't a problem with the equipment I used.)

Make sure the high voltage slider is set to 0 before turning it on, and then let the cathode temperature stabilize for a few minutes before turning the voltage slider up.

The bulb is glass, so make sure not to drop it and that it is in a stable place.

Setup
Set up the circuit as shown on this page. Once that is done, make sure the voltage slider on the high voltage power supply is set to 0 and then turn it on. After a few minutes the cathode temperature should be stable and now the high voltage slider may be turned up.

Once turned up, the bulb should glow(green in my bulbs case) and there should be a focused dot with two rings around it viewable on the top part of the bulb. Using the magnet that fits on the base of the bulb, move it around to center the dot in the middle of the top part of the bulb. For my experiment the bulb had clearly been used a lot before and the center was blackened so I aligned with that. However, a better way of doing this would have been to use some method of measuring how far from the top, bottom, left, right white edge of the top of the bulb you are and adjust from there. After already taking my data I decided to try using a wire and ruler just to get a rough idea how far from the center I had been, and it seemed to be only about a millimeter or two off so my data should be fine.

Taking Measurements
The bulb is old, so taking measurements is difficult. The rings are barely visible even at 5kV. Certain parts of the rings are easier to see at different angles, and at lower voltages this was the only way I was able to see anything. My method for measurement was to locate a part of the ring being measured, hold the part of the micrometer that doesn't move there, then find an angle I could see the other side and slide the other part of the micrometer till it aligned, then take the measurement. I also tried taking the measurements horizontally and then two other angles for each ring, and then averaging hoping that with multiple measurements of the same ring would help balance out the inaccuracy introduced in measurement by barely being able to see the rings. Looking at the data below, the majority of measurements on the same ring all landed in a fairly close range, so I believe my measurements are fairly close to the actual radius of the ring.

Finding the Corrected Diameter D
First, we need to correct the measured diameter Dm for the diameter D as shown in the drawing on the right. From the manual we have that

$$ D = 2*L*tan({\phi}) $$

but we don't know $$ {\phi} $$ directly. However we know by Pythagorean theorem that if R = Dm/2

$$ {{R^2}+{C-x}^2} = C^2 $$

Or that

$$ {x^2}-2x*c+{R^2} = 0 $$

and using the quadratic formula we find that

$$ x = \frac{2C-({(2C)^2}-4*{R^2})^{\frac{1}{2}}}{2} $$

We only use this solution as the other one does not make physical sense to use in the problem. (x shouldn't be bigger than C or it would mean the endpoint would be behind the starting point.)

Now that we have x, we know that $$ {\phi}$$ is

$$ {\phi} = tan(\frac{R}{L-x}) $$

And with that we may plug into

$$ D = 2*L*tan({\phi}) $$

to find

$$ D = \frac{2*L*R}{L-\frac{2C-({(2C)^2}-4*{R^2})^{\frac{1}{2}}}{2}}$$

Finding the lattice spacing d
I found the formulas

$$\lambda = d*sin(\phi) $$ on page 191

and

$$\lambda = \frac{h} $$ on page 194 of the textbook Modern Physics fifth edition by Paul Tipler and Ralph Llewellyn.

Now from my drawing in the previous section I know that

$$d*sin(\phi) = d\frac{D}{2*({L^2}+{.25*D^2})^{.5}}$$

From the looks of it the manual decided to take .5*D<<L, this isn't really true... but oh well. (I already did all my calculations using the manuals formula so I'll make the assumption. Now that I see this though it looks like my data is even a bit worse than my current calculations show.)

Assuming that D is small this becomes

$$d*sin(\phi) = d\frac{D}{2*L}$$

Combining formulas we have

$$ d\frac{D}{2*L} = \frac{h} = \frac{h}$$

or

$$ d = \frac{4{\pi}L{\hbar}c}{D{(2*m*e*{V_a}*{c^2})^{.5}}}$$

Values being used for calculations
All values found on chart on the inside cover of Modern Physics by Paul Tipler and Ralph Llewellyn

$$ {\hbar}c=3.16*{10^{-26}} J*m$$

$$ e = 1.602*{10^{-19}} C$$

$$ m = 9.1*{10^{-31}} kg$$

$$ c = 3*{10^8} m/s$$

$$ C = 66mm +1.5mm (to correct for glass thickness)$$

$$ L = 13.0cm +.15cm (to correct for glass thickness)$$

Data and Calculations Small ring
Using the formulas from above I find

and I find a line of best fit to be $$D = 5.35*\frac{1}{{V_a}^{1/2}}$$

with error in the slope being +-.11.

I forced the fit on both rings through the origin based on the assumption that as the accelerating voltage went to infinity, D would go to zero so the line should pass through the origin.

Now from above

$$ d = \frac{4{\pi}L{\hbar}c}{D{(2e{V_a}m{c^2})^{.5}}} $$

Plugging in

$$D =5.35*\frac{1}{{V_a}^{1/2}}$$

results in

$$ d = \frac{4{\pi}L{\hbar}c}{5.35{(2em{c^2})^{.5}}} $$

which once values are plugged in I find

d = .268(5)nm

Compared to the accepted value of .213nm I have an error of about 26%

Data and Calculations Small ring
Using the formulas from above I find

and I find a line of best fit to be $$D = 9.68*\frac{1}{{V_a}^{1/2}}$$

with error in the slope being +-.068.

Now from above

$$ d = \frac{4{\pi}L{\hbar}c}{D{(2e{V_a}m{c^2})^{.5}}} $$

Plugging in

$$D =9.68*\frac{1}{{V_a}^{1/2}}$$

results in

$$ d = \frac{4{\pi}L{\hbar}c}{9.68{(2em{c^2})^{.5}}} $$

which once values are plugged in I find

d = .148(1)nm

which compared to the accepted value of .123nm has error of 20%

Possible explanations for error
The true accelerating voltage is a bit lower than what is recorded, and this might account for the majority of my error. This would have meant slightly higher slopes resulting in smaller calculated values.

The other source of error would be measurements, it was really tough to see the rings. From viewing the size of the error bars on each D, there looks to be a high probability that I just couldn't get good enough data.

Using a magnet to center the beam probably didn't mess with the ring radius much, but it still might have done something.

I checked my calculations a few times, but there is still always the possibility I screwed something up. (and I did but this should now be corrected)


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