Electron Paramagnetic Resonance Experiment

[jameson2]

I've a few questions about an EPR experiment I did recently. It involved setting up a Helmholtz pair, placing a sample in a uniform field inside the pair, and observing results on an oscilloscope.

1. You can calculate the field at the centre of a Helmholtz pair using the formula for one Helmholtz coil twice. This formula is: $$B=\frac{\mu_0 N I r^2}{2(x^2 +r^2)^{(3/2)}}$$ This let's you get a value for the B/I ratio, which can be compared to an experimental value. I was wondering what would cause the two values to differ? Could it be that the coil has an actual thickness, rather than being infinitely thin as the derivation seems to assume? Or are there other factors?
2. At one point I had to have both a DC and AC component of the magnetic field (by having both an direct and alternating voltage). I was wondering about the AC field: since the experiment is looking at Zeeman splitting of energy levels, is the AC field meant to continually flip the electrons' spins, so that they are constantly making transitions between their two energy levels? (Therefore, being able to see a continuous up-and-down signal on the oscilloscope?)
3. Related to the last one: The voltage across the Helmholtz pair oscillates. So am I right in thinking that once every cycle of this voltage, the electrons will have their spin flipped over and back again, and hence you'll see two peaks on the oscilloscope for every cycle of voltage?
4. I had to measure the FWHH of the signal. This comes out in seconds. However I'm meant to give it in units of magnetic field. I'm really unsure about this part. I'm thinking I might use the formula $$h\nu=g \mu_B B$$, since I can get seconds in there by inverting the frequency (nu). It doesn't seem like a great answer though. I'd really like to know why it is useful to have it in this form, as well as how to do it.
5. What factors can affect the FWHH? Really no idea about this one...

 

[marcusl]

This sounds like a fun experiment. Was it for a physics lab class?

I'm confused about your fields, especially where you say that the field at the center reverses polarity on every cycle. A more usual configuration is to generate a static field (with a large DC current) that establishes the electron resonance frequency. Your RF system is tuned to that frequency. A small AC current is superimposed, to sweep the B0 field up and down a little. By tuning into and out of resonance, you can detect a changing signal at the AC frequency. Does that sound like your experiment?

 

[jameson2]

Yes, that is my experiment. I didn't want to put in a huge amount of detail in case it put people off reading my questions...

So the DC field is large and the AC just moves it up and down past resonance, so I can see the power absorption on the oscilloscope?

You can see by the amount of questions I have that I'm not totally comfortable with how it works, if anything seems wrong it's probably my bad.
Thanks for any help.

 

[marcusl]

I've a few questions about an EPR experiment I did recently. It involved setting up a Helmholtz pair, placing a sample in a uniform field inside the pair, and observing results on an oscilloscope.

1. You can calculate the field at the centre of a Helmholtz pair using the formula for one Helmholtz coil twice. This formula is: $$B=\frac{\mu_0 N I r^2}{2(x^2 +r^2)^{(3/2)}}$$ This let's you get a value for the B/I ratio, which can be compared to an experimental value. I was wondering what would cause the two values to differ? Could it be that the coil has an actual thickness, rather than being infinitely thin as the derivation seems to assume? Or are there other factors?

1. 
   You need to put two coils at x=+/-(r/2), and solve for the field at the center. There are corrections for finite coil thickness but they are quite small. How far off are you?
   
   

   [*]At one point I had to have both a DC and AC component of the magnetic field (by having both an direct and alternating voltage). I was wondering about the AC field: since the experiment is looking at Zeeman splitting of energy levels, is the AC field meant to continually flip the electrons' spins, so that they are constantly making transitions between their two energy levels? (Therefore, being able to see a continuous up-and-down signal on the oscilloscope?)
   [*]Related to the last one: The voltage across the Helmholtz pair oscillates. So am I right in thinking that once every cycle of this voltage, the electrons will have their spin flipped over and back again, and hence you'll see two peaks on the oscilloscope for every cycle of voltage?

   Well, partly but not exactly. The spins transition between the two energy levels for as long as they are "hit" by photons of energy equal to the difference between the states. In classical terms, the spins' directions "flip" (more exactly, they precess continually) so long as they are irradiated by RF of the correct frequency. The spins may flip over and over during the time during your sweep that the system is on resonance. In the original NMR apparatus (using the "Pound box" developed by Robert Pound), for instance, the Q of the RF cavity showed a slight dip as the sample inside absorbed microwave power. This power dissipated through "relaxation" mechanisms, primarily collisions and coupling from the spins to the bulk sample. I think the principles of EPR are similar.
   
   Here is an illustration I found online (see p. 9 of
   http://spectroscopy.lbl.gov/EPR-Robblee/EPR-Robblee.pdf")
   
   showing the modulation of the signal on the shoulder of the resonance curve. The signal modulation is at the same frequency as the field modulation. If you increase the amplitude so the field sweeps the frequency past the resonance peak, then you can convince yourself that you get an aysmmetric signal with a component at twice the modulation frequency. If, furthermore, you adjusted the static field to be right on resonance, then the detected signal would be symmetric at the second harmonic, with no fundamental.
   

   [*]I had to measure the FWHH of the signal. This comes out in seconds. However I'm meant to give it in units of magnetic field. I'm really unsure about this part. I'm thinking I might use the formula $$h\nu=g \mu_B B$$, since I can get seconds in there by inverting the frequency (nu). It doesn't seem like a great answer though. I'd really like to know why it is useful to have it in this form, as well as how to do it.
   [*]What factors can affect the FWHH? Really no idea about this one...

I think you relate seconds to field through the strength and frequency of the B0 field modulation.

Do some reading about relaxation and dissipation. Just like any oscillator or resonator (even mechanical pendulums), the Q and the width of the resonance curve depend on damping.

Yes, that is my experiment. I didn't want to put in a huge amount of detail in case it put people off reading my questions...

So the DC field is large and the AC just moves it up and down past resonance, so I can see the power absorption on the oscilloscope?

Basically yes, this is how EPR and NMR were done in the early days because its easy to sweep the field and also easy to build a fixed-frequency radio receiver. (In modern instruments the field is fixed and an agile RF or microwave transceiver does the real work.)

 

[jameson2]

For the B/I ratio, 0.00427 is the experimental value compared to 0.0048 from the theory. My thinking was basically like this: The original deriviation for a current carrying loop assumes an infinitely thin loop. But in a Helmholtz coil (320 turns in my case), it definitely has a spatial extent. So say I label the very center of the loop as the zero point on the axis. Then there will be some of the loops in the coil a little behind the zero point, and some a little in front. The coil is about 1cm thick, so I'd imagine this does have some effect, possibly meaning the maxima and minima of each loop in the coil do not line up exactly, hence the lower experimental value?

Also, I'm still stuck on another point. I see two resonance peaks per cycle of alternating voltage, and I can't quite get it straight in my head what this means. Is it that for half the cycle, the voltage induces one type of transition, then in the other half of the cycle induces the opposite transition (and the oscilloscope picks up the power absorption required for both these transitions-hence two peaks)?

 

[marcusl]

1. So you are about 10% off. What is the coil diameter? Comparable to 1 cm? I have references to thick Helmholtz coils at work so I can look next week, but I don't think thickness changes the field much if R>>1cm--probably by less than 1%. You need to look elsewhere for your errors, including the diameter and spacing, accuracy of current measurement, accuracy of your field meter, etc.

2. Refer to the illustration I mentioned before to see the action. Or you can look at it mathematically as follows: The absorption curve is Lorentzian, but we can approximate the top part of it as a quadratic (think an upside down U). It's not necessary to fit a quadratic but it makes the math easy. If the current is I, then the measured voltage from your EPR apparatus is proportional to

$$a-bI^2(t)$$,

assuming that your frequency is tuned right on resonance.

You are applying a sinusoidal current

$$I(t)=I_0 cos(\omega t)$$

so the response is proportional to $$cos^2(\omega t)$$.

You know from trigonometric identities that this can be written in terms of $$cos(2\omega t)$$, that is, the second harmonic.

 

[jameson2]

Ah, that makes sense.

The last thing I need is converting seconds to units of magnetic field. My best guess so far is, say I have 1 second (hypothetically), then frequency is 1, and then I just multiply this by h, divide by g and the Bohr magneton, and say it's that many teslas? Seems to just complicate things for no reason though.

Thanks for your help.

 

[marcusl]

No, the only thing you have to read out the resonance curve shape with is your field sweep. The number of seconds (out of the total sincusoidal sweep time) tells you how many gauss you swept. You can then convert that to frequency, and find the FWHM in frequency. You can also calculate the resonance Q (quality factor) as Q = f0 / FWHM.