# Heat variance in NMR

1. Aug 14, 2011

### babemagnetics

In the Nuclear Magnetic Resonance, do the applied magnetic and electromagnetic fields correspond linearly to the heat generated? If not, how do they vary?

Gracias

2. Aug 16, 2011

### babemagnetics

It seems from the below link that the relationship is linear, but it would be nice if someone could give me a confirmation.

http://pubs.acs.org/doi/abs/10.1021/ac60288a022" [Broken]

Last edited by a moderator: May 5, 2017
3. Aug 16, 2011

### babemagnetics

4. Aug 16, 2011

### babemagnetics

Hmm... I think I figured it out: The magnetic and EM fields applied are directly proportional to each other, and they are also directly proportional to the heat generated. I got this from the bottom of the following page:

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nmr.html#c1

It would be nice if someone could give me a confirmation of this.

Gracias

5. Aug 16, 2011

### Mike H

Sample heating in NMR is attributed to the interaction of the electric field component of the applied RF with any ions/electric dipoles in the sample. Certainly, there's a relationship between magnetic field strength and what RF frequencies you use (e.g., for a 9.4 Tesla magnet, you set your proton channel at ~ 400 MHz and so on), but otherwise, I'm not sure what you're suggesting.

If you're interested in solution NMR, I'd check out the following papers -

http://www-keeler.ch.cam.ac.uk/publications/tempaccuracy.pdf

and

http://www-keeler.ch.cam.ac.uk/publications/convectionandtemperature.html

- for a very nice and thorough demonstration of sample convection and temperature effects. If you look at Fig. 4 for the first paper, you see that the difference in sample temperature before and after irradiation jumps up fairly linearly at the maximum gas flow rate (~ 0.4 deg. C at 0.75 W, ~ 0.8 deg. C at 1.5 W, and ~ 1.6 deg. C at 3.0 W). But I'd suggest reading the entire paper to really get a good picture of the situation.

If your interests are oriented towards solids NMR, it will get a wee bit more complicated. But mostly people are inclined towards the paramagnetic compounds, whether coordination/chelation compounds or various inorganic oxides.

6. Aug 21, 2011

### babemagnetics

Now I am trying to figure out if NMR can be used to heat atoms that are not isotopes. (To my understanding, an isotope is an atom that has a different number of neutrons than the atom as it appears on the periodic chart). I know the wikipedia page on NMR says the atoms have to be isotopes, but is that just for getting emissions back from the nuclei? The following page seems to suggest that atoms, such as copper, do have a magnetic component to them, especially when they are in alloy form. Would the brass alloy (copper and zinc) be heatable with NMR? Here's that page: http://www.quora.com/Magnetism/Is-copper-magnetic" [Broken]

Is the requirement for NMR heating that the atom or molecule not have the same number of protons and electrons?

Key question: How is it that metals heat up so much in microwave ovens? Is that NMR? If not, what is it? Is it just because metals are more dense?

In the "Theory of nuclear magnetic resonance" section of the NMR wikipedia page, it says, to my understanding, that if the number of neutrons is odd, or if the number of protons is odd, then NMR will work for such an atom. Copper, for instance, has 29 protons, and 35 neutrons (both odd numbers). Therefore, wouldn't NMR be able to heat copper?

Furthermore, on the NMR wikipedia page, it says plain old hydrogen is the most sensitive atom for NMR (not any hydrogen isotope)!

If an atom does have to be an isotope (different number of neutrons than on the periodic chart) in order for NMR to cause heating, why is that the case? (Why only isotopes?)

Does anyone know the answers?

Gracias

Last edited by a moderator: May 5, 2017
7. Aug 22, 2011

### Mike H

The heating one observes of NMR samples and of what you place in a microwave oven can both be considered examples of dielectric heating, which I touched upon in my last post as the interaction of the electric field component of the RF (or microwave) radiation with ions/dipoles.

It sounds as if you are confusing the phenomenon of dielectric heating with what determines whether or not a nucleus is NMR-active. If a nucleus is spin-0 (such as carbon-12 or oxygen-16), then it will not be NMR-active. If it is spin-1/2 or greater, then it - at least in principle - will be possible to use it in NMR experiments.

And yes, people can and do NMR experiments with copper. See, for example -

http://www.pascal-man.com/periodic-table/copper.shtml

8. Aug 22, 2011

### babemagnetics

On that webpage I only see it talking about NMR with isotopes of copper. Can you do NMR on regular copper?

Gracias

9. Aug 22, 2011

### Mike H

See here -

http://www.webelements.com/copper/isotopes.html

Copper-63 and copper-65 are the naturally occurring isotopes, so in any copper-containing sample that has not been subjected to isotope enrichment, you will have a mix of these two according to the stated natural abundances.

10. Aug 22, 2011

### babemagnetics

Sorry to test your patience, but: Bottom line is non-isotope Copper (Copper as found on the periodic chart) is NMR active?

I'm going off the image of the proton, neutron, and electron configuration on this page: http://www.historyforkids.org/scienceforkids/chemistry/atoms/copper.htm" [Broken]

Oh, but now I see that 64Cu has a very short half life. Would it still have that short half life even if combined with another element, such as to form an alloy?

What is isotope enrichment?

Gracias

Last edited by a moderator: May 5, 2017
11. Aug 22, 2011

### Mike H

As you noted, 64Cu is radioactive and has a very short half-life. The half-life is not going to be changed to any experimentally appreciable extent by placing it in an alloy, or by changing the pressure or other external parameters. I am not a nuclear physicist, but my impression of the occasional talk about variations in nuclear decay rates is that it is probably something you can not worry about when planning NMR experiments.

Isotope enrichment refers to the process of preparing elemental sources or compounds that have a higher-than-naturally-occurring abundance of a stable isotope. For example, chemical suppliers will sell deuterium oxide (D2O). in varying abundances. I am honestly not sure if anyone does something similar with copper (I have mostly worked with fairly unexciting nuclei in my career - 1H, 2H, 13C, 15N, 19F, 31P - although in some fairly interesting contexts, including paramagnetic samples).

Perhaps if you could share a bit about what you have in mind to do, I might be able to provide some further guidance or - barring that - some good references.

12. Aug 22, 2011

### babemagnetics

Mike, you're awesome. Thank you for finally bringing my curiosity to a close. I stumbled across NMR and was left puzzled by the explanations on the internet. They seemed to suggest that only isotopes could be utilized for NMR. Also, I was curious about the relationships between the magnitudes used in NMR. I like to find an area of physics I'm not familiar with yet, and come to understand it, at least until something else grabs my attention. From other explanations on the internet about NMR it seemed like 64Cu should be eligible for NMR, but as much as I looked and looked, I couldn't find a mention of it being used with NMR. The short half life finally dawned upon me (can't believe it took me so long to realize that). And then I was curious about stabilizing half lives in general. I am contented on the topics of NMR and half life stabilization now.

Gracias!!!!

13. Aug 22, 2011

### Mike H

There is a lot of material online about NMR of varying quality, for better and worse. So I can understand your puzzlement.

It may help to remember that isotopes are simply variations of atoms of a particular chemical element with different numbers of neutrons. I think some people put across the idea that there is a primary variant (the one found most commonly in nature, usually) which is the "default" and that the rest are the isotopes, but I think that is a bit funny. The number of protons is what determines the element, after all, not neutrons. I don't discriminate - for example, all three isotopes of hydrogen are NMR-active. In the end, as long as you are dealing with a spin-1/2 or above nucleus whether it's naturally abundant or not, it will be NMR-active.