Can C14 Atoms Be Detected Using NMR Despite 0% Natural Abundance?

[HastiM]

Homework Statement

We want to image carbon C14 atoms using NMR with magnetic field of B=3 T. What frequency do we need to use? Use the following facts: C14 has 0% abundance, nuclear spin of I=3 and g=0.273

Homework Equations

The frequency is given by f= (2⋅I) ⋅ (g ⋅B ⋅μ)/(1836 ⋅h), where μ is the Bohr magneton and h is the Planck constant.

The Attempt at a Solution

Of course, there is no problem to determine the frequency f by plugging in all the relevant constants into the formula above. I have a question regarding the information of '0% abundance', because I do not know how it effects the outcome. In class we determined the analogous situation for carbon C13, which has an abundance of 1.1 %. We just used the formula above and ignored the abundance completely. But I could not find anything regarding the NMR of Carbon C14. Is it maybe possible that C14 is not detectable by NMR, due to the 0% abundance?

I would very much appreciate your help! [/B]

 

[TeethWhitener]

I have a question regarding the information of '0% abundance', because I do not know how it effects the outcome. In class we determined the analogous situation for carbon C13, which has an abundance of 1.1 %. We just used the formula above and ignored the abundance completely.

This is fine.

But I could not find anything regarding the NMR of Carbon C14. Is it maybe possible that C14 is not detectable by NMR, due to the 0% abundance?

As long as the sample has C14, you should be able to detect it, given the proper experimental setup. You could, for example, enrich the sample with C14 to do NMR. However, as you've noted, the natural abundance of C14 is extremely low (really only permits radiocarbon dating at very low concentrations), and C13 is 1) abundant in comparison, and 2) a spin-1/2 nucleus, which makes its relaxation process simpler and more amenable to high-resolution spectroscopy. In fact, I've never heard of anyone performing routine C14 NMR, whereas C13 NMR is ubiquitous.