Would it be possible to use radiocarbon dating on a tooth, lost by a person while they were still alive, to determine that person's birth year?

Imagine an adult who loses a tooth for whatever reason at age 40, but lives to age 100. Would that tooth suffice to approximate their birth year?

Edit to include:

In an effort to not waste people's time reading a long question, I erred on the side of too little information.

  • Yes, I'm very familiar with the Wikipedia article on radiocarbon dating.
  • I have read the linked similar question and all of its responses.

Which leads to these specifics:

  • Is one adult human tooth adequate for radiocarbon dating purposes?
  • Could such a tooth from a period of 200 to 700 years ago be dated to within a small enough window to make a difference, i.e. the window would be smaller than 500 years.

Delving into that, I found a number of articles. This one on radiocarbon dating on recent human remains was partially helpful: https://nij.ojp.gov/topics/articles/applying-carbon-14-dating-recent-human-remains

TL;DR: For people who acquired adult teeth between roughly 1965 and 2030, the technique will work, thanks to unusual amounts of C14 in the environment as a result of nuclear weapon testing.

This article suggests the ability to date samples of a similar recent vintage, though I did not purchase the article and read the whole thing: https://www.cambridge.org/core/journals/radiocarbon/article/abs/application-of-radiocarbon-dating-to-forensic-investigation-and-evaluation-of-formaldehyde-influence-on-radiocarbon-age/1B02FEF83B5D1F5E11F8E02D9550A57D

This is sadly not useful for me, with my teeth samples carefully preserved from someone who was born in 1552.

It looks like the answer to my question is "no" -- C14 dating is too imprecise to be useful for such a recent and short time span. But this is my non-expert reading of the various articles mentioned and linked.

  • 1
    $\begingroup$ Possible duplicate: biology.stackexchange.com/q/24084/5198 $\endgroup$ Commented Mar 7 at 4:03
  • 1
    $\begingroup$ @John Make your comment an answer and I'll mark it as the answer I was looking for. $\endgroup$
    – CXJ
    Commented Mar 21 at 21:38


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