Link to home
Start Free TrialLog in
Avatar of John Carney
John CarneyFlag for United States of America

asked on

How do we know that Carbon-14 has a half-life of 5,730 years?

How DO we know that Carbon-14 has a half-life of 5,730 years? Or the half life of any substance for that matter?

Thanks,
John
ASKER CERTIFIED SOLUTION
Avatar of sinfocomar
sinfocomar

Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
SOLUTION
Avatar of d-glitch
d-glitch
Flag of United States of America image

Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
Avatar of John Carney

ASKER

Thanks for all the input! I've read several of the links before and they're very informative about how the C-14 half-life is used. But I still have a 5-year old's annoying curiosity about why things are the way they are. For example, we all know that an object released into the air will fall to the earth because gravity makes the two objects draw toward each other, but I still want to know why it works that way. Why should two objects near each other be drawn to each other?

So some of my questions are:

How exactly do we count the number of atoms in a chunk of matter?
How do we identify the different elements so we can get an accurate count of each elements number of atoms? In my simple brain, counting requires some kind of mechanism - something like the mechanism that separates coins with different size holes - but what?
How many ppm of N-14 are in the atmosphere at this moment?
Assuming there is always some N-14 in the air, how do we know how many of the atoms were created by decay of the C-14?
Other than carbon dating, how can we know that a given layer of sediment goes back 50 or a hundred or 800 thousand years?

I'm sure my questions reveal my lack of sophistication in this area, but that's why I'm asking. :-)

Thanks,
John
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
>> sinfocomar

Great site.
Thanks, d-glitch and sinfocomar. Hopefully I'll have time tonight to study that site.
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
You guys have been very generous with your time and knowledge, so thanks!  Tom, if I understand your post correctly, the answer to my question is pretty simple. Here's my extreme oversimplification. Assuming fixed not amortized "depreciation" so to speak, is this logic correct:

If I start with a sample that has 11,460 atoms of C-14, and we're down to 11,459 atoms after 1 year, then I can project that we'll be down to 5,730  atoms of C-14  after 5,730 years, for a half-life of 5,730 years. If it takes 2 years to lose one atom, then we can assume a half-life of 11,460 years?  

John
Not exactly.  The number of disintegrations depends on the decay rate and the amount of material in the sample.  And these are average statistical phenomena, not clockwork mechanisms.

So if you have 11460 C-14 atoms to start with, you will have approx 5730 after 5730 years.
The average rate over that time will be 1 disintegration per year.

But the starting rate will be twice as fast as the final rate, because there is twice as much material at the beginning.

You need to use logs and exponents (and larger samples) to get more precise answers.
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
Perfect, thanks. I assume that somewhere in the answers/links I've gotten here is a link to the exact formula.

BTW, how does one grade a question like this?  :- )
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
SOLUTION
Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
Somewhere in your answers or links, I'm sure this is spelled out so that an ordinary dummy could figure it out, but I apparently am an extraordinary dummy so please take a look at the attached Excel file and tell me if my logic is correct on how half-life is determined.

Thanks,
John
C-14-Half-Life-table.xlsx
You are no dummy, no matter what you say.  An extraordinary dummy is a genius. :-)

This has been so even before Carbon Dating became fashionable. :-)
Your table and your logic are correct.

If you keep lose 0.5% of your sample every year (and keep 99.5%):

The half-life will be    log(2)/log(0.995)  =  (0.30103) /(0.002177)  =  138.28 years
I feel like I've used up half of your lives on this! Thanks, everybody.
Avatar of sinfocomar
sinfocomar

Glad to help. However you are now obliged to answer all Carbon dating questions at EE for the next 10000 years :)
Cheers!
I second the sinfocomar'r proposal.  :-)