John Carney

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

How

Thanks,

John

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

John

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>> sinfocomar

Great site.

Great site.

ASKER

Thanks, d-glitch and sinfocomar. Hopefully I'll have time tonight to study that site.

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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

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.

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.

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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? :- )

BTW, how does one grade a question like this? :- )

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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

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.

**:-)**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

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

ASKER

I feel like I've used up half of your lives on this! Thanks, everybody.

Glad to help. However you are now obliged to answer all Carbon dating questions at EE for the next 10000 years :)

Cheers!

Cheers!

I second the

**sinfocomar**'r proposal. :-)
ASKER

two objects near each other be drawn to each other?shouldSo some of my questions are:

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

Thanks,

John