Thread: Half-Life Question

1. Half-Life Question

The half life of carbon-14 is about 5760. If a bone sample only has 15% carbon-14 remaining, determine the age of the bone.

This question throws me off cause of the percent, if it was in mg's I could do it no problem.

The only part of the equation that I know is (1/2)^(x/5760) and I don't know where to put the .15 in cause it's not a measurable amount (ie. in mg's).

All I got is y=a(1/2)^(x/5760)

If anyone can show me what to do with the 15% it would greatly be appreciated. (I assume I need to find y and a, so I can solve for x)

Would it be something like .15=1(1/2)^(x/5760)?

2. Originally Posted by peekay

Would it be something like .15=1(1/2)^(x/5760)?
Yes, that looks about right, I'll double check and then edit

Half life: $\lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{5760}$

Exponential Decay $A = A_0e^{-\lambda t} \rightarrow 0.15 = e^{-\frac{ln(2)}{5760}t}$

I'm not sure if they are equivalent functions though - my method yields an answer of 15800 years (3sf)

3. yeah, i just did it with logs and the answer looks about the same.

did:

x/5760=log(.15)/log(.5)
x/5760=2.736
x=15764.92

Thanks for the confirmation!

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