# Homework Help: Number of C-14 atoms in old piece of wood

1. Sep 9, 2009

### kmoh111

1. The problem statement, all variables and given/known data
I'm working a homework problem that states:

A very old specimen of wood contained 1012 atoms of 14C in 1986.
• How many 14C atoms did it contain in the year 9474 B.C.?
• How many 14C atoms did it contain in 1986 B.C.?

2. Relevant equations

N(t) = N(0)exp^(-lambda*t)

3. The attempt at a solution

For C-14, lambda = ln2/T1/2 = 0693/5730 yrs = .000121 yr-1

Since we do not know N0, I took the ratio of C-14 atoms in 1986 (A.D.) to C-14 atoms in 9474 B.C.:

Let t1 reflect the time from t0 to 1986.
Let t2 reflect the time from t0 to 9474 B.C.
then the ratio of N(t1)/N(t2) is:

N0 * e-lambda*t1
-----------------------------------------------------
N0 * e-lambda*t2

N0 cancels and we know that N(t1) = 1012 atoms.

Solving for N(t2) gives us:

1012
----------------------------------------------------
e-lambda(t2-t1)

We know that the difference t2 - t1 is equal to 1986 + 9474 = 11460 yrs. Plugging the numbers and solving for N(T2), gives us

1012
----------------------------
e(-0.693(11460/5730))

N(t2) = 4 x 1012 ===> number of C-14 atoms in 9474 B.C.

Is this the correct approach? I used the approach for the second part of the question for number of C-14 atoms in 1986 B.C.. I got 1.6 x 1012 atoms.

Thanks for the help.

2. Sep 10, 2009

### queenofbabes

Well....*technically* $$N(t1) = N_0 * e^{-\lambda*t1}$$, so you shouldn't have cancelled out the N0, although it does not affect your final answer.

Last edited by a moderator: Sep 10, 2009
3. Sep 10, 2009

### kuruman

Perhaps an easier approach would be to note that the time elapsed, 11,460 years, is about 2 half-lives, 2.01 to be precise. What fraction of the original sample is left after 2 half-lives have gone by? If you know the fraction remaining, you can easily find the original amount.

In general, I find it easier to answer this kind of problem using the half-life as a unit of time because the fraction remaining after n half-lives is

f = 2-n.