Number of C-14 atoms in old piece of wood

[kmoh111]

Homework Statement

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.?

Homework Equations

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

The Attempt at a Solution

For C-14, lambda = ln2/T_1/2 = 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<sup>(-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.</sup>

 

[queenofbabes]

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

 

[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.