Formula for calculating age by using carbon-14 (source: HowStuffWorks) $$ t = \frac{\ln (N_\mathrm f/N_\mathrm o)}{-0.693}t_{1/2} $$ The page doesn't explain what the $-0.693$ represents. What is $-0.693$? Secondly, it says that $N_\mathrm o$ is the original level of $\ce{C-14}$ and $N_\mathrm f$ is the measured value. How do you get the original value? Is it the same for all living organisms?

Wikipedia says that the age of an object (e.g. a fossil) can be calculated by comparing the amount of $\ce{C-14}$ in the fossil and the amount of $\ce{C-14}$ in a sample of living tissue. The half-life of $\ce{C-14}$ ($\pu{5,730 years}$) can be used to find out how much time has passed since the death of the organism because you should be able to calculate how much time has passed by measuring how much $\ce{C-14}$ has decayed. But how can you calculate how much has decayed if you don't know how much was there in the first place?


Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.

What is −0.693?

Radioactive decay processes follow first-order kinetics. A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power. So consider the following reaction:

$$\mathrm{A} \longrightarrow \mathrm{products}$$

The rate can be expressed as the rate of change in the reactant concentration with a change in time such that

$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$

This simply means that [A] is being consumed (i.e. converted to products) as the reaction proceeds with time. The corresponding rate law can be written as

$$\mathrm{rate} = k[\mathrm{A}]$$

where $k$ is a rate constant. We can set these two expressions equal to each other to give

$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$

At this point we must apply a bit of calculus. Let us write the previous equation in differential form to give

$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or} ~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$

Integrate over $t=0$ to $t = t$ to give

$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$

which results in

$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$


$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$

where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.

Now that we have our first-order integrated rate law, we can now discuss half-life. Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to half its original value (where the original value is simply [A]$_0$. So

$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$

Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~ t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$

Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$

We can then simplify this equation such that

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~ t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2) ~~~\longrightarrow~~~ t_{\frac{1}{2}} = \dfrac{0.693}{k}$$

So to answer your question, the 0.693 value comes from the $\ln(2)$ term. As ringo pointed out below (and I will include here to be thorough), you can now solve your first-order integrated rate law for $t$ and substitute in your half-life equation such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt ~~~\longrightarrow~~~ \dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t ~~~\longrightarrow~~~ \dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}} { -\frac{0.693}{t_{\frac{1}{2}}} } =t $$

or more cleanly...

$$t = \dfrac{\left(t_{\frac{1}{2}}\right) \left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)} {-0.693} $$

How do you get $N_0$?

You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant. Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!). Therefore you could use its activity for $N_0$. Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.

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    $\begingroup$ Very interesting about $\ce{^{14}C}$ dating is that there are calibration curves. Baffled me somewhat, but makes sense. $\endgroup$ – tschoppi May 20 '15 at 15:26

Carbon-14 dating works because in a living organism, carbon-14 is constantly decaying, but is also constantly being replenished by the organism, meaning its concentration in the living tissue essentially does not change. Though the concentration of the element carbon itself is different for all organisms, the percent abundance of carbon-14 in all living organisms will be $\approx 0.00000000001\%$ (one part per trillion). In a deceased organism, carbon-14 can no longer be replaced, so the supply of carbon-14 slowly diminishes over time ($t_{1/2}=5730\:\mathrm{years}$). The concentration of carbon-14 in a deceased organism at any given time can therefore be used to find how long ago the organism stopped replenishing its supply of carbon-14.

The decay of $\ce{^14C}$ to $\ce{^14N}$ is given by the equation:

$$\ce{^14_6C->^14_7N +e-}$$

The equation you see is derived from the integrated rate law. This is a first order reaction with respect to both $\ce{^14C}$ and $\ce{^14N}$, meaning there is one of each atom in the rate determining step of the reaction. This means that the change in the concentration of each of these atoms can be used to determine the age of a sample which it decays can be modeled by the equation:


or as you have it in your question:


The rate constant for the equation, derived from the time taken for the concentration of the concentration to be half of the initial ($N_f=\frac{N_0}{2}$) is:






$$\ln(2)\approx 0.693$$


Solving the equation for any time $t$:






Substitute in what we solved for:





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