Plotting t against F with a value of l=1 gives the graph on the right. The equivalent thickness for the medium in radiation attenuation is known as "half-value thickness".

Similarly, in a population which grows exponentially with time there is the concept of "doubling time".

Since we also know the ratio of C-14 to C-12 originally, we can find the time that has passed since carbon exchange ceased, that is, since the organic material "died".

In the case of the Dead Sea scrolls, important questions required answers. Did they really date from around the time of Christ? Using Libby's radiocarbon dating technique, the scrolls have been dated, using the linen coverings the scrolls were wrapped in.

Carbon 14 (C-14) is a radioactive element that is found naturally, and a living organism will absorb C-14 and maintain a certain level of it in the body.

This is because there is carbon dioxide (CO exchange, and so the ratio of C-14 to the far more common carbon isotope, C-12, will begin to decrease as the C-14 atoms decay, yielding nitrogen (N-14) with the emission of an electron (or "beta particle") plus an anti-neutrino.

However, now the "thin slice" is an interval of time, and the dependent variable is the number of radioactive atoms present, N(t). If we have a sample of atoms, and we consider a time interval short enough that the population of atoms hasn't changed significantly through decay, then the proportion of atoms decaying in our short time interval will be proportional to the length of the interval.

We end up with a solution known as the "Law of Radioactive Decay", which mathematically is merely the same solution that we saw in the case of light attenuation.

To show this, we needed to make one critical assumption: that for a thin enough slice of matter, the proportion of light getting through the slice was proportional to the thickness of the slice.

Exactly the same treatment can be applied to radioactive decay.

Now, take the logarithm of both sides to get $$-0.693 = -5700k,$$ from which we can derive  k \approx 1.22 \cdot 10^.

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