(Carbon Dating) Carbon dating is a technique used by archaeologists, geologists, and others who want to estimate the ages of certain artifacts and fossils they uncover. The technique is based on certain properties of the carbon atom. In its natural state the nucleus of the carbon atom 12c has 6

Question

(Carbon Dating) Carbon dating is a technique used by archaeologists, geologists, and others who want to estimate the ages of certain artifacts and fossils they uncover. The technique is based on certain properties of the carbon atom. In its natural state the nucleus of the carbon atom [latex]{ }^{12} \mathrm{C}[/latex] has 6 protons and 6 neutrons. An isotope of carbon [latex]{ }^{12} \mathrm{C}[/latex] is [latex]{ }^{14} \mathrm{C}[/latex], which has 2 additional neutrons in its nucleus. [latex]{ }^{14} \mathrm{C}[/latex] is radioactive. That is, it emits neutrons until it reaches the stable state [latex]{ }^{12} \mathrm{C}[/latex]. We make the assumption that the ratio of [latex]{ }^{14} \mathrm{C}[/latex] to [latex]{ }^{12} \mathrm{C}[/latex] in the atmosphere is constant. This assumption has been shown experimentally to be approximately valid, for although [latex]{ }^{14} \mathrm{C}[/latex] is being constantly lost through radioactive decay (as this process is of ten termed), new [latex]{ }^{14} \mathrm{C}[/latex] is constantly being produced by the cosmic bombardment of nitrogen in the upper atmosphere. Living plants and animals do not distinguish between [latex]{ }^{12} \mathrm{C}[/latex] and [latex]{ }^{14} \mathrm{C}[/latex], so at the time of death the ratio [latex]{ }^{\circ}{ }^{12} \mathrm{C}[/latex] to [latex]{ }^{14} \mathrm{C}[/latex] in an organism is the same as the ratio in the atmosphere. However, this ratio changes after death since [latex]{ }^{14} \mathrm{C}[/latex] is converted to [latex]{ }^{12} \mathrm{C}[/latex] but no further [latex]{ }^{14} \mathrm{C}[/latex] is taken in.
It has been observed that [latex]{ }^{14} \mathrm{C}[/latex] decays at a rate proportional to its mass and that its half-life is approximately 5580 years. [latex]{ }^{†}[/latex]That is, if a substance starts with [latex]1 \mathrm{~g}[/latex] of [latex]{ }^{14} \mathrm{C}[/latex],

then 5580 years later it would have [latex]\frac{1}{2} g[/latex] of [latex]{ }^{14} C[/latex], the other [latex]\frac{1}{2} g[/latex] having been converted to [latex]{ }^{12} \mathrm{C}[/latex].

We may now pose a question typically asked by an archaeologist. A fossil is unearthed and it is determined that the amount of [latex]{ }^{14} \mathrm{C}[/latex] present is [latex]40 \%[/latex] of what it would be for a similarly sized living organism. What is the approximate age of the fossil?

Accepted Answer

Let [latex]M(t)[/latex] denote the mass of [latex]{ }^{14} \mathrm{C}[/latex] present in the fossil. Then since [latex]{ }^{14} \mathrm{C}[/latex] decays at a rate proportional to its mass, we have

[latex]\frac{d M}{d t}=-\alpha M ext {, }[/latex]

where [latex]\alpha[/latex] is the constant of proportionality. Then [latex]M(t)=c e^{-a t}[/latex], where [latex]c=M_{0}[/latex], the initial amount of [latex]{ }^{14} \mathrm{C}[/latex] present. When [latex]t=0, M(0)=M_{0} ;[/latex] when [latex]t=5580[/latex] years, [latex]M(5580)=\frac{1}{2} M_{0}[/latex], since half the original amount of [latex]{ }^{14} \mathrm{C}[/latex] has been converted to [latex]{ }^{12} \mathrm{C}[/latex]. We can use this fact to solve for [latex]\alpha[/latex] since we have

[latex]\frac{1}{2} M_{0}=M_{0} e^{-\alpha \cdot 5580}, \quad ext { or } \quad e^{-5580 \alpha}=\frac{1}{2}[/latex].

Thus

[latex]\left(e^{-\alpha} ight)^{5580}=\frac{1}{2}, \quad ext { or } \quad e^{-\alpha}=\left(\frac{1}{2} ight)^{1 / 5580}, \quad ext { and } \quad e^{-\alpha t}=\left(\frac{1}{2} ight)^{t / 5580}[/latex],

so

[latex]M(t)=M_{0}\left(\frac{1}{2} ight)^{t / 5580}[/latex]

Now we are told that after [latex]t[/latex] years (from the death of the fossilized organism to the present) [latex]M(t)=0.4 M_{0}[/latex], and we are asked to determine [latex]t[/latex]. Then

[latex]0.4 M_{0}=M_{0}\left(\frac{1}{2} ight)^{t / 5580}[/latex],

and taking natural logarithms (after dividing by [latex]M_{0}[/latex] ), we obtain

[latex]\ln 0.4=\frac{t}{5580} \ln \left(\frac{1}{2} ight), \quad ext { or } \quad t=\frac{5580 \ln (0.4)}{\ln \left(\frac{1}{2} ight)} \approx 7376 ext { years. }[/latex]

The carbon-dating method has been used successfully on numerous occasions. It was this technique that established that the Dead Sea scrolls were prepared and buried about two thousand years ago.

Question 11.2.6: (Carbon Dating) Carbon dating is a technique used by archaeo...

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