User:Jclerman/Dating calc

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1 Carbon exchange reservoir in the real world
2 Note
- 2.1 Computations of ages and dates

[edit] Carbon exchange reservoir in the real world

Radiocarbon Variations and Absolute Chronology (Twelfth Nobel Symposium) (edit. by Olsson, I. U.) (Wiley, New York, 1970).

Libby's original exchange reservoir hypothesis assumes that the exchange reservoir is constant all over the world. The calibration method also assumes that variation in ¹⁴C level is global, such that a small number of samples from a specific year are sufficient for calibration. ^[1]

The following variances in the carbon exchange reservoir are seen in the real world, they are assumed insignificant or are accounted for in calibration:

Erosion and immersion of carbonate rocks (which are older than 60,000 years and so do not contain ¹⁴C) causes an increase in ¹²C and ¹³C in the exchange reservoir, which depends on local weather conditions and can vary the ratio of carbon that living material ingests. This is believed negligible since most erosion will flow into the sea.^[2]
Volcanic eruptions eject large amount of carbonate into the air, causing an increase in ¹²C and ¹³C in the exchange reservoir and can vary the exchange ratio locally. This explains the often irregular dating achieved in volcanic areas.^[3]
¹⁴C is known to behave chemically in a way different from ¹²C and ¹³C (due to different atomic mass), such that it is possible one isotope will be involved in decomposition reactions out of ratio with other isotopes, but the chemical behavior effects are extremely minor.^[4]
The earth is not affected evenly by cosmic radiation, the magnitude of the radiation depends on land altitude and earth's magnetic field strength at any given location, causing local variation in ¹⁴C production. This is accounted for by having calibration curves for different locations of the globe. However this cannot always be performed, as tree rings for calibration are only recoverable from certain locations.^[5]

These effects were first confirmed when samples of wood from around the world, which all had the same age (based on tree ring analysis), showed variance from the expected per minute decay frequency, assuming they had the same ¹⁴C ratios. This meant the dating of the samples varied by as much as 700 years. Calibration techniques and tree samples continue to increase the accuracy. Samples are accurate to 'at worst' 700 years.^[6]

[edit] Note

[edit] Computations of ages and dates

Equivalent methods to calculate ages are described in the following.

[edit] Method 1. Comparison table

For radioactive decay, the relationship between fraction remaining (f) and the number of half-lives (n) elapsed is:

n	1	2	3	4	5	6	7	8	...	n
f	1/2	1/4	1/8	1/16	1/32	1/64	1/128	1/256	...	...
	1/2¹	1/2²	1/2³	1/2⁴	1/2⁵	1/2⁶	1/2⁷	1/2⁸	...	1/2ⁿ
years	5730	11460	17190	22920	28650	34380	40110	45840	...	...

Given any intermediate f value, its comparison with the n and age rows provides a rough idea of the age range, sidestepping any calculations. For example, if f = 1/10, n lies between 3 and 4 half lives and the age t lies between 17,190 and 22,920 yrs.

For precise age calculations, see the next methods.

[edit] Method 2. Simple equation

From the above it follows that the fraction remaining after time n is: f = 1/2ⁿ, which is equivalent to f = (1/2)ⁿ or f = 0.5ⁿ. This makes the calculation of the age, if the fraction remaining is known, quite simple.

Solving the above relationship for n using the properties of logarithms:

f = 0.5ⁿ becomes

ln f = n·ln 0.5 and

$n = \frac{\ln \left({f}\right)}{\ln \left({0.5}\right)}$

Example 1: for a sample containing 0.25 of the original C-14:

$n = \frac{\ln \left({0.25}\right)}{\ln \left({0.5}\right)}$ ; solving this gives n = 2 half lives and

2 half lives * 5730 yrs/half life = 11,460 yrs.

Example 2: for a sample that contains 0.06780 of the original C-14:

$n = \frac{\ln \left({0.06780}\right)}{\ln \left({0.5}\right)}$ ; solving this gives n = 3.883 half lives and

3.883 half lives * 5730 yrs/half life = 22,250 yrs.

[Suggestion, include Example 3, for an age near 50,000 yrs]

[edit] Method 3. From the laboratory to a calendrical date

Radiocarbon dates are obtained, in the laboratory, after the following succesive steps:

laboratory measurements of three specific (radio)activities: activity of the sample to be dated (S), activity of a modern standard (M), and activity of a background sample (B),
calculation of the net sample activity, as fraction of modern, i.e. (S-B)/M; this is the value denoted below as ${\frac{N}{N_0}}$
calculation of the sample raw age (t) using the formula given below, and
evaluation of the calibrated calendrical date by inputting t into a calibration curve.

The radioactive decay of carbon-14 follows an exponential decay. A quantity is said to be subject to exponential decay if it decreases at a rate proportional to its value. Symbolically, this can be expressed as the following differential equation, where N is the quantity and λ is a positive number called the decay constant:

$\frac{dN}{dt} = -\lambda N.$

The solution to this equation is:

$N = Ce^{-\lambda t} \,$ ,

where $C$ is the initial value of $N$ .

For the particular case of radiocarbon decay, this equation is written:

$N = N_0e^{-\lambda t}\,$ ,

where, for a given sample of carbonaceous matter:

N 0

= number of radiocarbon atoms at

t = 0

, i.e. the origin of the disintegration time,

N

= number of radiocarbon atoms remaining after radioactive decay during the time

t

λ =

radiocarbon decay or disintegration constant.

Two related times can be defined:

half-life: time lapsed for half the number of radiocarbon atoms in a given sample, to decay,
mean- or average-life: mean or average time each radiocarbon atom spends in a given sample until it decays.

It can be shown that:

t 1 / 2

= $\frac{\ln 2}{\lambda}$ = radiocarbon half-life = 5568 years (Libby value)

t a v g

= $\frac{1}{\lambda}$ = radiocarbon mean- or average-life = 8033 years (Libby value)

Notice that dates are customarily given in years BP which implies t(BP) = -t because the time arrow for dates runs in reverse direction from the time arrow for the corresponding ages. From these considerations and the above equation, it results:

For a raw radiocarbon date:

$t(BP) = \frac{1}{\lambda} {\ln \frac{N}{N_0}}$