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Friday, October 13, 2017

Carbon dating and the Math

One would have to be a hermit not to have heard about carbon dating.  This is the dating, for instance, of a piece of wood in an old building or a piece of charcoal in an archaeological dig.

At a first approximation, the physics is pretty straight forward.  An atom consists of a nucleus with electrons whizzing around the nucleus.  Which element the atom is depends on the number of electrons and the number of electrons, in turn, depends on the number of protons in the nucleus.  In a normal, unionized atom, the number of electrons and protons are equal and the atom is neutrally charged.

The glue that holds these positively charged protons together in the nucleus (remember like charges repel each other) are the neutrons.  Don't ask me how they do this.  The explanation is way above my pay grade.  Very roughly speaking, there are the same number of neutrons as protons but this can vary.  Carbon, for instance, can exist in a state with 6protons and 6 neutrons for an atomic mass number of 12. It can also exist in a form with 6 protons and 8 neutrons for a mass number of 14.

These two types are called isotopes of Carbon.  There is a third one but it is not needed for this explanation.

Some isotopes are stable, some are not (why is also above my pay grade).  In the case of Carbon, 12 is stable, 14 is not. 

Carbon 14 disintegrates into Nitrogen 14 with the ejection of an electron from one of it's neutrons.  The neutron becomes a proton so the atom is now a new element with 7 protons and 7 neutrons, hence 14N.

No one knows when any individual Carbon 14 atom is going to disintegrate.  There is a very small probability at any one moment but when you have a lot of 14C, you can predict how many atoms will change to 14N in any given time period.  This results in something interesting which has been observed experimentally.  If you know how much of the radioactive element you have, you will observe that half of it will break down in a given time, referred to as it's half life.  The half life of various radioactive isotopes varies from tiny fractions of a second to many millions of years.

In the case of 14C, it's half life is 5730 years give or take 40 years.

In 5730 years you will have half left, in another 5730 years, a quarter of the original amount, in one more half life, one eighth of the original amount and so forth.

So now we need the math for this.  We will work out what I call the straight forward formula and then we can change it around (solve for other parts) so that each component of the formula becomes the subject.

First a note on mathematical notation.

What is meant when you see a symbol.

xA means multiply the A by x.  If A is 2 and x is 3 then xA is 6

Ax means multiply A by itself x times.  If A is 2 and x is 3 then Ax is 8.  In words, A is raised to the xth power.

However, in the symbols Ax,  x is not an operator.  ie, it doesn't say to do anything.  It is a label.  It means the xth A.  For instance you could have A1, A2, A3 etc.  This is the first, second and third A.  Or Ao and At which for our purposes will mean A at time zero and A at a specified future time.

There is a special one in Chemistry.  I'll use Carbon since this is what we are talking about.  For instance 14C.  This means the carbon atom with 14 nucleotides.   ie, The sum of neutrons and protons adds up to `14.  There also exist 12C and 13C.  Of course both have 6 protons or it wouldn't be Carbon.  The number of neutrons varies.

And one more in Math.  If the subscript is after the word log such as log5 then it means log to the base 5.  If only log is used, it is understood it is to the base 10.   That is to say, log = log10 and if ln is used it is to the base 'e'.  Don't worry about it, we don't need 'e'.  I only mention it because it is on your little hand held computer and you might wonder.

Lets go back to the basics.  Every half life period, (h) the amount is halved. In the case of Carbon, the half life is 5730 years but half lives for other isotopes varies hugely.   Lets call the amount we start with as Ao (A at time zero) and the amount we are left with as At (A at some time t in the future).  The amount we will have left after one half life is:

1.   A1 = Ao(1/2)1

After two half lives
2.    A2 = Ao(1/2)2

After three half lives
3.    A3 = Ao(1/2)3
Remember 1/2 times 1/2 is 1/4.   Multiply once more by 1/2 and you have 1/8.  When you see a times sign between fractions, replace it in your mind with "of".  then 1/2 x 1/2 becomes one half of one half.

The 1,2 and 3 are the number of half lives that have gone by.

4.  So An - Ao(1/2)n  or in words, to find the amount of a substance after n half lives have gone by, multiply Ao, the initial amount, times 1/2 raised to the nth power.

Note that in the notation Ax,  x means the amount at time x expressed in half lives.

Also note that even if the n is not a whole number and therefore would take a wee bit of higher math (knowing logarithms), to solve, your computer does this with no problem.  Your high school computer can solve, for instance, 63.22 without raising a sweat.

Suppose we start with one gram of a radioactive substance and one half life has gone by.  We simply multiply 1gram times 1/2

Suppose 4 half lives have gone back.  We multiply the one gram times (1/2)4.  that is to say by 1/2 times 1/2 times 1/2 times 1/2 which equals 1/16th times the original amount.

Now suppose we know what the half life (h) of a particular isotope is.  Say it is 10 years, for simplicity.  Say 30 years have gone by.  Obviously 3 half lives have past.  In other words n, the number of half lives equals the time elapsed (t) divided by the Half life (h).  In this case n = 30/10 = 3.

5.   n=t/h.

And, as I said, it doesn't have to be a whole number.  If the half life is 10 years and 75 years have gone by then n = 75/10 = 7.5.  With simple math we would have a problem raising a number to a fractional exponent but your computer has no such problem so don't sweat it.

You can see where this is leading.  Since n=t/h, we can substitute t/h into the formula where we see n.

The radioactive decay formula then becomes

6.  At = Ao(1/2)t/h
or in words, to find the amount of radioactive material remaining after time t, multiply Ao, the initial amount, times one half raised to the power of t/h.

Good heavens!  I forgot to tell you where the radioactive Carbon comes from.  If it's half life is only 5730 years, in about 50,000 years there will be so little of it that carbon dating is out of the question and the world has been here for over 4b years.  Clearly, 14C must be being created somewhere.  the 'Where',, is in the upper atmosphere.  As cosmic rays hit the upper atmosphere, they are so energetic that they cause some nuclear reactions and one of these is to change some14N into 14C.  It is a very small amount but enough to be detected in living material with modern methods so we have a clock we can use.  When an organism dies it stops taking up carbon and the clock starts to tick.  If we  analyze it sometime in the future, we can know when it died (up to about 50,000 years).

Now we can do what a mathematician calls solving for Ao or for t or for h.  In other words we re-arrange the formula so that each of these terms in turn become the subject of the formula (ie. is by itself on the left and everything else is  on the right). I'll tell you what each variation of the formula is good for as we rearrange them.

The basic principle of solving for a factor (one of the letters) in a formula is that we can do anything we want to one side as long as we do the same to the other side.  After all if I have a formula that 7 = 3+4, if I multiply both sides by, say, 5, the formula is still correct.  Of course we don't just do random things to both sides of the formula. The trick is to do something that gets us closer to the solution we are looking for.

One other thing.  At one point in the procedure I am going to have to take a log of both sides.  Even if you don't understand logarithms, this should pose no emotional problem since I am doing the same to both sides.  Then, however, you are going to have to take my word for a 'log identity'.  If you are into logarithms, you will understand why the identity holds but if not, don't sweat it.  It is true.  This identity is:

logabc = clogab.  Incidentally, the inverse of the left side of this formula is ac =b.  That may give you a clue why the identity works.

In words:   log to the base 'a' of 'b' raised to the 'c'th power equals c times the log to the base a of b.

So let's start.  I want to end up with a formula for each of the terms, in turn, on the left side of the equation.

The original equation is

At = Ao(1/2)t/h

Let's divide each side by (1/2)t/h.  Note that this cancels out the (1/2)t/h on the right side and leaves it on the left in the denominator*.  It is more conventional to have the subject of the formula on the left so we will exchange them.  After all if 7 = 3+4 then 3+4 = 7.  Our formula then becomes

* The bottom part of a fraction.

Ao = At divided by (1/2)t/h. Don't know how to get my computer to write this so I will leave you to write it down on a piece of paper.

So what is this formula good for.  It was noted early on in the use of carbon dating that there were some discrepancies.  With artifacts for which the exact date was known, the Carbon date did not agree.  The hypothesis was that the rate of 14C production in the upper atmosphere might not have been constant over the years.  So cores were drilled into very old trees, the rings were separated and carbon dated.  The above formula was used to work out the concentration  of carbon 14 which had been present for each year  that a ring was laid down.  And indeed it was found that the true curve diverged by a small but significant amount over time from the theoretical curve.  When the true curve was used, the dates all fell into place.

Now let's work on t and h.  The first thing I will do is to divide both sides by Ao.  This cancels Ao on the right side and leaves us with

At/Ao = (1/2)t/h

Now I'll take the log of both sides

log (At/Ao) = log[(1/2)t/h]

Remember our identity.  I can take t/h to the front of the right side so

log(At/Ao) = t/h(log1/2)

Now it is simple.  I simply divide both sides by log1/2 and we have t/h by themselves on the right side.  You take it from here.  Isolate t and h.  If you do it right you will find that

t = [hlog(At/Ao]/[log(1/2)]


h = [tlog1/2}/[log(At/Ao}

How about the formula for t.  This is pretty obvious.  Now that we have the needed correction of the production of 14C over the past , we can date any object that was once alive up to about 50,000 years.  This is carbon dating.

How about h.  We can't actually wait around for 5730 years to see when we have half of a quantity of radioactive carbon left.  We can, thought, observe the rate of disintegration on a shorter time span.  Using the h formula we can work out the half life of each radioactive isotope and some of them are multi millions of years.

It is never that easy

There are always complications.  Charcoal, for instance, if it is in ordinary soils or even in a cave can be colonized by micro-organisms.  If in active soil, the micro-organisms will have a modern carbon signature.  One has to first clean the charcoal of the modern material in order to get the correct date for the charcoal

Add to that, that we have been spewing carbon into the atmosphere from fossil fuel.  This is old carbon and hence contains no Carbon 14.  On the other side we have had nuclear tests in the air.  They have added Carbon 14 to the air.  For future anthropologists, they will have to take this into account.

Other types of radioactive dating have their own special requirements.  For instance when a rock melt cools, crystals form and just as a solution of salt and sugar, as it crystallizes, will  produce crystals of pure salt and pure sugar, the  crystals in a melt are of one type of molecule.  If one of these is a radioactive species and it's end product is known you can measure the concentraton of both and calculate when the rock  was melted. 

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