{ Special Intro Note:

OK, this post has been a pain to transfer from my original writing into WordPress. A lot of the complication originated from the use of super- and sub-scripts in the mathematics, it’s also meant that I got pretty irritated in reworking it so many times that now I haven’t gone back to ensure that the math is correct anymore.

I’m saying all this because I want to say, “don’t take my word for anything here if you believe I’ve made a mistake, let me know and I’ll check it out when I have less frustrated eyes.”}

}

Exponential decay is a concept integral to determining the age of natural substances when initial and final concentrations are known for a specific unstable radionuclide. There are largely stable ratios of individual elements and their isotopes in the environment. Some of these isotopes are stable, and some are not. Unstable isotopes will break down over time. Although one could never predict how long any given atom will exist before it breaks down, the extraordinary number of atoms in any sample and the predictability of their decay into new atoms allows us to compute how long a substance has existed since new atoms were being incorporated (i.e. the time of death). By looking at these ratios over time, we can make very accurate measures of the age of a sample.

Different decay rates of various isotopes provide an array of measuring sticks for us to use.

Some examples:

^{32}**P** has a short **half-life of 14.29 days **and therefore has to be made synthetically for lab use.

^{35}**S** is formed from cosmic ray spallation of ^{40}Ar in the atmosphere. It has a **half-life of 87 days**.

^{14}**C** is formed cosmogenically by the reaction ^{14}N + ^{1}n → ^{14}C + ^{1}H. It has a **half-life of 5,730 +/- 40 years**.

^{40}**K** It has a **half-life of 1.248×10 ^{9} years**.

Some elements have a fairly straightforward decay path. For instance, Cosmic rays (high-energy protons and atomic nuclei) from outside of the solar system bombard the atmosphere striking atoms. When this occurs, the atoms are fractured into some Helium, some protons, and some neutrons. When these neutrons strike nitrogen ^{14}N it displaces a proton and is converted to ^{14}C. ^{14}C then decays back to ^{14}N by one neutron breaking down into a proton and electron which is emitted from the atom (beta decay) with a half-life of 5730 years (note that during formation, a neutron takes the place of a proton. Then in decay, the neutron breaks down resulting in a proton, which stays and the emission is an electron, which is why the atomic mass does not change from 14 in either direction).

### n + ^{14}N –> ^{14}C + p^{+}

^{14}C –> ^{14}N + e^{–}

**Speaking Mathematically…**

Exponential decay occurs in a general exponential function

In other words, as **x** increases, **f(x)** decreases and approaches zero. This is exactly the type of relation we want to describe half-life. In this case, we want a = ½, so that we have the relationship

**Rewrite in terms of half-life.** Of course, our function does not depend on generic variable **x**, but time, **t**.

- Simply replacing the variable doesn’t tell us everything, though. We still have to account for the actual half-life, which is, for our purposes, a constant.
- We could then add the half-life
**t**into the exponent, but we need to be careful about how we do this. Another property of exponential functions in physics is that the exponent must be dimensionless. Since we know that the amount of substance depends on time, we must then divide by the half-life, which is measured in units of time as well, to obtain a dimensionless quantity._{1/2} - Doing so also implies that
**t**and_{1/2}**t**be measured in the same units as well. As such, we obtain the function below.

**Incorporate initial amount.** Of course, our function f(t)f(t){\displaystyle f(t)} as it stands is only a relative function that measures the amount of substance left after a given time as a percentage of the initial amount. All we need to do is to add the initial quantity N0.{\displaystyle N_{0}.} N_{0}. Now, we have the formula for the half-life of a substance.

**Solve for half-life.** In principle, the above formula describes all the variables we need. But suppose we encountered an unknown radioactive substance. It is easy to directly measure the mass before and after an elapsed time, but not its half-life. So let’s express half-life in terms of the other measured (known) variables. Nothing new is being expressed by doing this; rather, it is a matter of convenience. Below, we walk through the process one step at a time.

Divide both sides by initial amount **N0**.

- Take the
**logarithm, base 1/2**of both sides. This brings down the exponent.

- Multiply both sides by
**t1/2**and divide both sides by the entire left side to solve for half-life. Since there are logarithms in the final expression, you’ll probably need a calculator to solve half-life problems.

**Example Problems **

- If you start with a sample of 600 radioactive nuclei, how many would remain un-decayed after 3 half lives?

- What is meant by ‘decay constant’?

**Warm-up Problem.**You receive a shipment of^{32}P in the lab on the first of the month. When it arrives, you perform an experiment using 10mL of this reagent. 57 days later, you wish to repeat this experiment using the same amount of**radioactive P**. About how many mL of your stock will you use (Don’t calculate this using the equations above, just work it out logically for an approximate answer)?

** **

** **

**Worked Examples **

- 300 g of an unknown radioactive substance decays to 112 g after 180 seconds. What is the half life of this substance?
**Solution:**we know the initial amount N0=300 g, final amount N=112 g, and elapsed time t=180 s.- Recall the half-life formula

### t_{1/2} = t / log_{1/2}(N(t)/N_{0})

- Half-life is already isolated, so simply substitute and evaluate.

### t1/2=180s log1/2(112g / 300g)

### = 127s

- Check to see if the solution makes sense. Since 112 g is less than half of 300 g, at least one half-life must have elapsed. Our answer checks out.

**2. **A nuclear reactor produces 20 kg of uranium-232. If the half-life of uranium-232 is about 70 years, how long will it take to decay to 0.1 kg?

**Solution:**We know the initial amount N0=20 kg,- Rewrite the half-life formula to solve for time.

### t=(t1/2)log1/2(N(t)N0)

- Substitute and evaluate.

### t=(70 years)log1/2(0.1 kg20 kg)≈535 years

- Remember to check your solution intuitively to see if it makes sense.

Of Note:

In sourcing this material (especially the maths), I came across something I did not expect. That is… a very good article explaining C14 dating in Answers In Genesis, a source that typically does not curate science in a remotely responsible manner. However, the article linked above does an excellent job in describing the steady-state production of C14 in the atmosphere and the process by which it is used to date carbon-containing remains.

In the end, Answers in Genesis quickly departs to an attack on a straw man, suggesting that only Carbon is used for dating the Earth, but this is (willfully?) mistaken in several ways.

“[B]ecause the half-life of carbon-14 is just 5,730 years, radiocarbon dating of materials containing * carbon yields dates of only *thousands

**, God’s eyewitness account of history. ” (my emphasis)**

*of years, not the dates over millions of years that conflict with the framework of earth history provided by the Bible*First, C14 is only used to date materials in which Carbon was incorporated (e.g. organisms) during life – i.e., it is not the way non-living material is dated. Second, C14’s 5730-year half-life allows dating of materials to approximately 40,000 years, at which point there is so little C14 remaining, that this method’s accuracy is reached. Further, at such low levels, background contamination from other sources (e.g. bacteria) compromises accuracy.

I think if I were to use an analogous argument: ‘*[memory]*** yields dates of only [dozens]**** of years, not the dates over [thousands] of years that conflict with [my notion of the universe beginning with my birth].’ **one might see the fallacy.

>Taken in part from: http://www.wikihow.com/Calculate-Half-Life