A radioactive sample loses atoms at a rate proportional to the number it has, which is the decay equation y′ = −k·y with k > 0. Its half-life T and its rate are tied by k·T = ln 2, so each converts to the other, rounded because ln 2 is irrational. A sample that has a quarter of its original amount left has been through two half-lives, and the lab prints both that age and the first step at which the exact Euler sequence has fallen that far.
Core ideas
Three ideas: what the sign convention is, how a half-life and a rate are the same information, and how a remaining fraction becomes an age.
Write y′ = −k·y with k > 0, so that k is the size of the rate and the minus sign carries the direction. The solution is y₀·e^(−kt) by the same separation as before. The lab takes the signed number, so the rate k = 3/25000 of this lesson is typed as −3/25000.
The sample has halved when e^(−kT) = 1/2, which is k·T = ln 2. Give either T or k and the other follows: T = ln 2/k and k = ln 2/T. Because ln 2 is irrational, a conversion from a measured half-life produces a rounded rate, and a rational rate produces a rounded half-life.
After m half-lives a fraction 1/2 multiplied m times is left. So a quarter means m = 2 and an eighth means m = 3. A fraction that is not a power of 1/2 falls between two of them, and its age needs a logarithm.
The material
The sign convention, the two conversions, and an age worked from a quarter of the original amount, with the lab's exact steps beside the rounded half-life.
Carbon dating rests on one observation: a radioactive atom has the same chance of decaying in any given year, whatever its age, so the number of atoms that decay in a year is proportional to the number still there. That is a rate proportional to the size, the equation of “Exponential Growth” with the direction reversed.
A quantity decays at rate constant k > 0 when y′ = −k·y. Its half-life T is the time in which any amount falls to half.
The sign is written outside k so that k can be read as a size. In the lab the rate box holds the signed coefficient, −k.
The solution is y = y₀·e^(−kt), by the same separation. It has halved when e^(−kT) = 1/2, and taking the natural logarithm of both sides gives −kT = −ln 2. That is the same relation as the doubling time of the previous lesson, with the same independence from the start.
A measured half-life for carbon-14 is commonly quoted as about 5730 years. Turning it into a rate gives ln 2/5730, which is irrational, and Euler's factor would then be rounded. The lab instead uses the fraction k = 3/25000 per year, which is 0.00012: a rate that keeps every Euler step exact, and a half-life that prints as ≈ 5776.23, close to the measured one and not equal to it. That difference is the cost of choosing a rational rate, and it is stated here so that the lab's age is not read as a measurement.
If a fraction f of the original is left, then e^(−kt) = f, and the age is t = ln(1/f)/k. For the fractions that are powers of one half this needs no new calculation, because each halving costs one T.
The last line is a check that needs no logarithm: it brackets the age before any rounded figure is computed, and a rounded answer outside the bracket is wrong.
With a step of one century the Euler factor is 1 − 100·(3/25000) = 247/250, an exact fraction, and the sequence (247/250)ⁿ is geometric as before. The lab prints the first step at which it has fallen to one half or below. The step is a whole number of centuries, so it lands near the half-life ≈ 5776.23 on the grid and not on it.
Switch the lab to the second preset, whose target is 1/4 and whose step is 200 years. The half-life tile still reads ≈ 5776.23, because that tile depends on the rate and not on the target. The age of a sample with a quarter left is twice it, ≈ 11552.5 years, a figure the reader computes from the tile and the lab does not print. The step tile gives the grid's version of the same crossing: step 58 (t = 11600), which is within one 200-year step of ≈ 11552.5.
The third preset is the same arithmetic with round numbers: a rate of 1/4 per unit time, a start of 100 and a target of 50. The factor is 3/4; (3/4)² = 9/16 is still above one half and (3/4)³ = 27/64 is below it, so the lab prints step 3 (t = 3) while the half-life tile reads ≈ 2.77259, which is 4·ln 2, rounded.
What dating assumes should be said as plainly as what it computes. It assumes the sample began with a known amount, that nothing entered or left it except by decay, and that the rate constant has not changed. The lab computes what follows from those assumptions; whether they hold for a particular sample is not something it can tell you.
Interactive
A geometric sequence beside an exponential
Method
Five moves. The bracket in the third is the one that catches an arithmetic slip before it is reported.
Decay is y′ = −k·y with k > 0. If you are given a half-life T, the rate is k = ln 2/T, rounded; if you are given k, the half-life is T = ln 2/k.
Divide what is left now by what was there at the start. A quarter is 1/4, fifteen percent is 15/100.
Find the powers of 1/2 on either side of the fraction. A quarter is exactly two half-lives; fifteen percent lies between two and three.
For a power of one half the age is the number of half-lives times T. For any other fraction it is ln(1/f)/k, rounded and marked ≈.
Name the rate constant, the starting amount and the closed system. The age is exactly as good as those three.
The rate is the lab's k = 3/25000 per year. The exact lines are fractions, and the lines with a logarithm in them are rounded and say so.
The exact sequence (122/125)ⁿ is a geometric sequence that never reaches zero, and the lab prints the first step at which it is at or below 1/4. That step is a whole number of 200-year steps, so it sits within one step of the true age and is not the true age.
Had the sample held fifteen percent, the bracket from the table says its age lies between 2T and 3T, which is between ≈ 11552.5 and ≈ 17328.7 years. A computed age outside that range would be a slip.
You should be able to write decay as y′ = −k·y, derive k·T = ln 2, convert in both directions with the result marked rounded, bracket an age between two half-lives, and say what the dating assumes.
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