I understand that C14 decays at a given rate. I also interpret this to mean that 100% of the atoms of C14 in an object will all decay at the same right, individually.

So if I have 4 C14 atoms, will they all be gone in a full-life? or will two be 100% in a half-life while the other 2 are 0%?

Please explain how the decay of C14 works.

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If you like this question you might also enjoy reading physics.stackexchange.com/q/7584/2451 –  Qmechanic Sep 1 '11 at 15:53
possible duplicate of How a half-life works –  MichaelT yesterday
Half-life is just a matter of statistics. Decays are random and happen with a certain probability in a given time window. –  Brandon Enright yesterday

Half-life is used to describe exponential decay. What you're describing would be linear decay.

In one half-life period, on average, half of the C14 atoms would decay. So one would expect that if you start with four C14 atoms, you would after one half life have two, and after another half life only one would remain.

However, note that this process has a random component. You cannot predict exactly when an individual atom will decay. However once you have a larger number of atoms, you can make accurate predictions of how many will be left after a certain time period.

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Basically, nuclear disintegration is probabilistic in its very nature. What it means is that one cannot say with conviction that, say, one atom kept on a table will disintegrate in, say, the next 1 minute. All one can say is that among a given sample of, say, 100 nuclei, 10% of it will disintegrate in the next 1 minute.

Nuclear disintegration follows what is known as first order kinetics which means that rate of reaction is directly proportional to the quantity of reactant present. In other words,

$$d/dx(C) = -k C$$

where C is the current concentration of reactant and k is proportionality constant.

From this calculation, what one can get is a term called half-life, which means that after this time has elapsed, half of the concentration gets disintegrated (I'm using disintegrated and reacted interchangeably, since the reaction in nuclear disintegration is disintegration).

This means that a sample of 100 atoms after one half-life would remain 50 $=100 * (1/2)^1$, which after 2 half-lives would become 25 $=100 * (1/2)^2 = 50 * (1/2)^1$ and so on...

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