# Why does radioactive matter decay in series of half-lives?

I was studying types of nuclear decay, and I came across the concept of "half-life". And I started wondering why does nuclear matter decay in half-lives? In other words, why does the rate of nuclear decay decrease as a function of time?

Is it because as more atoms decay, the surface area of the radioactive matter decreases, and the number of atoms "exposed" decrease? Does the outer layer of atoms somehow "shield" the inner atoms and stabilize them, preventing them from decay?

Or is there another mechanism at play?

• But the rate stays the same over time (the foundation of radiocarbon dating). After $t_{1/2}$, half of what you started with is gone. At $2t_{1/2}$, you are at half of what you had after $t_{1/2}$ (so a quarter compared to $t = 0$). The rate is the same: 50% of what you have right now is gone after another $t_{1/2}$. Radioactive decay is spontaneous and statistical, there is no shielding etc. – TAR86 Jan 11 '20 at 22:14
• The decay is an intrinsic property of an atom so it does not matter if they are in rock or a vacuum they still decay. The rate at time $t$ is proportional to the amount unreacted $dx/dt$ where $x$ is the number, and changes in time as $dx/dt=-kx$ where $k$ is a number specific to each type of atom called the rate constant or given as $\tau = 1/k$ the decay lifetime.The minus sign means the species decay. The half life is a measure of decay similar to the lifetime, the lifetime measures time to $1/e$ of the initial amount, the half life to 0.5. – porphyrin Jan 12 '20 at 10:14

There are fewer decays because there are fewer atoms to decay

The simple reason why the number of decays (strictly, the number of decays per unit time) decreases in simple radioactive decay is because there are fewer atoms left to decay.

Nuclear decay is probabilistic. The probability of any given unstable atom decaying is constant (independent of time or the environment). So the reaction rate (the total number of decays per second) per mole is constant. If this were a chemical reaction we would call it a first-order reaction (since it isn't a chemical phenomenon we can call it a first order process but the mathematics are the same).

In any first order process there is a characteristic time when half the original material will have gone (we call this the half life of the process). So, if we measure time in half-lives at t=1 we will have, say, 1 mole of the material; by t=1 we will have 0.5 moles; at t=2 we will have 0.25 moles of material. If we count the number of emissions per second they will follow the amount of material left. So the radioactivity will decay at the same rate as the material.

We don't need to invoke any interactions between the atoms of the material to explain this.

PS it gets a little more complicated if one radioactive compound decays into another. Then the total radioactivity will depend on two competing decays with different half-lives (but the radioactive emissions may well be different as each has a characteristic energy and expensive instruments can tell them apart giving two separate emission counts). But the mathematics remains the same.

It is a general principle, not limited to nuclear chemistry, but is common for many areas, e.g. for the reaction kinetic of the 1st order.

All processes, where the value time rate is proportional to the value, have value time evolution in the form of the exponential function.

$$\frac {\mathrm{d}x}{\mathrm{d}t}= -k \cdot x$$

$$x= x_0 \cdot \exp { (-k \cdot t )}$$

Radioactive atomic kernels have constant probability of the decay over the given time interval. The actual decay rate is then for large kernel numbers proportional to this probability and the number of these kernels, described by the above equations in the differential and integral form.

When there is twice as many radioactive atoms of the same kind there is twice as high decay rate.

The shape of the exponential function shows its value decreases to a half by constant time, no matter what the value is. This constant time is the decay half life.

where

$$t_{1/2}=\frac {\ln{2}}{k}$$

• I upvoted this nice answer! Any physical system, with one energy storage element and one decay pathway, must follow the ordinary differential equation, with initial condition, that you give. So it is ubiquitous in the real world: Newton's law of cooling, the RC low pass filter, radioactive decay, etc. – Ed V May 26 '20 at 2:18

## Decaying are like coin tossers

It helps me to compare two different mental pictures.

The first picture, which is not at all like radioactive decay, is of lighting a fuse. It starts burning at the near end and burns toward the far end at some number of millimeters per second. Each section's chance of burning depends completely on the section before it: the second millimeter of fuse can't burn before the first one, and the third can't burn before the second, etc.

The second picture is a lottery based on coin flips. A billion people each have a coin to toss. Everyone who gets tails sits down, and the rest proceed to the second round of play. The last person standing wins the prize. Each person's chance of getting tails is completely independent of anyone else's toss, and every toss has a 50% chance of tails. You can expect that roughly half the people will sit down in each round. The actual number will be closest to 50% when the number of people is still large.

Atoms of radioactive isotopes are like the coin tossers. Each atom will decay at some random time, completely unaffected by what the others are doing. Of course, many rounds of "the radioactive decay lottery" are played per second, so instead of a coin toss you might imagine the atoms rolling a die with very many sides and sitting down if they get a "1". The more stable atoms have die with more sides, but each atom's moment of decay is still unpredictable.

Another way of stating a half life might be this: "given how unstable this isotope is, how long would you have to watch a given atom to have a 50% chance of seeing it decay?"

## Just a repeated percentage change

I think a half-life pattern is just a convenient way of stating a recurring percentage decrease.

For example, what happens if you start with 1 million atoms and each year you lose 0.1% of whatever you have left?

I wrote a little computer program to simulate that. I had it run 10,000 rounds and print a message each time it reached a halving - that is, when the remaining amount reached 500,000 or less, then when it reached 250,000 or less, etc. Here's what I saw:

reached 499900.2346477281  at round 693  - 693 rounds elapsed
reached 249900.24460085342 at round 1386 - 693 rounds elapsed
reached 124925.1909144911  at round 2079 - 693 rounds elapsed
reached 62450.132251566276 at round 2772 - 693 rounds elapsed
reached 31218.83576633967  at round 3465 - 693 rounds elapsed
reached 15606.30332502206  at round 4158 - 693 rounds elapsed
reached 7801.594694162155  at round 4851 - 693 rounds elapsed
reached 3900.019018238128  at round 5544 - 693 rounds elapsed
reached 1949.6204223478458 at round 6237 - 693 rounds elapsed
reached 974.6157066056886  at round 6930 - 693 rounds elapsed
reached 487.2106204235443  at round 7623 - 693 rounds elapsed
reached 243.55670347259502 at round 8316 - 693 rounds elapsed
reached 121.75405321597741 at round 9009 - 693 rounds elapsed
reached 60.864879771978956 at round 9702 - 693 rounds elapsed
...


So every 693 rounds, the amount halves. This remains consistent after 10,000 rounds and after 1,000,000 rounds. (Eventually it breaks down, probably because the program has limited mathematical precision.)

If I have it lose 2% each round instead of 0.1%, the half-life is 35 rounds.

Based on this, I think you could state any half life in terms of percentage loss over a period of time. But "percentage loss per year" would hardly make sense for isotopes whose half life is a fraction of a second, and "percentage loss per second" would be incredibly small for isotopes with a half life in the billions of years. Speaking in terms of half lives gives us a single number to compare between those two isotopes.