# Radioactive decay-Why does it depend on numbers left

As textbooks describe the rate law for radioactive decay as a first order reaction dN/dt=kN,why is it so(mine does not give the reason)? How is the radioactive decay of one atom depeendent on the other? The equation states that the rate of decay slows down as the number of non disintegrated nuclei decreases.How plausible is this. Each atom functions as an individual system with its decay a total certainty. How does its own decay impact the decay of other nuclei around it? In my opinion a nuclear decay of mass should not extend to infinite time. Could someone please clear this up? Thanks!

• The radioactive decay rate of any given atom is not dependent on the number present. The decay rate of the $N$ atoms depends on $N$, which makes intuitive sense. Indeed, what would the derivative $dN/dt$ mean in the context of a single atom? Mar 9, 2016 at 15:37
• Duplicate of physics.stackexchange.com/questions/30185/… and links therein. Mar 9, 2016 at 19:00

In my opinion a nuclear decay of mass should not extend to infinite time.

This is (more or less) the same as asking the following

If you have a million coins and toss each of them once a year, and throw out any that land on heads. How many years would the remainder last?

If you were to draw a graph of the number of coins remaining over time, it would look like an exponential decay and approach the time axis at a very low angle.

As you increase the number of coins (say trillions) the curve would extend further.

The mathematical curve that models the behaviour for arbitrarily large numbers would extend to infinity

So yes, as that mathematical curve drops near the order of magnitude of 1, your lump of plutonium will cease to contain any atoms of PU-239 - since you cant have a fraction of a plutonium atom. Nevertheless, the simple mathematical model extends as far right as you like and never reaches 0.

Why does it depend on numbers left?

Another way to look at this

At the beginning I have 1 trillion coins. the probability of each landing on heads that year is 50%. I expect to have about 500 billion land on heads (= decay) so my decay rate is 50%.

Many years later, I might have 1000 coins left. The probability of any coin landing heads remains 50%, it is not affected by how many coins are present or how many landed on heads in prior years. Yet I expect that year about 500 will land on heads (decay) - a rate of 50%.

However, 500 is a lot less than 500 billion.

The numbers decaying in a specific year depends not on the total number of coins but on the number left available for the coin-tossing process (coins that never landed on heads = undecayed atoms).

That is - the change in numbers of undecayed atoms in a period depends on the numbers of undecayed atoms at the start of the period. Hence the equation.

• Hah, that's funny - I started my answer in the morning, and didn't notice another answer had been posted until I posted mine. I guess we thought of the same analogy. Mar 9, 2016 at 18:08

Try this. Put 100 pennies in a cup. Toss them onto a table. Any that are tails up have "decayed" - remove them and put them aside. Put the remaining pennies in the cup. Repeat the process over and over.

In the first "unit of time" (the first round), your pennies should have decayed at a rate of approximately 50 pennies / unit of time, since there were 100 pennies, and around 50% of them decayed. The rate $dN/dt$ is proportional to the number there, and the decay rate: $$dN/dt = k N = 0.5 \cdot 100 = 50 ~\rm pennies~ /~ unit ~of ~time$$ In the second round, you started with approximately 50 pennies, and you should find that approximately 25 of them "decayed". Again, the rate that they decayed is proportional to the number that remained: $$dN/dt = k N = 0.5 \cdot 50 = 25 ~\rm pennies~ /~ unit ~of ~time$$ And so it is for radioactive nuclei. Each nucleus has a certain probability of decaying in a certain amount of time. Thus the overall rate of decay of a group of nuclei is proportional to the number that currently remain.