59
$\begingroup$

Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period?

$\endgroup$
  • 9
    $\begingroup$ The first two paragraphs here may help: en.wikipedia.org/wiki/Poisson_distribution $\endgroup$ – Not_Einstein Feb 16 at 18:44
  • $\begingroup$ Related: physics.stackexchange.com/q/102222/2451 and links therein. $\endgroup$ – Qmechanic Feb 16 at 19:05
  • 2
    $\begingroup$ RE: the OP mentions "by the same proportion" -- Well it isn't exactly the same proportion. If you start out with a billion radioactive atoms then after one half-life about one-half remain. There is a very low probability that exactly one-half remain. If you repeat the experiment many times you'd see that there is a distribution about 1/2 of the original number of atoms. $\endgroup$ – MaxW Feb 16 at 22:49
  • 2
    $\begingroup$ Actually, the half life represents the same proportion of substance because radioactive disintegrations follow first-order kinetics. In other words, the half life in independent of the initial concentration of substance. $\endgroup$ – Sam Feb 17 at 4:51
  • 16
    $\begingroup$ @user207421 Decay of the total amount of the substance is exponential precisely because decay of the individual atoms is random. $\endgroup$ – Vince O'Sullivan Feb 17 at 13:37
161
$\begingroup$

An example that might help:

Start with a big pile of coins. Flip them. Remove the heads. About half remain.

Take the remainder and flip them. Remove the heads. About half remain.

Take the remainder and flip them. Remove the heads. About half remain.

The analogy: An atom has a 50% chance of decaying in some particular interval $T_{1/2}$. After each of those intervals, half are left.

$\endgroup$
  • 3
    $\begingroup$ Thanks! It seems so simple now that you've explained it. $\endgroup$ – Saharsh Aanand Feb 16 at 18:49
  • 9
    $\begingroup$ The 'rule' you (@Saharsh) describe only works for large numbers. But atoms are so small that even microscopic amounts of atomic matter contain billions of atoms. $\endgroup$ – Gert Feb 16 at 18:58
  • 20
    $\begingroup$ @Saharsh I've heard of this demo being done at a university hall, with hundreds of students playing the part of radioactive atoms. Each student has their own wristwatch, but the watches are not synchronized. Each minute they flip a coin. Those that get heads leave the hall, the others stay behind and flip their coin again in the following minute. It's a simple but effective way to demonstrate how the half-life works. $\endgroup$ – PM 2Ring Feb 16 at 19:06
  • 13
    $\begingroup$ @jamesqf I read about that a while ago, before smartphones were common, and when wristwatches were more common. :) I suppose these days you could do it with each person using a stopwatch app on their phone. $\endgroup$ – PM 2Ring Feb 17 at 5:24
  • 7
    $\begingroup$ @user253751 Exactly, and we don't want that for this demo, which is why I suggested using a stopwatch app. Each person starts their stopwatch at a (relatively) random time. $\endgroup$ – PM 2Ring Feb 17 at 13:22
21
$\begingroup$

The word random in this context does not mean totally without order.
What it means is that one cannot predict exactly when a particular unstable nucleus will decay although there is an underlying probability of decay of an unstable nucleus in a specified interval of time.

An interval of time which is often used for a particular species of unstable nucleus is the half-life.
The probability that an unstable nucleus will decay in a time interval equal to one half life is $\frac 12$.
If an unstable nucleus does not decay in that time interval then the probability that it will decay during the next time interval of the same length is still $\frac 12$ . . . etc.

You will note that this is similar to the toss of a coin with two outcomes heads and tails each with a probability of $\frac 12$.
However an insight to the randomness of coin tossing might be shown in that although the probability of tossing a head is $\frac 12$, then if one tossed a coin $100$ times there is a fairly low probability, $0.07959$, of the outcome being exactly $50$ heads and $50$ tails.
So what you have is a number of possible outcomes, number of heads + number of tails $=100$, for which you can predict the probability of them happening but you cannot say for certain, probability $=1$, which of those outcomes will actually occur.

In the context of radioactive decay on average half a sample of unstable nuclei will decay in one half life and then on average half the remaining unstable nuclei will decay during the next interval of one half life etc.
With samples of billions and billions of unstable nuclei the statistical fluctuations about “one half will decay during a time interval of one half life” will be small.
As the number of unstable nuclei becomes smaller the statistical fluctuation about one half will get larger.
Just think about what you would predict about the decay of $3$ unstable nuclei in an interval of one half life. They could all not decay for ten half lives although the probability of the happening is small.

$\endgroup$
  • 1
    $\begingroup$ First section... "Random" means exactly that - we can't say for a single atom whether it will decay "now" - "in 5 years" - "in x seconds" - that is the random part.... The half life time is just statistical $\endgroup$ – eagle275 Feb 17 at 10:47
  • 1
    $\begingroup$ @eagle275 I have added a word to my second sentence. $\endgroup$ – Farcher Feb 17 at 13:13
  • 1
    $\begingroup$ And 3 nuclei will surely not decay to 1.5 nuclei :) $\endgroup$ – Carsten S Feb 17 at 15:30
  • $\begingroup$ @eagle275 Certainly not and that is why chose an odd number to illustrate that after one half life there may be 1 or 2 or 3 or even no unstable nuclei still left. $\endgroup$ – Farcher Feb 17 at 15:58
  • 1
    $\begingroup$ And then you want to involve quantum mechanics - that say as long as you watch the atom with great accuracy it will most likely not decay - jk $\endgroup$ – eagle275 Feb 17 at 16:21
8
$\begingroup$

You're asking the wrong question. There is no magical notion by which it decays by half. That is why "half-lives" vary so much. "Half-lives" are simply the chosen method of measurement. Your question is analogous to asking "why all cars travel in hourly increments?" just because we measure speed in km/h.

The notion implicit is using half-lives, is that given a large enough sample (1 mol = 6 E23) the rate of decay is close enough to constant. ie: That if say in any second, the "chance" of decay is X% for any atom, then over such a huge sample X will present as a constant. For example, if we said that a person having open-heart surgery has a 0.1% chance of dying on the table, we would not expect that to be accurate over a small sample. We couldn't say "well only 10 of us are having the operation today, so I'm safe." But over many trillions of such operations, we'd expect the 0.1% to hold true.

In summary, a half-life is simply a different way of presenting a constant rate of decay. (Keeping in mind that a constant RATE of decay, delivers progressively smaller decay as the amount of un-decayed material reduces.)

$\endgroup$
  • 1
    $\begingroup$ A constant rate of decay means that $\dot N=-k$ for some constant $k$. So then $N(t)=N_0-kt$. This isn't what radioactive decay looks like. I think you mean a constant rate of decay per molecule $\endgroup$ – Aaron Stevens Feb 18 at 4:30
  • 2
    $\begingroup$ I really dont understand the downvotes this answer is getting, it explains an aspect of the question that the other answers don't address. People downvoting should also add a comment explaining their opinion. +1 $\endgroup$ – WhiteMaple Feb 19 at 7:52
  • $\begingroup$ @WhiteMaple me neither. Upvoted in sympathy. May not be the best answer but it's not a wrong answer. $\endgroup$ – nigel222 Feb 19 at 16:45
0
$\begingroup$

It's a consequence of the fact that the nucleus doesn't know how many other nuclei are in your lump of material. A 3kg block of uranium has to decay at the same rate as three 1kg blocks of uranium. Which means a 1kg block has to decay at 1/3 of that rate. A 1kg block is the same as 3 1/3kg blocks, so a 1/3kg block has to decay at 1/3 of that rate too.

Now suppose you have a 3kg block of uranium (or whatever-ium), and it takes a year to decay into 1kg of whatever-ium (and 2kg of other stuff - let's pretend you have a system to take that away because we're only talking about the uranium here). Since you have 1kg, it must decay at 1/3 of the rate that it did at the start. It takes the same time to decay 2/3kg from a 1kg block as it does to decay 2kg from a 3kg block. That means after a year, you only have 1/3kg left. And decaying 2/9kg from a 1/3kg block takes the same time as decaying 2/3kg from a 1kg block. So after another year, you have 1/9kg left. And so on.

We say that whatever-ium has a third-life of one year.

We can extrapolate with maths. We know it has a ninth-life (1/3 squared) of two years. We know it has a 57.3%-life (square root of 1/3) of half a year. We know it has a half-life of 0.63092975357 years (you need to use logarithms to work this out).

We measure things in half-lives because it's convenient. We could equally well use third-lives or quarter-lives or fifth-lives or two-thirds-lives.

$\endgroup$
0
$\begingroup$

A couple of answers above hit it well. Here is a slightly different perspective.

From an visual standpoint, consider a pointillist painting. If you look at any single dot up close, the painting makes no sense. Stand back, and order falls into place.

The term “random” does not mean without order. It means that nothing we know up to this point with this particular perspective enables us to predict its function going forward in time.

$\endgroup$
0
$\begingroup$

You may view radioactive decay as a random process. The half-life period tends to some constant value. It is because of a large number of atoms. This is not just true for half-life period but for any ratio period.

Consider an example below:

We define $T_{1/10}$ : Time after which 1/10th of isotope mass left of the radioactive substance.

Initial number of atoms N = 10 atoms

After completing first $T_{1/10}$ period number of un-decayed atoms be $X$ and it would follow binomial distribution $Binom(N,p)$.

$X \sim Binom(10,0.1)$ and P(X) the probability for $X$ atoms remaining for each of $X$ can be given in the following table.

X    P(X)
0    0.3486784401
1    0.3874204890
2    0.1937102445
3    0.0573956280
4    0.0111602610
5    0.0014880348
6    0.0001377810
7    0.0000087480
8    0.0000003645
9    0.0000000090
10   0.0000000001

Expected value (mean) for number of remaining atoms is 1. If we want to estimate how many atoms would remain with 99% confidence then we get values to be from 0 to 4 atoms i.e. $P(0 \le X \le 3)$. These values are (-100% to 300%) deviation from the mean value. This means remaining mass can have -100% to 300% fluctuation from expected value.

With N = 1000 atoms, Expected number of remaining atoms = 100 Possible remaining atoms from 75 to 125 with probability $P(75 \le X \le 125) = 0.9928133 $ has deviation from expected value $\pm 25\% $

With N = 10000 atoms, Expected number of remaining atoms = 1000 Possible remaining atoms to make similar effect as above are from 920 to 1080 with probability $P(920 \le X \le 1080) = 0.9928133 $ has deviation from expected value $\pm 8\% $

With N = 100000 atoms, Expected number of remaining atoms = 10000 Possible remaining atoms to make similar effect as above are from 9745 to 10255 with probability $P(920 \le X \le 1080) = 0.9929232 $ has deviation from expected value $\pm 2.55\% $

Here you can observe degree at which deviation is decreasing as initial number of atoms are increased. This deviation is also equivalent to deviation in expected value of mass.

At larger scale values of N are very high which are of the order of $6.022×10^{23}$ for each mol. Hence not exactly the same proportion decays every time but the proportion value saturates to a fixed value in the participation of very large number of radioactive atoms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.