0
$\begingroup$

This question already has an answer here:

Why do radioactive isotopes have a half-life? I know that they decay in order to become stable but why would it take out enough subatomic particles to be half? Or am I approaching this question the wrong way: the remaining radioactive substance is going to be half in x amount of time and the other half will be stable atoms.

$\endgroup$

marked as duplicate by Ben Crowell, stafusa, sammy gerbil, rob May 21 '18 at 18:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ I can't understand what the question is. This seems to have been written in less than one minute, with no attempt to think it through. $\endgroup$ – Ben Crowell May 21 '18 at 3:35
  • $\begingroup$ @BenCrowell the OP appears to think that half-life means you have one-half of a particle left. $\endgroup$ – ZeroTheHero May 21 '18 at 5:52
  • $\begingroup$ Crossposted from chemistry: chemistry.stackexchange.com/questions/97261/… $\endgroup$ – Tyberius May 21 '18 at 17:53
  • $\begingroup$ @Tyberius good catch. IMO however this is more of a physics and a chemistry question so it's the CSE that should be closed. $\endgroup$ – ZeroTheHero May 21 '18 at 17:59
6
$\begingroup$

It is observed empirically that the rate at which the number of radioactive elements decreases is proportional to the size of the sample: $$ -\frac{dN}{dt}=k N\, , \qquad \Rightarrow \qquad N(t)=N_0e^{-kt} $$ By dimensional analysis, $k$ has units of inverse time. The half-life is the time $T$ after which the sample is one-half of its initial size: $$ N(T)= N_0e^{-kT}=\frac{1}{2}N_0\, . $$ This is a statistical law: it does not predict the moment at which individual particles decay, but is a statement about the aggregate sample. Basically, if you wait one minute, you will find on average $N(60)=N_0 e^{-k\times 60}$ particles in your sample.

The decay of your sample is still governed by the same constant $k$ (which is not so easy to compute from first principles as it involves quantum tunneling and some very delicate calculations): the remaining atoms in your sample are not “less radioactive”: because there are fewer of them, you will have to wait longer to register a decay.

The process is probabilistic and can be compared to playing the lottery. Suppose you have a lottery with one of 1000 different possible numbers on a ticket, and there is a draw every day. Obviously if you have 100000 players, you expect that, on the first day there will be roughly 100 winners.

On the second day, the winners cannot play again and everyone of the remaining 99900 or so players gets a new ticket. The odds have not changed, but this time you expect roughly 99.9 winners. Of course, you cannot have “fractional winners” so basically at the end of the second day you expect you will have something like 99 or 100 winners. It may be that you get a little more, maybe a little less, but statistically it should be close to 99 and 100.

On the third day, the winners of the first two days are taken out, and the remaining players get a new ticket and so on. The “half-life” here would be the amount of time after which you have 50000 players.

Obviously if you started with 200000 players rather than 100000, you’d expect to have around 200 winners on the first day, around 199.8 winner on the second etc. This doesn’t change the half-life: it would be the amount of time after which you have 100000 players, i.e. half of your original amount.

The “half-life” for a single player would be the amount of time this person has to wait so she/he has as 50% change of having been eliminated by winning. An individual player never knows if she/he will win, but might rightfully be impatient if she/he is “unlucky” and is still in the game well after several “half-lives”.

Now suppose you start with just 100 players, then you might not get any winners at all for the first few days... after all, since there are fewer players and the possible number of lottery number, the odds of finding the winning ticket amongst one player is 1/1000, so with 100 players the odds are just 100/1000=1/10. These are just odds so you might get 3 winners in the first 10 days... that’s unlikely but not statistically impossible. In this situation you’d have a hard time “experimentally” measuring the half life because the fluctuations are significant for a sample that is so small. Nevertheless, the concept still makes statistical sense and of course, the odds for an individual player have not changed: still 1/1000 if this player holds a single ticket.

$\endgroup$
2
$\begingroup$

A single radioactive nucleus has a constant decay rate--it is independent of time. Thus, there is is a fixed time over which is has a 50% chance of decaying (called the half life). Again, this time is independent of how long the nucleus has existed. If you dig up a 4,000,000 year old U-238 from Earth and compare it with some made in a reactor this morning--their decay probability per unit time is identical. Now if you have a macroscopic quantify of a pure isotope, on average half of it will have decayed after 1 half life of time passes. Of course there are statistical fluctuations, but since Avogadro's number is so much bigger than the square root of Avogadro's number, you can safely ignore them and say 1/2 has decayed.

$\endgroup$
0
$\begingroup$

Either the radioactive atom is there or it is not there. The concept of half-life means nothing when doing experiment on a single atom. One has to have a large number of atoms and half-life is defined as the time taken for half of the atoms to decay. Deacy process is statistical there is a probability for an atom to decay after a certain time. Eventually all will decay.

$\endgroup$
0
$\begingroup$

Radioactive isotopes are unstable decompose into other smaller nuclei but just because they are unstable all of them dont decompose at once. There is a statistical relation between the number of particles at any time 't' N = Ae^-kt where A is the number of particles at t = 0 and k is a constant. This being a statistical law it is valid only in large quantites. Now to measure how fast or slow the decomposition is we have different quantites we have different quanties. One of them is half time. HALF TIME is the amount of time in which the given number of particles get halved. That doesnt mean that the rest of them are stable. It just so happens that the rest of the particles are not decayed yet (It just wasnt their turn yet).

$\endgroup$
  • $\begingroup$ For certain isotopes, it is possible to get a large fraction of them to "decay at once" (meaning within a few microseconds). This is called a fission bomb. $\endgroup$ – JEB May 21 '18 at 2:44
0
$\begingroup$

Start with 1000 coins, and pretend they are all turned so that they are heads-up. Toss them up high and let them fall on the floor, then remove all that land tails-up. The tnumer of surviving coins will be approximately 500, because each coin has a 50% chance of being removed. Repeat, and the number will reduce to about 250, and so on. If you do this process once every year, you could say that the "heads" coins have a half-life of one year. Each coin has an equal chance of being removed in each cycle.

If you"re pretty good at high school algebra, you can generalize this principle to the continuous case. It's something like calculating interest that is compounded monthly, daily, hourly, -- and continuously.

That's what happens with radioactive atoms. In any given time interval there is a certain probability that each atom will decay. The half life is the time interval over which an atom has a 50% chance of decaying.

$\endgroup$

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