4 ms·
>If you have two atoms in a box, you must wait approximately half of the time to have a 50% chance to see the decay. (It's not exactly one half, there are some
by contravariant 8y ago
>If you have two atoms in a box, you must wait approximately half of the time to have a 50% chance to see the decay. (It's not exactly one half, there are some technical details here, but one half is a good approximation.)
Really? As far as I know radioactive decay is mostly believed to be a Poisson process (or rather the time it takes for a single atom is believed to be exponentially distributed), in which case the time it will take is exactly one half, on average.
- a1369209993 8y agoAre the two of you using the same answer to the question "Does a X% chance of 2 events (in given interval) contribute X% or 2X% to the chance of seeing one event?"? That is, do you mean [time for a 50% chance of seeing ≥1 decay] or [time to see, on average, 0.5 decays].
- contravariant 8y agoIn my case I'm talking about the (average) time until the first decay.
- fspeech 8y agoPoisson distribution is the limit of having infinite particles around.
- contravariant 8y agoIt is, although the only difference is that the rate of this particular process changes after the first decay. Which makes it more of a continuous time Markov process, but the general idea is similar.
- gus_massa 8y agoIf you put two atoms A and B in an box and wait 1/2 of the half life, then the probability that the atom A survives is 1/sqrt(2). The probability that the atom B survives is also 1/sqrt(2). So (assuming independence) the probability that both survive is exactly 1/2. So you are right! (I'd prefer to say that it's a property of the exponentially distribution, not the Poisson process.)