Question

At t = 0 we begin to observe two identical radioactive nuclei that have a half-life of 5 min. At t = 1 min, one of the nuclei decays. Does that event increase or decrease the chance that the second nucleus will decay in the next 4 min ?

Answer 1

Are the nuclei related in any way? For instance, is one in a box on the moon and is the other in a box in Japan?

Answer 2

No further information is provided. I think we may assume they both are at the same place and part of one single sample...

Answer 3

\[\dfrac{dN}{dt} = -rN\]

Answer 4

I'd like to think that there are only two atoms of that radioactive nuclei in my room.
I'm wondering if the knowledge of the decay of one of them give any new info about the decay of the other atom... Because the decay rate depends on the size of the sample hmm

Answer 5

I would say that the the decay of one of the nuclei would have NO effect on the decay of the other. Isn't half life a probabilistic phenomenon? If you have two coins and one is flipped, does the fact that coin 1 turned up "heads" have any effect on how coin 2 will behave when flipped?

Answer 6

I'm pretty sure that decay is independent probabilistic events because they are indistinguishable particles.

I think it's tempting to say that because you had 2 particles and now you have half as many after one decays, then because of the definition of half life ( the amount of time it takes for approximately half of the sample to decay) we have issues. However I think that is sorta retro-actively trying to use the predictions of the probability.

We could go through the mathematical derivation of the formulas for half life and all that and see if it ends up somehow working out though maybe. At least we'd get a more precise idea of what's being said.

Answer 7

Ahh, that is very interesting! Thanks for bringing up the coin example, I'm still thinking...

Answer 8

Thanks ganeshie8 :-)

Answer 9

reminds me of shrodinger's cat, so there is no way to tell "when" the other nuclei decays xD

thank you both :)

Answer 10

the same way there is no way to tell when the next coin shows heads

Answer 11

u r welcome and thanks for the medals. This has to be the most "philosophical" discussion in which I participated in "openstudy.com" LOL

Answer 12

So if you solve the separable differential equation:
\[\frac{dN}{dt}=-rN\]
\[N=N_0e^{-rt}\]
Half life, \(t_{1/2}\) is the time it takes for your original amount \(N_0\) to decay by half to \(\frac{N_0}{2}\), so we can solve for r in terms of the half life,

\[\frac{N_0}{2} = N_0 e^{-rt_{1/2}}\]
\[r = \frac{\ln 2}{t_{1/2}}\]
Plugging in,

\[N=N_0 2^{-t/t_{1/2}}\]
Now we can start looking at our scenario when \(N_0=2\)
\[N=2*2^{-t/t_{1/2}}=2^{1-t/t_{1/2}}\]
Supposedly when \(t=1\) does this give rise to the same probability as if we started at \(N_0=1\) at \(t+1\)?

\[N'=1*2^{-(t+1)/t_{1/2}}\]

So they appear to give slightly different results since after you've observed a particle decay you have updated your knowledge of the situation, but the probability of events happening has really remained unchanged because they are independent events I think.

This sort of thing seems to be distinct from the 'Schrodinger's cat' sorta form of weirdness though.

Answer 13

I'm fairly certain that this sort of reasoning is improper since you're looking at decay probability as a continuous quantity in your differential equation and then you're turning around and saying "wait it was actually discrete" by talking about 2 particles in which case your probability distribution is broken. You're trying to play "law of large numbers" sorta reasoning on only a single case of 2 things. I think it's not a consistent use of the model.

Answer 14

yeah the "decay rate" depends on the size of the sample
\[N' =N_0e^{-rt}\]

eventhough the rate at which the decay happens decreases with the decrease in the size of the sample, the probability of decay remains constant ?

Answer 15

The point of Schrodinger's cat is that you're literally saying that the cat is in a superposition of being simultaneously alive and dead until you open the box.

On the other hand, in this scenario you've presented we're observing the particles the whole time and the probability distribution over time will average out to this exponential decay if you plot Particle Decays vs Time regardless of when the particles are ejected and is just a fundamentally random as a result of independent decay.

Even though the term "half life" seems to convey the idea that the decay rate somehow depends on the total mass, I think this is a misconception. It just happens that if you have more substance, more of it decays and the independent probabilities can be added.

To show this, look at if you had two independent piles of radioactive metals,

\[N= N_0 2^{-t/t_{1/2}}\]\[M= M_0 2^{-t/t_{1/2}}\]
We expect that since particle decay is entirely independent of whether the particles are part of one metal or two different metals, that the events remain the same, so what happens if we had started with the combined mass \(M_0+N_0\)? Surely we should get \(M+N\). So we prove this:

\[(M_0+N_0)2^{-t/t_{1/2}} = M_0 2^{-t/t_{1/2}}+ N_0 2^{-t/t_{1/2}} = M+N\]

Nothing weird. The separation of the molecules is COMPLETELY unimportant! There's no quantum weirdness here.

Answer 16

This means the distribution of particle decay vs time is independent of initial mass.

Answer 17

I think I have a cute way of explaining this.

Imagine you have a bunch of guys flipping coins in a competition and if you flip heads, you're out.

If you start with 16 guys and they all flip a coin, then you expect half of them to get eliminated leaving 8 guys in the next round.

Now you expect half of these guys to get out in the next round, etc...

Same idea for decay.

Answer 18

In fact you can see how the "competition" doesn't even have to be held in the same room on the same day either, you can just have all the guys flip coins until they get heads and then whoever got the longest string of tails wins since they're all independent events. This is analogous to separating the mass into separate boxes! Makes no difference.

Answer 19

This is really getting interesting...
suppose i collect all the radium in the universe and watch it decay till only 10 particles are left. Then, since the rate of decay has decreased, I can tell that the probability for one of these to decay in the next second is lower, compared to the probability in the start of decay process...

Answer 20

What's wrong with my above reasoning ?

Answer 21

The rate of decay is a constant.

Answer 22

The rate of decay decreases as the amount of remaining material decreases right ?

Answer 23

Half life is the rate of decay, and it is constant. I think the problem is there's a bit of confusing terminology going on.