Radioactive decay, relation between binomial to expon. dist

In summary: Basically, the formula ##P(k) = (\mu t)^k e^{-\mu t}/k!, \: k = 0, 1, 2, \ldots## is supposed to be the probability that ##k## decays occur in a time interval of length ##t##; here ##\mu = n \lambda## is the decay rate parameter when ##n## radioactive (un-decayed) particles are present and ##\lambda## is decay rate parameter for a single particle. That means that for a small time interval of length ##\Delta t > 0## a single un-decayed particle has decay probability of ##\lambda \Delta t + O((\Delta t)^2) \approx
  • #1
laserbuddha
3
0
You can model the probability for radioactive decay as a Poisson distribution. This is the probability for radioactive decay within a specific time interval. (I probably got some of it wrong here).

P(k,μ)=λ^k⋅exp(-μ)/k!

Is there a way to use this formula to derive the other formula for radioactive decay?

N(t)=N0⋅exp(-λt)

The only thing I've found, is all connected to differential equation, but nothing about how it is connected to Poisson distribution.
 
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  • #3
Vanadium 50 said:

Yeah, I've read it. Don't know if you get my question.What I'm wondering is, what is the reasoning (physics) or mathematical connection between these two formulas. I get the basic premise and reasoning for both formulas, but I don't get the relationship between the two. As I stated before if I start with

P(k,μ)=λ^k⋅exp(-μ)/k!

What is the reasoning for endup with

N(t)=N0⋅exp(-λt)
 
  • #4
laserbuddha said:
You can model the probability for radioactive decay as a Poisson distribution. This is the probability for radioactive decay within a specific time interval. (I probably got some of it wrong here).

P(k,μ)=λ^k⋅exp(-μ)/k!

Is there a way to use this formula to derive the other formula for radioactive decay?

N(t)=N0⋅exp(-λt)

The only thing I've found, is all connected to differential equation, but nothing about how it is connected to Poisson distribution.

The answer to your question AS WRITTEN is no, you cannot use one formula to derive the other. The reason is that they are describing different aspects of the situation.

The second one has no randomness in it; it gives the average number of particles remaining in a sample (where "average" = average in the statistical sense, which is the mean of a probability distribution).

What do you think the first formula represents? In other words, the quantity ##\lambda^k e^{-\mu}/k!## represents the probability of some event; what IS that event? What is the relation between ##\lambda## and ##\mu##, if any?
 
  • #5
Ray Vickson said:
...
What do you think the first formula represents? In other words, the quantity ##\lambda^k e^{-\mu}/k!## represents the probability of some event; what IS that event? What is the relation between ##\lambda## and ##\mu##, if any?

First of all, don't know if this is the right forum (someone moved this from the physics forum). So perhaps I need to give some background. Both functions are related to radioactive decay.

I've also written the first expression wrong, should be:
P(k,μ)=μ^k⋅exp(-μ)/k!

This is the probability for k atoms to decay, μ=np, n total number of atoms p probability of decay.

The second expressions is how many atoms are left, with regard to time. λ decay rate and N0 the number of atoms from start.

Ray Vickson said:
...
The second one has no randomness in it; it gives the average number of particles remaining in a sample (where "average" = average in the statistical sense, which is the mean of a probability distribution).

My speculation is that the two functions could be related somehow, e.g. something similar to how poisson distribution and exponential distribution are related.
From the first expression you get the number of atoms left, by e.g. calculating the number of decays for the mean. And this is for a specific time interval.
While the second function is related to the time between. Don't know if any of this makes sense. I'm just interested to see if someone has seen something similar about this.
 
  • #6
laserbuddha said:
First of all, don't know if this is the right forum (someone moved this from the physics forum). So perhaps I need to give some background. Both functions are related to radioactive decay.

I've also written the first expression wrong, should be:
P(k,μ)=μ^k⋅exp(-μ)/k!

This is the probability for k atoms to decay, μ=np, n total number of atoms p probability of decay.

The second expressions is how many atoms are left, with regard to time. λ decay rate and N0 the number of atoms from start.
My speculation is that the two functions could be related somehow, e.g. something similar to how poisson distribution and exponential distribution are related.
From the first expression you get the number of atoms left, by e.g. calculating the number of decays for the mean. And this is for a specific time interval.
While the second function is related to the time between.Don't know if any of this makes sense. I'm just interested to see if someone has seen something similar about this.

laserbuddha said:
First of all, don't know if this is the right forum (someone moved this from the physics forum). So perhaps I need to give some background. Both functions are related to radioactive decay.

I've also written the first expression wrong, should be:
P(k,μ)=μ^k⋅exp(-μ)/k!

This is the probability for k atoms to decay, μ=np, n total number of atoms p probability of decay.

The second expressions is how many atoms are left, with regard to time. λ decay rate and N0 the number of atoms from start.
My speculation is that the two functions could be related somehow, e.g. something similar to how poisson distribution and exponential distribution are related.
From the first expression you get the number of atoms left, by e.g. calculating the number of decays for the mean. And this is for a specific time interval.
While the second function is related to the time between.Don't know if any of this makes sense. I'm just interested to see if someone has seen something similar about this.

Basically, the formula ##P(k) = (\mu t)^k e^{-\mu t}/k!, \: k = 0, 1, 2, \ldots## is supposed to be the probability that ##k## decays occur in a time interval of length ##t##; here ##\mu = n \lambda## is the decay rate parameter when ##n## radioactive (un-decayed) particles are present and ##\lambda## is decay rate parameter for a single particle. That means that for a small time interval of length ##\Delta t > 0## a single un-decayed particle has decay probability of ##\lambda \Delta t + O((\Delta t)^2) \approx \lambda \Delta t## to first order in small ##\Delta t##. The probability that one of the ##n## particles decays during ##\Delta t## is ##n \lambda \Delta t## to first order in ##\Delta t##. By making ##\Delta t## small enough, the probability of more than one decay during ##\Delta t ## is negligible, it being of higher than first order in ##\Delta t##.

However, all that is merely an approximation; it may be a good one in many cases, but in principle is not exact. Why? Well, after the first decay, you will be left with ##n-1## particles, so the relevant decay rate for the next decay has a slightly smaller rate parameter ##(n-1) \lambda##. Then, after the second decay the new rate is ##(n-2) \lambda##, etc.

Working out the exact expression for the probability of ##N-k## decays in an interval of length ##t## has been done be several authors; an example is in
wwwhome.math.utwente.nl/~boucherierj/onderwijs/158052/158052sheetshc2.ppt
(which, unfortunately, a powerpoint slide show). The formula for the probability there are ##j## particles left at time ##t##, given ##N## particles at time 0, is given on slide 8 (which uses the symbol ##\mu## instead of what I called ##\lambda##). You will see that the distribution is NOT Poisson: it is binomial with parameters ##N## and ##p = e^{-\lambda t}## (I use ##\lambda## instead of the author's ##\mu##). So, in the large-##N## small ##\lambda t## limit, with ##\mu = \lambda t N## held constant, we obtain essentially the Poisson distribution for ##k=N-j## itself; that is, the probabilities for the number of decays follow pretty closely to a Poisson distribution! However, that is NOT EXACT, just very often a really good approximation (but sometime not so good at all).

What about the other issue: that of times? In a pure Poisson process with constant rate ##\mu## the times between successive events (individual decays) are independent and identically-distributed random variables with exponential distribution of rate ##\mu##. That means that If ##T_i## is the time between the ##i##th and ##(i+1)##th decay, then
[tex] P\{ t < T_i < t + \Delta t \} = \mu e^{- \mu t} \: \Delta t [/tex]
to first order in small ##\Delta t##. In other words, the probability density function of ##T_i## is ##f(t) = \mu e^{-\mu t} ## for ##t > 0## and is ##0## for ##t < 0##. The expected time to the next event is ##E(T_i) = 1/\mu##.

Why does that work? Well, look at the Poisson probability of no decay in time interval ##0 \longrightarrow t##; that is ##P_0(t) = e^{=\mu t}## (obtained by putting ##k = 0## in the formula ##(\mu t)^k e^{-\mu t}/k!##). However, 0 arrivals occur between times 0 and ##t## if and only if the first arrival occurs after ##t##, so
[tex] P_0(t) = P\{ T_i > t \} = e^{-\mu t}.[/tex]
That implies that the probability density function of ##T_i## is ##f(t) = -\frac{d}{dt} P \{ T_i > t \} = \mu e^{-\mu t}##. Voila! The exponential distribution!

To prove that the successive inter-event times are independent takes a bit more work and must look at some additional properties of Poisson distribution, but it all works out eventually.

However: none of that applies exactly to radioactive decay! Again, the reason is the variable rates involved. The time to the first decay is, indeed, exponentially distributed with rate ##\lambda n## and expected value ##1/ \lambda n##. However, after the first decay the time to the next decay is exponentially distributed with rate ##\lambda (n-1)##, etc. The two times are, indeed, independent, but are no longer identically distributed.

OK: I think I have said enough for now.
 
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What is radioactive decay?

Radioactive decay is the process by which an unstable atomic nucleus loses energy and emits radiation, eventually transforming into a more stable nucleus. This process is random and cannot be predicted, but it follows a specific pattern known as the decay rate.

How is the decay rate of a radioactive substance measured?

The decay rate of a radioactive substance is measured in terms of half-life, which is the time it takes for half of the initial amount of the substance to decay. This measurement can be used to determine the rate of decay and the remaining amount of the substance.

What is the relationship between binomial distribution and exponential distribution?

Binomial distribution and exponential distribution are two different probability distributions that are often used to model radioactive decay. The binomial distribution describes the probability of a certain number of successes or failures in a fixed number of trials, while the exponential distribution describes the probability of the time between events occurring.

How can the binomial distribution be used to model radioactive decay?

The binomial distribution can be used to model radioactive decay by considering each decay event as a "success" and the time between decay events as a "trial." This allows us to calculate the probability of a certain number of decays occurring in a specific amount of time.

Why is the exponential distribution considered a more accurate model for radioactive decay?

The exponential distribution is considered a more accurate model for radioactive decay because it takes into account the randomness of the process and the fact that each decay event is independent of previous events. This allows for a more realistic prediction of the decay rate and the remaining amount of the substance.

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