# Thread: Poisson Distribution - Particle detector

1. ## Poisson Distribution - Particle detector

Hello everyone,

I have a problem solving the following question:

"While a particle detector is recording the passage of a particle, it is “dead” — i.e., insensitive to the
passage of further particles. This can become a problem when rates are high. Assume that a
counter has a dead time of 200 ns, and is exposed to a beam of 1 × 10^6 particles per second,
so the mean number of particles hitting the detector in a 200 ns window is $\mu$ = 0.2. Find the
average number of particles recorded in a 200-ns period (hint: no particles are recorded if zero
pass through; otherwise one particle is recorded), and hence the efficiency of the counter."

My initial attempt was a failure (I've attached it in a .pdf file)

Could someone please explain to me what I have done wrong?

Question.pdf

2. ## Re: Poisson Distribution - Particle detector

the average will be given by

$E[X] = \sum \limits_{k=0}^\infty~(\text{# particles recorded | k particles emitted})P[\text{k particles emitted}]$

since we record only 1 particle in all cases except that of $k=0$ this becomes

$E[X] = \sum \limits_{k=1}^\infty~P[\text{k particles emitted}] = 1 - P[0] \approx 0.1813$

This seems like a case of having your detector window too short.

3. ## Re: Poisson Distribution - Particle detector

Hi Romsek,

Could you help me understand what was wrong with my original method?
I've consulted a couple tutors and they said they don't know what was incorrect.

Thank you.

4. ## Re: Poisson Distribution - Particle detector

Originally Posted by mbary
Hi Romsek,

Could you help me understand what was wrong with my original method?
I've consulted a couple tutors and they said they don't know what was incorrect.

Thank you.
A glance at your answer tells me you don't really understand the Poisson distribution, or maybe even probability distributions at all.

You write

"The Poisson Distribution is expressed as the following formula: $\sum_{n=1}^\infty~\dfrac{\lambda^x}{x!}*e^{-\lambda}$

Putting aside the mismatch between $n$ and $x$ the distribution isn't expressed as that formula. All that formula gets you is $1-e^{-\lambda}$ and says nothing about what the probability of some number of events is.

Then you use an incorrect formula for finding the mean.

Sorry to be harsh but in short your answer doesn't make it look like you have a good handle on what you're doing.

,

6. ## Re: Poisson Distribution - Particle detector

You're completely right, it is a completely new concept to me and I'm tying to figure it out by myself, failing miserably as you've pointed out.
English isn't my native language either so I could have potentially written it wrongly.

1−e

7. ## Re: Poisson Distribution - Particle detector

Originally Posted by mbary
You're completely right, it is a completely new concept to me and I'm tying to figure it out by myself, failing miserably as you've pointed out.
English isn't my native language either so I could have potentially written it wrongly.

1−e

yes but that's entirely coincidence.

let me ask you these 2 questions

1) When I say that say N is a random variable that is Poisson distributed. What does that say? What is N in general?

2) What is the probability P[N=n] ? The formula for it. You've already sort of written it.

8. ## Re: Poisson Distribution - Particle detector

It means that N is the number of times an event occurs in an interval at a constant rate, without the events affecting each other? So it's the count of occurrences?

I am not entirely sure regarding the second question? Is it the formula where I mismatched x and n?

I'm sorry, but quite frankly I have not used advanced maths in years, apologies for my lack of knowledge.

9. ## Re: Poisson Distribution - Particle detector

You are right on the first question except that it is a constant average rate, not a constant rate. Events occur randomly.

Yes, N can take values that are the number of events that occurred in some given time.

So for example let's say I've got a piece of uranium giving off alpha particles. I could say the number of alpha particles given off in an hour is Poisson distributed with some rate.

For each number k, we can then calculate the probability that k particles were given off in that hour.

That gets us to the second question. What is that probability for a given k.

This is where the formula you wrote comes in

$P[\text{k particles were given off in an hr}] = P[k] = \dfrac{\lambda^k e^{-\lambda}}{k!}$

where $\lambda$ is the average number of particles given off in an hr.

The $k$ items are just events of any sort in general and units can be anything as long as $\lambda$ is appropriately chosen.

If you understand this post you should be able to go back to post #2 an understand what I've said there.

10. ## Re: Poisson Distribution - Particle detector

I believe I do?

You tried finding what is the probability of P[0], that is when NO particles are detected due to the 200 ns "dead time".

Am I correct?

11. ## Re: Poisson Distribution - Particle detector

Originally Posted by mbary
I believe I do?

You tried finding what is the probability of P[0], that is when NO particles are detected due to the 200 ns "dead time".

Am I correct?
it turns out that way but not for the reason you think.

We're trying to find the expected value of the number of particles that get recorded in that window.

Ordinarily this would be $\sum_{k=1}^\infty k P[k]$

but your detector doesn't work that way. It only records 1 particle regardless of how many appear in that window. So we end up with

$\sum_{k=1}^\infty 1 \cdot P[k]$

since the index starts at one, and we know that $\sum_{k=0}^\infty P[k] = 1$

we have that $\sum_{k=1}^\infty 1 \cdot P[k] = 1 - P[0]$