# Probability of Thermoemission

1. Apr 14, 2010

### irycio

Hi! We'd been thinking with a bunch of friends for a couple of days, but eventually came up with nothing. Hence my question.
The exercise seems to be quite simple:

Given that the probability of one electron being (thermo)emitted form a surface of a metal in an inifinitively short period of time is A*dt, A being constant, and that each two emissions are statistically independant, calculate the probability of n electrons being emitted over a period of time t.

First idea was just to integrate A*dt from 0 to t to get the probability of one electron being emitted, but with A being constant one would eventually end up with the probability >1, which is total rubbish. Unfortunately, no other ideas showed up since tuesday ;). I myself would definityely expect a probability denisty function that would asymptotically drift towards a value of one, never to reach it.
However, the other idea that also makes sense to me is the probability being constant for the whole time, with the value of A. In example, if the probability of winning in a lottery is 1/700, then it doesn't matter how many times you try, it will always remain the same (playing 4 times gives you 4 chances over 2800, which is 1/700 again). Having said that, though, I don't understand the "dt" part :).

Certainly, the probability of n electrons being emitted is p^n, p being the probability of one electron being emitted.

2. Apr 14, 2010

### nickjer

3. Apr 15, 2010

Yeah, I was considering it as well, but ended up having no idea, what is the correlation between the probability i'm given (A*dt) and the number of expected emissions $$\Lambda [\tex] Poisson distribution expects me to provide :(. 4. Apr 15, 2010 ### nickjer If you are looking at the Poisson process page I gave you then in the very first equation they show you, A would be their lambda. Since it is the expected number of processes per unit time. 5. Apr 15, 2010 ### irycio I'm not really sure whether you're right. I do like the idea of Poisson process being used, but I'm not fairly concerned if A I'm given is their lambda. Maybe it is, but I just don't see the relation between probability of one electron being emitted and the average number of them being, again, emitted. Could you, or anyone else, clarify this to me, please? 6. Apr 15, 2010 ### nickjer The average of the poisson process is [tex]\lambda t$$. It is the same for your process. Since you are told the probability of a single electron being emitted is $$A dt$$, then given a time $$t$$, you would expect $$A t$$ electrons emitted.

So if A = 0.01 emissions/sec. Then in 200 seconds you would expect 2 emissions.

You can even derive the Poisson distribution from the binomial distribution. If you treat the probability of tossing a heads to be a very small probability, and do many many coin tosses. Then you can treat each coin toss as a unit of time, and the probability of tossing a heads as the probability per unit time. Using the binomial distribution of a coin toss you can derive the poisson distribution.

Also, check out: