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Radioactive decay

  • Thread starter skrat
  • Start date
  • #1
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Homework Statement


Analyzing 1000 events (each event is one radioactive decay of an unknown sample), we notice that the time between two consecutive events is larger than 1 second in 30% of the cases while in 5% it is longer than 2 seconds. Can we, at 5% risk level deny the hypothesis, that the characteristic decay time is 1 second long?

Homework Equations




The Attempt at a Solution


I really don't know how to begin here. I really don't. I do assume that I will have to use either Student distribution or Chi-squared distribution. But I have no idea how to start and what to do :/

Please help!
 

Answers and Replies

  • #2
34,051
9,910
Which distribution does a radioactive decay have?
Which fraction of events do you expect to be larger than 1 second, and how likely is it to have 300 out of 1000 then?
 
  • #3
744
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Events should be normally distributed.
Ammm If I understood your question correctly: 300 events should take longer than 1 second. How likely? Am, I would guess with 95% probability.
 
  • #4
34,051
9,910
Events should be normally distributed.
The times of the events? Are you sure?
Ammm If I understood your question correctly: 300 events should take longer than 1 second. How likely? Am, I would guess with 95% probability.
300 is your observed number, I am asking for the prediction. And the 95% is something you will have to compare your result with later.
 
  • #5
744
8
Well radioactive decay is a completely stochastic process, knowing this I would say it is normally distributed around characteristic time. But I guess it isn't. My next option, which is a result of guessing, would be that the number of decays observed over a given time interval obeys Poission statistics (distribution). If that is the case, than I would highly appreciate a two sentenced explanation, if not than I am lost. :/

Again, I thought that since all events are stochastic and since measurements show that in 30% of the consecutive decays the time is longer than 1 second, it is also my prediction that 30% of number of decays over a given time interval will always be longer than 1 second.
 
  • #6
34,051
9,910
Do you know the concept of half-life? What does it suggest if there is a fixed time during which half of the remaining particles decay?
 
  • #7
744
8
Ok, I think I understand what I am supposed to do. Please check if this seems to be good.

##N=1000## is the number of decays. Experiment tells us that ##N(1 s<t<2 s)=300## and ##N(t>2 s)=50## therefore ##N(t<1 s)=650##.
Of course $$\frac{dP}{dt}=\frac 1 \tau e^{-\frac t \tau}$$ So the theory for first interval (##t<1s##) predicts $$F_1=\int _0^1\frac 1 \tau e^{-\frac t \tau}=1-e^{-1}=0.6321$$ and similary ##F_2=\int _1^2\frac 2 \tau e^{-\frac t \tau}=e^{-1}-e^{-2}=0.232## and ##F_3=\int _2^\infty\frac 2 \tau e^{-\frac t \tau}=e^{-2}=0.135##.

This now leaves me with $$\chi ^2 =\frac{(650-632)^2}{632}+\frac{(300-232)^2}{232}+\frac{(50-135)^2}{135}=73.9$$ while the data from the tables say that $$\chi ^2(3-1) ^{ 5 \text{%}}=5.9915 $$

So I guess the answer is no?
 
  • #8
34,051
9,910
I think the 30% "longer than one second" include the 5% "longer than 2 seconds".

The three values are not uncorrelated (because each event has to be in one of the classes) so you cannot add their ##\chi^2##-contributions like that, but the last term alone is sufficient to draw the same conclusion.
 

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