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Probability of decay of a nucleus

  • #1
Krushnaraj Pandya
Gold Member
697
71

Homework Statement


I want to know why probability disintegration per second of a radioactive nucleus does not depend on time lived by it.

Homework Equations


N/N(initial)=e^(-λt)

The Attempt at a Solution


According to the above equation, the probability should increase with the passage of time...I'd really appreciate some help, thank you
 

Answers and Replies

  • #2
dRic2
Gold Member
637
122
According to the above equation, the probability should increase with the passage of time
No.

##N = ## number of nucleus (still "alive").

##\lambda = ## the probability that a nucleus decays in the time interval ##\Delta t## between the instant ##t## and ##t + \Delta t##

##\Delta N## the change in the number of nucleus.

So ##\Delta N = -\lambda N \Delta t## means that the change in the number of nucleus (the number of decays) is equal to the probability that a nucleus decays in the time interval ##\Delta t## times the total number of nucleus "alive" (seems reasonable).

If you take the limit ##\Delta t → 0## you get a very simple differential equation and the solution is the one you provided.

This equation
$$ N = N_0 \exp(-\lambda t)$$
shows that the population of nucleus decrease exponentially, not that the probability of decay does.
 
  • #3
Krushnaraj Pandya
Gold Member
697
71
No.

##N = ## number of nucleus (still "alive").

##\lambda = ## the probability that a nucleus decays in the time interval ##\Delta t## between the instant ##t## and ##t + \Delta t##

##\Delta N## the change in the number of nucleus.

So ##\Delta N = -\lambda N \Delta t## means that the change in the number of nucleus (the number of decays) is equal to the probability that a nucleus decays in the time interval ##\Delta t## times the total number of nucleus "alive" (seems reasonable).

If you take the limit ##\Delta t → 0## you get a very simple differential equation and the solution is the one you provided.

This equation
$$ N = N_0 \exp(-\lambda t)$$
shows that the population of nucleus decrease exponentially, not that the probability of decay does.
I think I'm missing a subtle difference, the probability of nucleus to decay in one mean life is {1-[N/N(initial)]}= 1-(1/e) and to decay in two mean lives is 1-(1/e^2) from the same equation I mentioned- clearly the probability seems to change with time
 
  • #4
dRic2
Gold Member
637
122
My bad, I was wrong. I think the confusion comes from my wrong use of words.

The law of decay says that "the probability per unit time that a nucleus decay is a constant and that constant is called ##\lambda##". It just says that ##\lambda## is a constant.
The probability that any nucleus will decay is time dependent as you said because it is represented by the ratio ##N/N_0## which is a function of time.
 
Last edited:
  • #6
Krushnaraj Pandya
Gold Member
697
71
Ah, just what I needed- thank you very much
PS: I went back and checked on my book.

Here's some scans from my book. Hope it will help you
Just what I needed, thank you very much :D
btw is that an undergraduate level textbook, if so- which year is it in (as in 1st or 2nd year of coursework)
 
  • #7
Charles Link
Homework Helper
Insights Author
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It looks like it is already solved, but here's some additional inputs:
The equation ## \Delta N=-N \lambda \Delta t ## describes the system very well in a probability sense. Basically it says the probability of decay in time ## \Delta t =\lambda \Delta t ##. If we want to know the probability ## p ## that it survives for time ## t ##, that is ## p(t)=(1-\lambda \Delta t)^{t/\Delta t} ##. ## \\ ## One result that comes out of the calculus of the exponential function is ## e^x=(1+\frac{x}{N_1})^{N_1} ## as ## N_1 \rightarrow +\infty ##. ## \\ ## Now let ## \Delta t=\frac{1}{N_1} ## and let ## N_1 \rightarrow +\infty ##. ## \\ ## Then we have, (with ## x=-\lambda ##), ## p(t)=(1-\frac{\lambda}{N_1} )^{N_1 t}=e^{-\lambda t} ##. (I'm distinguishing ## N_1 ## from ## N ## here, because ## N_1=\frac{1}{\Delta t } ## allows ## \Delta t \rightarrow 0 ## as ## N_1 \rightarrow +\infty ##, while ## N ## represents the number of particles). ## \\ ## The average number of particles if ## N ## is large can be computed. If we start with ## N_o ## , we will have ## N(t)=N_o e^{-\lambda t} ## after time ## t ##.
 
  • #8
Krushnaraj Pandya
Gold Member
697
71
It looks like it is already solved, but here's some additional inputs:
The equation ## \Delta N=-N \lambda \Delta t ## describes the system very well in a probability sense. Basically it says the probability of decay in time ## \Delta t =\lambda \Delta t ##. If we want to know the probability ## p ## that it survives for time ## t ##, that is ## p(t)=(1-\lambda \Delta t)^{t/\Delta t} ##. One result that comes out of the calculus of the exponential function is ## e^x=(1+\frac{x}{N_1})^{N_1} ## as ## N_1 \rightarrow +\infty ##. ## \\ ## Now let ## \Delta t=\frac{1}{N_1} ## and let ## N_1 \rightarrow +\infty ##. ## \\ ## Then we have, (with ## x=-\lambda ##), ## p(t)=(1-\frac{\lambda}{N_1} )^{N_1 t}=e^{-\lambda t} ##. (I'm distinguishing ## N_1 ## from ## N ## here, because ## N_1=\frac{1}{\Delta t } ## allows ## \Delta t \rightarrow 0 ## as ## N_1 \rightarrow +\infty ##, while ## N ## represents the number of particles). ## \\ ## The average number of particles if ## N ## is large can be computed. If we start with ## N_o ## , we will have ## N(t)=N_o e^{-\lambda t} ## after time ## t ##.
Much clearer now, thank you :D
 
  • #9
dRic2
Gold Member
637
122
btw is that an undergraduate level textbook, if so- which year is it in (as in 1st or 2nd year of coursework)
This book is more for engineers than (I'm assuming) a physics undergrad. Anyway I'm using it right now in my first year as a grad student
 

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