# Probability of decay of a nucleus

Gold Member

## 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

dRic2
Gold Member
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.

Gold Member
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

dRic2
Gold Member
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:
dRic2
Gold Member
PS: I went back and checked on my book.

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Gold Member
Ah, just what I needed- thank you very much
PS: I went back and checked on my book.

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)

Homework Helper
Gold Member
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 ##.

Gold Member
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