Probability of decay of a nucleus

[Krushnaraj Pandya]

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]

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.

 

[Krushnaraj Pandya]

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]

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.

 

[dRic2]

PS: I went back and checked on my book.

Here's some scans from my book. Hope it will help you

 

[Krushnaraj Pandya]

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)

 

[Charles Link]

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 $t$.

 

[Krushnaraj Pandya]

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

 

[dRic2]

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