# Probability of decay of a nucleus

• Krushnaraj Pandya
In summary: 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

## 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

Krushnaraj Pandya said:
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.

dRic2 said:
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

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:
PS: I went back and checked on my book.

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

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Ah, just what I needed- thank you very much
dRic2 said:
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)

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 ##.

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

Krushnaraj Pandya said:
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

## 1. What is the probability of a nucleus decaying?

The probability of a nucleus decaying is unpredictable and depends on various factors such as the type of nucleus, its energy state, and the surrounding environment. It is typically expressed as a half-life, which is the time it takes for half of the nuclei in a sample to decay.

## 2. How is the probability of decay determined?

The probability of decay is determined by the decay constant, which is a measure of how many nuclei in a sample decay per unit time. It is calculated by measuring the number of decays that occur in a given time period and dividing it by the total number of nuclei in the sample.

## 3. Can the probability of decay be changed?

The probability of decay cannot be changed or manipulated by any external factors. It is an inherent property of a nucleus and remains constant throughout its lifetime.

## 4. What affects the probability of decay?

The probability of decay can be influenced by the type of nucleus, its energy state, and the presence of external factors such as temperature and pressure. It can also be affected by the presence of other particles, such as neutrinos, which can trigger or inhibit decay.

## 5. Is the probability of decay the same for all types of nuclei?

No, the probability of decay varies for different types of nuclei. It depends on the stability of the nucleus, which is determined by the ratio of protons to neutrons. Generally, heavier and more unstable nuclei have a higher probability of decay compared to lighter and more stable nuclei.

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