Calculate Probability of Decay Using Decay Constant ##\lambda##

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To calculate the probability of decay for a nucleus given its decay constant ##λ##, the relevant equation is derived from the exponential decay formula, where the probability of decay from time 0 to ##t## is expressed as ##P = 1 - e^{-\lambda t}##. The discussion emphasizes a frequentist approach, suggesting that the probability can be understood by observing a large number of identical nuclei and determining the fraction that decayed within the time interval. It is clarified that the decay probability is time-dependent and does not rely on external factors, as the decay process is inherently random and independent of the nucleus's age. The conversation also touches on the derivation of the exponential distribution of decay, highlighting the constant rate of decay characterized by the decay constant ##λ##. Understanding these concepts is essential for grasping the nature of radioactive decay and its probabilistic framework.
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Homework Statement


Knowing the decay constant ##\lambda## of a nucleus, find the probability of the decay of the nucleus during the time from 0 to ##t##.


Homework Equations





The Attempt at a Solution


I don't know where to start from. I know that the decay is first order and the number of particles remaining at any time ##t## is given by ##N(t)=N_0e^{-\lambda t}## but I have no clue how to set up the equations for finding the probability. Please give a few hints to begin with.

Any help is appreciated. Thanks!
 
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You're probably overthinking it. Take the frequentist approach. Start off with a bunch of identical set-ups and see how many nuclei decayed in time ##t##. The fraction that decayed is the probability.
 
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Hi vela! :)

vela said:
You're probably overthinking it. Take the frequentist approach. Start off with a bunch of identical set-ups and see how many nuclei decayed in time ##t##. The fraction that decayed is the probability.

Let there be ##N_0## nuclei initially. The remaining nuclei at time ##t## is ##N_0e^{-\lambda t}##. The nuclei that decayed in time ##t## are ##N_0(1-e^{-\lambda t})##. The probability is then
$$P=\frac{N_0(1-e^{-\lambda t})}{N_0}=1-e^{-\lambda t}$$
Is this correct?
 
Yup. As you can see, at t=0, the probability is 0 that it has decayed, and as ##t \to \infty##, the probability approaches 1, as you'd expect.
 
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vela said:
Yup. As you can see, at t=0, the probability is 0 that it has decayed, and as ##t \to \infty##, the probability approaches 1, as you'd expect.

Thanks a lot vela! :smile:

You are right, I was really over thinking the problem. I usually go blank even on the simplest probability problems. :redface:
 
This is an alternative approach:
The lifetime of a nucleus obeys exponential distribution, and it can be derived from the the property that nuclei do not age. The nucleus has the same probability of decaying during the next dt time interval any time of its life-span: it is λdt .

A is the event that the nucleus does not decay before t. B is the event that it does not survive t+dt. F(t) is the distribution function of the lifetime τ of the nucleus. F(t) = P(0<τ<t) is the probability that it decays before time t. The probability that the nucleus is alive at time t is P(A)=1-F(t).

AB is the event that the atom is alive at time t but decays during the following dt time: P(AB)=P(t<τ<t+dt)=(dF/dt) dt. P(B|A) is the conditional probability that the atom decays before t+dt with the condition that it is alive at time t. P(B|A)=λdt.

P(BA)=P(B|A) P(A)--> (dF/dt) dt=λdt (1-F(t))--> F ' = λ(1-F), F(0)=0. The solution is

F(t)=1-e-λt, the probability that the nucleus decays during the time from 0 to t.

ehild
 
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Thank you ehild for the alternative method but I am having a hard time in comprehending even the first line. How do you get ##\lambda dt##, what does it supposed to represent? :confused:

I know this should be obvious but probability is one of my weakest points. :(
 
No it is not obvious, and Probability Theory is very hard...Sometimes it is difficult to figure out what are the elementary events.


The probability that the nucleus decays in the next very short time interval is proportional to the length of the interval, but does not depend on the age of the nucleus if it is still alive. Lambda is that proportionality factor.


ehild
 
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ehild said:
The probability that the nucleus decays in the next very short time interval is proportional to the length of the interval,
Umm...why only on time interval? Why not on some other factors, say, number of nuclei present or decayed? How do you know beforehand that the probability has to depend on time interval? :confused:

Sorry if these are stupid questions.
 
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Pranav-Arora said:
Umm...why only on time interval? Why not on some other factors, say, number of nuclei present or decayed? How do you know beforehand that the probability has to depend on time interval? :confused:

Sorry if these are stupid questions.

Well, you know from experience that the the number of the non-decayed nuclei decreases exponentially with time. That law can be derived from the property of radioactive decay that it is independent of any environmental factors. It does not depend on the other atoms, is not influenced by temperature, it is also independent on the time since the nucleus exists. It is determined only by factors inside the nucleus, or it is completely accidental. There is a nucleus and it can decay any time. Time is continuous. You can not give the probability that the decay happens exactly at 12.00 h. You can say that it happens between t and t+Δt. There is some probability that it decays during the next minute. With twice of that probability it decays during the next two minutes. So we say that the probability that the nucleus decays during the subsequent Δt time interval is proportional to Δt.

ehild
 
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  • #11
Pranav-Arora said:
Umm...why only on time interval? Why not on some other factors, say, number of nuclei present or decayed? How do you know beforehand that the probability has to depend on time interval? :confused:

Sorry if these are stupid questions.
The differential equation N satisfies,
$$\frac{dN}{dt} = -\lambda N,$$ reflects what ehild just said. The rate at which nuclei decay depends only depends the number present and a constant ##\lambda##. If you rewrite it slightly, you get
$$\frac{dN}{N} = -\lambda\,dt.$$ This says the fraction that will decay on average between time ##t## and ##t+dt## is ##\lambda\,dt##. The rate is constant with time.
 
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  • #12
Thank you ehild and vela for the helpful explanations. :smile:

The solution by ehild is starting to make sense to me, I need to work more on the Probability section. :)
 

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