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Is activity in radioactive decay a differential?

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Acid92
#1
Apr16-12, 03:04 PM
P: 42
According to my textbook, the decay constant is the probability that a radionuclide will decay in any second (hence the unit s^-1) and so the total number of radionuclides decaying at any second, i.e. the activity, is λN but this is also the rate of change of N thus

dN/dt = -λN

Surely dN/dt can't be a differential since λN is the number of radionuclides decaying in a time interval of 1 second and not an instant?
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Naty1
#2
Apr16-12, 03:29 PM
P: 5,632
The decay constant λ, "lambda" is the inverse of the mean lifetime...so it's unit is the reciprocal of time, but not one second....

See total activity A here:

http://en.wikipedia.org/wiki/Radioac...ve_decay_rates
Acid92
#3
Apr16-12, 03:39 PM
P: 42
Quote Quote by Naty1 View Post
The decay constant — λ, "lambda" is the inverse of the mean lifetime...so it's unit is the reciprocal of time, but not one second....

See total activity A here:

http://en.wikipedia.org/wiki/Radioac...ve_decay_rates
Yes Ive looked at that but Im interested in its definition as the probability of decay (I looked around the internet and noticed this term being used but without the explanation Im looking for) as this is what my textbook introduces it as. My original post is how my textbook introduces it and the activity differential but it doesnt make sense to me...

Even that doesn't make sense to me, why isn't it (total activity) the number of decays at instantaneous time and not per unit time if we are talking about differentials?

truesearch
#4
Apr16-12, 03:49 PM
P: 349
Is activity in radioactive decay a differential?

The equation dn/dt = -λN stated in words means 'the number of decays per second is proportional to the number of nuclei present'.
This is the natural law of decay (or growth)
If you have 1000 nuclei in a sample and there are 10 decays per second it sounds obvious that in a sample of 2000 nuclei there would be 20 decays per second....this is 'natural.
In this case λ = 1/100 which means that the probability of a nucleus decaying is 1 in 100
λ is the probability of a decay.
Acid92
#5
Apr17-12, 11:08 AM
P: 42
Quote Quote by truesearch View Post
The equation dn/dt = -λN stated in words means 'the number of decays per second is proportional to the number of nuclei present'.
This is the natural law of decay (or growth)
If you have 1000 nuclei in a sample and there are 10 decays per second it sounds obvious that in a sample of 2000 nuclei there would be 20 decays per second....this is 'natural.
In this case λ = 1/100 which means that the probability of a nucleus decaying is 1 in 100
λ is the probability of a decay.
If λ is the probability of decay per unit time then the total decay per second will be λN, I understand that but then surely the activity can only be stated as

ΔN/Δt = λN NOT dN/dt which is a differential

If we want dN/dt, surely you would need the probability of decay in an instant (the limit as Δt tends to 0?) unless the probability of decay is independent of time intervel.

I don't understand how you say the total activity per unit time is dN/dt because this is a differential and gives the rate of change of N instantaneously and not per unit time as λN would be.
HallsofIvy
#6
Apr17-12, 11:15 AM
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PF Gold
P: 39,549
You seem to be misunderstanding the prase "total number of radionuclides decaying at any second"- and perhaps it is phrased badly. Precisely because nucleotides is random, that cannot actually be the number of radionuclides that decay. It has to be an "average" which is then saying that we are treating the rate at which radionuclides decay as if it were a constant. The slope of a straight line is the derivative.
Acid92
#7
Apr17-12, 11:45 AM
P: 42
HallsofIvy, Im not quite sure what youre saying.

Ok let me put this as simply as possible: Is the activity the average total number of decays per second, in other words the change in N per second, yes or no?

If yes, why is this (ΔN/Δt) the same as the instantaneous activity dN/dt?

If I were to draw a graph of N against t, wouldnt λN give me the gradient of the secant line over an intervel of 1 second and not the actual gradient at any point?


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