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I have learnt that in radioactivity $λ=\ln2/t_{1/2}$, thus showing the link between λ and radioactive half time. I have also learnt how to apply it in calculations (such as $R={R_0}e^{-λt}$) etc.

Now I just learnt that average life-span of radioactive substances is calculated as $1/λ$.

This leads me to the following two questions:

  1. What does $λ$ actually stand for? My book teaches how to use it, but never defines the meaning beyond stating that it is the "constant of radioactive decay".
  2. Is there a way we can mathematically show why $1/λ$ is equal to average life-span, perhaps linking it back to $λ=\ln2/t_{1/2}$?
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Say we have a sample with a bunch of the original stuff, and you can see when each of them decays. One of them will take $t_1$, another $t_2$, another... If we calculate the average of those, we get a time called 'average life', usually denoted by the greek letter $\tau$:

$\tau = \frac{\sum_i t_i}{N}$

If the sample is big enough, the number of decays we see in a time interval $dt$ is (*)

$ N\frac{dt}{\tau} $,

or, in terms of change in the number of undecayed nuclei,

$ dN=-N \frac{dt}{\tau} $.

If you solve the differential equation, you get:

$ \frac{dN}{dt} = -\frac{N}{\tau} \Rightarrow N(t)=N_0 e^{-t/\tau} $

Which, comparing to yours,

$ N(t)=N_0 e^{-t/\tau}=N_0 e^{-\lambda t} \Rightarrow \lambda = \frac{1}{\tau} $

(*) I know this is overly simplified, but you would probably have to work from Poisson statistics up to really justify this.

Edit: to better answer question 1, let's have a look at the activity by derivating the number of particles not yet disintegrated, $ N(t) $:

$ A(t) = - \frac{dN}{dt} = - \frac{d}{dt} \left( N_0 e^{-\lambda t} \right) = \lambda N_0 e^{-\lambda t} = \lambda N(t) $

So $ \lambda $, besides being the reciprocal of the average life, also links activity and number of radioactive particles.

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  1. $\lambda$, the decay constant of radioactivity is a measure of rate of change of concentration of the sample per unit sample concentration i.e, it is the radioactivity rate for one unit of sample. It depends on the nature of the material.
  2. Calculate area under $R$ vs $t$ curve: $A = \int_{0}^{\infty} Rdt = R_0/\lambda$

This solution defines average time: $1/\lambda$, as the effective time taken by the processes.

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  • $\begingroup$ Thank you for the answer. So λ essentially expresses the rate at which the process of total radioactive breakdown of the sample progresses? $\endgroup$ – Pregunto Oct 10 '18 at 14:26
  • $\begingroup$ @Pregunto Not progress, Since it is a constant but you can say that it is the unit of progression of decay. $\endgroup$ – Aman pawar Oct 10 '18 at 16:52

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