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The law of decay is $$\frac{dN}{dt} = -\lambda N ,$$ where $\lambda$ is the decay constant. I read something about $\lambda$ but not really clear.... Someone told that is a probability, someone that is a constant that can be greater than one, and I am a lot confused.

So, if it is the "probability of decay of a particles per unit time" how can be greater than one? And from the expression: $\frac{dN}{dt N} = - \lambda$ seems that can't be greater than one because is a fraction of total particles that decay..... but I read in many case for example $\lambda = 100/s $ so what?

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    $\begingroup$ Sometimes called a half-life (not the video game). And, of course it can be greater than one, with the next question being one what? Per nanosecond, per year, per millenia, per ??? $\endgroup$
    – Jon Custer
    Commented Oct 29, 2021 at 21:06
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    $\begingroup$ Does this answer your question? What does the decay constant mean? $\endgroup$ Commented Oct 29, 2021 at 21:10

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The differential equation $$\frac{dN}{dt}=-\lambda N$$ becomes more instructive when we write it as $$-\frac{dN}{N} = \lambda\ dt.$$

Then you have a dimension-less quantity (actually a probability) on each side of the equation. Replacing the infinitesimal differentials by finite differences we get $$-\frac{\Delta N}{N} \approx \lambda\ \Delta t.$$ which is a valid approximation if both sides of the equation are much smaller than $1$.

Let's stick to your example $\lambda=100\text{ s}^{-1}$ and pick $\Delta t=10^{-4}\text{ s}$. Then you get $-\frac{\Delta N}{N} \approx \lambda\ \Delta t = 0.01$ which means the number $N$ decreases by $1$ % during this $10^{-4}$ s.

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    $\begingroup$ I like this answer because it addresses the implied question "what if the dimensionless $\lambda\ dt$ is greater than one?" The answer is "that's not a very good differential; go smaller and integrate." $\endgroup$
    – rob
    Commented Oct 29, 2021 at 21:34
  • $\begingroup$ So if I take $\Delta t$ equal to 1 year, accordino to the answer, I'm wrong because I need infinitesimal variation? $\endgroup$
    – Michealrr6
    Commented Oct 30, 2021 at 6:40
  • $\begingroup$ @Michealrr6 Yes, that would be wrong. After $1$ year the right side is $\lambda\ \Delta t=3.2\cdot 10^9$. And the left side is $-\frac{\Delta N}{N}=1$ because $100$ % has decayed. $\endgroup$ Commented Oct 30, 2021 at 7:24
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$1/\lambda$ is the mean lifetime for a single particle to transition from the initial state to the final state. It is not a probability; is is a (physical) parameter that characterizes the probability distribution for the lifetime.

With respect to $\frac{dN}{dt N} = - \lambda$ and $\lambda >1$ -- re-write the expression as $dN = -\lambda N dt$. This is a statement that the amount of change in the count is proportional to $\lambda$, and proportional to the current number present and scales linearly with the time increment.

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  • $\begingroup$ +1 the experimental view. $\endgroup$
    – anna v
    Commented Oct 30, 2021 at 4:01
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The equation can be rewritten as $\lambda = -\left[\frac{dN}{N}\right]/dt$ (this is somewhat abuse of notation, but gives a more intuitive understanding). $\frac{dN}{N}$ is a portion of the total amount; you can think of it as a percentage. So you might have, for instance, a decay rate of 1% per second. This is an instantaneous percentage; a decay rate of 1% per second means that as the time interval goes to zero, the ratio of the percentage decayed to time elapsed goes to 1%. So, for instance, you might have 1.01% decay in one second, 0.1001% decay in 0.1 seconds, 0.010001% decay in 0.01 seconds, etc. You can sort of think of it as the probability of a particular particle decaying in a second, except that there's a compounding effect; solving the differential equation gives $N = N_0e^{-\lambda t}$, so a decay rate of 1% per second means that in a second, you have a decay of $e^{0.01}=1.01$.

If the decay rate is more than one, that just means that you have a shorter time period to get any given amount to decay. For instance, a decay rate of 200% per second means that you're getting 0.2% decay per millisecond. If you're confused about how that works if we have an elapsed time of one second, remember that decays are multiplied, not added, so over one second we don't have 0.2%*1000=200% decay; rather, we have $0.998^{1000}$, or 13.5% left over.

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