My textbook says that $\lambda$ is the probability per unit time that 1 particle will decay in one second. This makes absolutely no sense to me - I can see that it is related to probability but cannot see how it is the probability. We have: $$N = N_0 e^{-\lambda t} \tag A$$ Now, if the probability of a particle decaying is $p$ then we can say that at $t=1$, $N=(1-p)N_0$. Therefore:

$$1-p= e^{-\lambda}\tag B$$ Rearranging we get: $$\lambda = \ln\biggl(\frac{1}{1-p}\biggr) \tag C$$

What is wrong with this reasoning?


I just want to add an example - this is what initially confused me. Say we have a probability per unit time of decay of $\frac16$. We would therefore expect the number of particles to go from $N$ to $\frac 56 N$ in one second which, using $(A)$, implies that $\frac 56 = e^{-\lambda}$ (I think this is the dodgy step?). This means that $\lambda = ln \biggl(\frac 65 \biggr) \approx 0.1823 $ and not $\frac 16 \approx 0.1667$.

  • 3
    $\begingroup$ Upvoting because this is a good example of where OP has a problem and has made a good effort to work through it alone. I wonder whether "λ is the probability per unit time that 1 particle will decay in one second" is correct? λ is defined to my mind as the probability per unit time of decay. Can you verify that your first sentence is exactly what the textbook says? $\endgroup$ – Matt Jan 7 '16 at 12:05
  • $\begingroup$ My textbook says: "For each second that it exists, there is a certain probability that the nucleus will decay. This probability is called the decay constant, $\lambda$" $\endgroup$ – bnosnehpets Jan 9 '16 at 13:16
  • $\begingroup$ Could you post a direct quotation from the textbook? The phrase "probability per unit time that 1 particle will decay in one second" doesn't make sense. Probability per unit time is a thing, and the probability of decay in one second is a thing, but I have no idea what probability per unit time to decay in one second means :P $\endgroup$ – DanielSank Jan 10 '16 at 21:36

It is not a probability. It has a different name. It is also called a decay constant ($1/\lambda=\tau$, where $\tau$ is a lifetime), the state (or level) width, etc. but not probability. The right relationship is $\lambda = -\dot{p}/p$.

  • $\begingroup$ For a single particle isn't the probability of decay in a particular time interval constant? Or a step function? Either the particle is present, in which case the probability of decay in some time interval is $p$, or the particle has gone missing, in which case the probability of a second decay is zero. Your final relationship is correct for an aggregate population with the same $\lambda$. $\endgroup$ – rob Jan 7 '16 at 22:53
  • $\begingroup$ @rob: It is the definition of probability - it concerns one single particle. For ensembles one writes numbers of decayed or "alive" particles (see eq.(A) of OP, for example). $\endgroup$ – Vladimir Kalitvianski Jan 7 '16 at 23:05

First, careful with units: the argument of an exponential or a logarithm must always be dimensionless. Keeping $t$ in your (B) $$ 1-p = e^{-\lambda t} \tag{B'} $$ and taking the logarithm of both sides gives

$$ \ln (1-p) = -\lambda t \tag{C'} $$

Now we use the Taylor expansion of the natural logarithm around unity, where the function has zero value and unit slope, $$ \ln (1 + \epsilon) \approx \epsilon \qquad\text{when }|\epsilon| \ll 1, $$ to write $$ -p \approx -\lambda t. $$ This equation becomes exact in the limit of very short time intervals or very small probabilities. You can interpret this as saying "the probability of observing a decay is proportional to how long you wait, as long as you don't wait too long," with $\lambda$ as the constant of proportionality.

In this writing $\lambda$ is the "decay probability per unit time" rather than "the probability that one particle will decay in one second"; those two statements are equivalent only when $\lambda^{-1} \gg \rm 1\,s$. Calculus is all about infinitesimals.

With respect to your example problem: 1/6 is too big. Try with $p=0.01$.

  • $\begingroup$ I understand that a logarithm has to be dimensionless which is one of the reasons why I thought my working must be wrong. The difference between (C') and (C) is that (C') contains a $t$. I don't see how (C') is correct because that implies that as $t$ varies $\lambda$ varies. Should it not be $ln(1-p)^t = -\lambda t$ and therefore $ln(1-p) = - \lambda $? $\endgroup$ – bnosnehpets Jan 9 '16 at 13:39
  • $\begingroup$ @bnosnehpets Edited. Note that you can't raise a number to the $t$ power any more than to the $\lambda$ power: exponents must be dimensionless. What varies with time is not $\lambda$, but $p$. $\endgroup$ – rob Jan 10 '16 at 21:34

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