# Average lifetime of particles [duplicate]

I have heard that If I have $$10^{24}$$ particles (for instance) and I observe them for 1 years, I can say that they have an average lifetime at least of $$10^{24}$$ years.

How this is derived?

So with $$\mathrm{N}$$ particles and $$\mathrm{T}$$ time of observation how I can estimate the average lifetime (or half life)?

We can start by setting the number of decays to be less than one: $$N(t) = N_0e^{-t/\lambda} > N_0 - 1$$ where $$N$$ is the number of undecayed particles, $$N_0$$ is the number of undecayed particles at $$t=0$$, $$t$$ is the time of observation, and $$\lambda$$ is the average lifetime of the particle. In other words, the number of particles left is greater than the number we start with minus one--that is, all of them. We then solve for the average lifetime $$\lambda$$. $$\lambda > \frac{-t}{\ln\left(\frac{N_0 - 1}{N_0}\right)} = \frac{-t}{\ln\left(1 - \frac{1}{N_0}\right)}$$ Because $$|1/N_0| \ll 1$$, we can use $$\ln(1 + x) \approx x$$ to simplify to $$\lambda > \frac{-t}{-1/N_0}$$ $$\lambda > tN_0$$ So, if the observation time $$t$$ is one year and the number of particles $$N_0$$ is $$10^{24}$$, then the average lifetime of the particle is at least $$10^{24}$$ years.

The radiation from radiaoactive substances does exponential decay, so it's about fitting the numbers to this equation $$N_t = N_0 e^{-\lambda t}\tag1$$ and trying to find the decay constant $$\lambda$$

Then the half life is $$T_{1/2} = \frac{ln2}{\lambda}\tag2$$

If all $$10^{24}$$ remain, the lifetime could be infinite, but if just one was about to decay (but we were unlucky and missed it) then $$e^{-\lambda t} = 1-1\times 10^{-24}$$, taking logs of both sides $$\lambda \times 1 = 1\times 10^{-24}$$ and so $$T_{1/2} = 1\times 10^{24}$$ approximately, or more, in years.

If you had the same number of particles initially and $$10^{23}$$ where left, for example, after 6 months, then in years $$0.1 = e^{-\lambda \times 0.5}$$ taking logs of both sides gives $$\lambda = 4.6$$ and the half life, from 2) is $$0.15$$ years.

The decay constant $$\lambda$$, the average lifetime $$\tau$$ and the half-life $$t_{\rm1/2}$$ are connected, $$\lambda = \frac 1 \tau = \frac{\ln 2}{t_{1/2}}$$.

I have tried to explain in simple terms what is involved in the estimation of half-life from one observation.

The probability of one unstable nucleus not decaying in a time interval of $$t$$ is $$P_{\text{ no decay, one}}(t) = 2^{-t/t_{1/2}}$$.

If there are $$N$$ unstable nuclei of the same type then the probability of all $$N$$ nuclei not decaying in a time interval of $$t$$ is $$P_{\text{no decay,N}}(t) = 2^{-Nt/t_{1/2}}$$.

You have chosen to look at $$N = 10^{24}$$ nuclei for a time $$t=1\,\rm year$$ and found that none of them decayed.

$$P_{\text{ no decay,}10^{24}}(\text{1 year}) = 2^{-10^{24}/t_{1/2}}$$

Now it is down to choosing a half-life and finding out the probability of the no decay scenario.

Assume that the half-life is $$10^{24}\, \rm years$$ then the probability that in one year none of nuclei decayed is $$2^{-10^{24}/10^{24}} = 2^{-1} = 0.5$$ and so such a half-life is consistent with the experimental data.

Note however the probability changes as your guess as to what the half-life is changes, being $$2^{-0.1}\approx 0.99$$ if the half-life is $$10^{25}\,\rm years$$ and only $$2^{-10}\approx 0.001$$ if the half-life is $$10^{23}\,\rm years$$.

From your one observation there is a very good chance that the half-life is greater than $$10^{23}\,\rm years$$ but there is no upper bound.