0
$\begingroup$

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)?

$\endgroup$
2

3 Answers 3

4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.