3
$\begingroup$

Decay of single isotope follows exponential law.

BUT decay of nuke fallout / spent nuclear fuel will follow hyperbolic law. It is easy to explain if we assume that half-lives are log-uniformly distributed - then dominating half-life of the mix is proportional to it's age, we get double negative feedback and that produces hyperbola.

And indeed - I took Decay Data Evaluation Project data for about 1,400 isotopes and for interval from 0.1 seconds to hundreds of years there is roughly log-linear distribution (repository link).

enter image description here

I guess we could get this linearity quite naturally - for example probability of escape of gas molecule drops exponentially as it becomes heavier and we can apply similar assumptions to nuclear liquid-drop model.

But what about 2 steps at the right? Is rapid drop at the left and steep upward curve at the right caused by selection bias (it is hard to measure extremely small and big half-life) or does it mean anything?

Does any modern radioactivity theories allow to explain details on this graph?

$\endgroup$
4
  • $\begingroup$ Maybe I am doing something wrong, but when I try running "get_nucleide_data.sh" on the repository you provide, I just get "HTTP request sent, awaiting response... 404 Not Found" because "lnhb.fr/DDEP_WG/DDEPdata.htm" gives a 404 error. The two steps in your plot are almost certainly artifacts, but without the source data it is hard to say. $\endgroup$ Commented Jan 28, 2023 at 16:57
  • $\begingroup$ @David Bailey Sorry, looks like files were moved. I committed dataset to my repo, run "git pull" to load it from github. $\endgroup$
    – Vashu
    Commented Feb 3, 2023 at 6:54
  • $\begingroup$ When I download that data and code, it generates a plot very similar to that in my answer, but with only 221 entries instead of 1400, and with no steps as in your plot above. $\endgroup$ Commented Feb 3, 2023 at 7:27
  • $\begingroup$ @DavidBailey I got full graph with xundl_201001.all.zip from NNDC site: github.com/vashu1/data_snippets/tree/master/fallout_7_10_rule/… Yeah, those steps on graph look like some artifact, I need to compare it with latest dataset. $\endgroup$
    – Vashu
    Commented Feb 3, 2023 at 11:23

1 Answer 1

3
+50
$\begingroup$

The data is showing power law behaviour, and although empirical power laws are often poorly understood, in this case there is a plausible rough explanation based on theoretical relationships between decay energy and lifetime for alpha and beta decays.

Here are NuDat 3.0 data for 1797 radionuclides plotted in the same manner as in the question. (I get a 404 error when I try to access the data using the question's repository code.)

Sorted lifetimes distribution plot

This plot is very similar to the plot in the question, but the two steps at long lifetimes in that plot do not appear and are almost certainly artifacts. It is hard to say more about the steps without the question's actual source data.

Instead of a cumulative plot, this probability density plot of the same data better shows the behaviour we are trying to understand:

Half-Life Probability Distribution

The fuzzy jaggedness of the data (shown in grey) is because the empirical probability density is calculated directly from the data and is not smoothed or fit. The distribution falls off roughly as as the inverse of the half-life, i.e. $t_{h}^{-1}$.

A theoretical explanation of why we would roughly expect a $1/t_h$ can be found in Corral, Font, and Camacho's paper on "Non-characteristic Half-lives in Radioactive Decay". Alpha and beta decays need to be considered separately.

Alpha decay is a quantum tunnelling process, and the decay energy $Q$ and half-life $t_{h}$ are related by the Geiger-Nuttal Rule, which we can write as: $$ \ln{\frac{t_{h}}{A}} = \frac{BZ}{\sqrt{Q}} = \frac{B}{\sqrt{U}} $$ where $Z$ is the radionuclide's atomic number, $Q$ is the decay's energy, $U\equiv Q/Z^2=B^2/(\ln^2 t_h/A)$, and $A$ and $B$ are coefficients that are approximately constant. The (unnormalized) probability density can then be written $$ D_\alpha = \frac{dN_{\alpha}}{dU}\left|{\frac{dU}{dt}}\right|=f\left(\frac{B^2}{\ln^2 t_h/A}\right)\left(\frac{2B^2}{\ln^3 t_h/A}\right)\frac{1}{t_h} $$ We don't know the form of $f$, but because it only depends logarithmically on half-life, it is a slowly varying function for pretty much any reasonable choice. The only non-logarithmic half-life dependence in the expression is the final $1/t_h$ factor, so this is expected to dominate.

Beta decay is a four-fermion interaction that roughly follows Sargent's Rule: $$ t_{1/2} \sim \frac{C^5}{Q^5} $$ and the probability density can be written as $$ D_\beta \sim \frac{dN_{\beta}}{dQ}\left|{\frac{dQ}{dt}}\right|=g\left(\frac{C}{t^{1/5}}\right)\frac{1}{t_h^{1+1/5}} $$ If $g$ is a slowly varying function of $t_h$, then this will give a $1/t^{1.2}$ power law, but the validity of the "slowly varying" assumption is less obvious because the coefficient $C$ varies by orders of magnitude for different radionuclides. According to the Corral-Font-Comacho paper, however, the Pickands–Balkema–De Haan statistical theorem implies that $g$ is likely to have a Pareto power-law tail, and the combined result is that $D_\beta$ is likely to have roughly inverse power law behaviour with an exponent between $1$ and $1.2$.

So it is reasonable that the observed distribution has a power law distribution with exponent around $1$. As shown in the figure, the exponent seems to vary from $\sim 0.9$ to $\sim 1.2$. It is not immediately obvious if the exponent $<1$ for short half-lives is due to physics or some sort of selection bias.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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