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April 20, 2011

Radioactive Schwarzium

By: Aaron Datesman

One of the program activities we undertook at the nuclear power industry summer camp I attended in 1996 was an experiment to measure the activity of a radioactive sample. It’s a simple demonstration to perform: there’s a sample of some material with a short half-life, a proportional counter, and a stopwatch. At the end of three hours, we plotted out the data on a sheet of logarithmic graph paper, drew a line, and read off the half-life. Easy.

There’s a quirk, though, a little scientific twist which I failed to understand well at the time. Because the decay process is probabilistic rather than deterministic, the measurement suffers from a systematic uncertainty which cannot be removed. As it happens, if you measure 900 decays in an hour, what you can say with certainty about the next hour is only that you expect to see between 840 and 960 counts.

Because there were N decays and because N stands for an iNteger and because you probably can’t be a scientist if you can’t count up integers correctly, it’s sort of weird to be told that there’s an irreducible error present in this experiment. But it’s there, and it’s natural that it should be there. An analogy which is pretty easy to understand involves coin flips: if you flip a fair coin eight times, it’s pretty likely that you won’t get four heads and four tails, although that is the most likely outcome. The result of a large number of random outcomes, taken together, will tend toward a deterministic outcome (half heads, half tails, half-life) – but some scatter will always remain.

It’s noteworthy that, if you measure 16 rather than 1000 counts in an hour, your knowledge is even worse in a relative sense – the number of counts you expect to see in the next hour lies between 8 and 24. That’s an error of +/- 50%. This uncertainty has really fascinating implications for the health effects of low exposures to ionizing radiation, which I hope to expand on over the next several posts.

As preparation for the mathy-ness to come, I have a little thought(crime) experiment. It may be lethal for you. Sorry about that. This is the experiment:

You are locked in a room and cannot escape. The door operates on a timer, and will open in five minutes. In the room, along with you, there are four atoms of radioactive Schwarzium (Sw). Sw has a half-life of five minutes.

The decay of radioactive Sw is tremendously energetic. If more than two atoms of Schwarzium decay in your vicinity, you will suffer acute radiation sickness and die.

Are you safe? Can you get out of the room before the third and fourth atoms of Schwarzium decay?

— Aaron Datesman

Posted at April 20, 2011 10:46 PM
Comments

Ooh, I love GEDANKENEXPERIMENT.

Clearly, by definition of "half-life", the expectation value of the number of decays that have happened when the door unlocks is two. But I'm ill and too feverish to calculate the overall probability of survival.

Posted by: Cloud at April 20, 2011 11:29 PM

what am i wearing

Posted by: hapa at April 20, 2011 11:43 PM

I spy a Bill Murray reference.

Posted by: Amandasaurus at April 21, 2011 12:17 AM

'Never be afraid to die, because you are born to die'


Walter Breuning, the world's oldest man and the second oldest person, has died aged 114 from natural causes. He died in a Montana hospital after being admitted at the beginning of February with an undisclosed illness.

He attributes his long life to eating only two meals a day - because 'that's all you need' - working as long as you can, embracing change, helping others and of course, accepting death. He said the last one was a lesson he learned from his grandfather, proclaiming: 'We're all going to die. Some people are scared of dying. Never be afraid to die. Because you're born to die.'

http://www.dailymail.co.uk/news/article-1377132/Worlds-oldest-man-Walter-Breuning-dies-aged-114-natural-causes.html#ixzz1K85UNJwB

Posted by: mistah charley, ph.d. at April 21, 2011 12:46 AM

Aaron, is the chance that you'll clear up the mess from your last post higher than 11/16?

Posted by: bobs at April 21, 2011 01:30 AM

"Easy" to do what you're told. On that note, I have no intention of discussing tophat fashion, even though I'm quite dim (or maybe my education sucks).

The math-nalogies could be a little clearer. Scatter, I get (reality is fuzzier than averages). I'm pretty lost otherwise.

900 "counts" (whatever that means) divided by two is 450. I can similarly get 8 from 16 with that formula, as you seemed to do later on in your post. Both in reference to "the next hour." Why does the scatter from one include its half but the other doesn't even come close?

I'm sorry I can't entertain your thought experiment yet. Some of the logic just isn't connecting on this end.

Posted by: Lamda free, home of the brave at April 21, 2011 01:48 AM

Bobs - maybe. I didn't realize there was a mess there!

Posted by: Aaron Datesman at April 21, 2011 06:15 AM

All I know is that some of me will be alive and some of me will be dead until somebody notices me.

Posted by: Benjamin Arthur Schwab at April 21, 2011 11:15 AM

Could someone please see to it that the grant gets transferred to someone still capable of carrying out the research? Oh, and dispose of Schroedinger's cat.

Posted by: davidly at April 21, 2011 12:11 PM

I must be oversimplifying this, but it sure seems like you ought to have a 50% chance of surviving.

Posted by: John Entrekin at April 21, 2011 12:46 PM

John, I think that's the point.

Posted by: Amandasaurus at April 21, 2011 01:33 PM

25% chance one will go off, 12.5% chance two will go off,IMHO. Of course my guess depends on WHAT causes the nucleus to decay, some outside influence, some internal clock, or a random moment.

Posted by: Mike Meyer at April 21, 2011 01:41 PM

This is a fine thought experiment, but this becomes totally irrelevant when you're talking about realistic quantities of radioactive material. For example, Tokyo's water supply is apparently reading somewhere around 200 Becquerels per liter worth of radioactive iodine right now - meaning a liter of water contains about 300,000 I-131 atoms. If you toss 300,000 coins, you're going to find that they'll get pretty damn close to 50% heads and 50% tails - or, so close that the probability distribution isn't really an issue - your error falls off as the square of your sample size, and when you're talking about atoms in biologically relevant quantities, the sample size is really, really large.

Posted by: saurabh at April 21, 2011 04:31 PM

Saurabh - Yeah, that's what I thought, too. Except that it's not so in every interesting case. Keep reading the posts over the next couple of days. I know you'll enjoy them....

Posted by: Aaron Datesman at April 21, 2011 06:22 PM

mistah charley, your customary wisdom reminds me of something said by a man who ruined his reputation by being ross perot's first vice presidential candidate--admiral jim stockdale, who once commented I believe basically that it's good to remember that the worst thing they can do is kill you, and that's really not that bad either.

if a person has to go from radioactive schwarzium, why, it could be worse

better that than radioactive bs

Posted by: N E at April 21, 2011 10:11 PM

There could very well be some Schwarzium in that inevitable cloud of Cesium, Xenon, Kripton, Strontium, Radon, Uranium gas, Plutonium particulates, and of course Iodine. Since Japan has yet to get a lid on the leak issues and now has its very own deadzone to deal with, I'd say WE have as many "roll-o-the-dice" on these percentages as could possibly be wanted "comming down the pike" so to speak. Lady Bad Luck, thy name be "Down Wind".

Posted by: Mike Meyer at April 21, 2011 11:09 PM

saurabh - "your error falls off as the square of your sample size, and when you're talking about atoms in biologically relevant quantities, the sample size is really, really large."


The SE(mean) declines with the the square root of the sample size, not the square. Otherwise, relatively small samples would have no error worth speaking of.

Posted by: Barry at April 22, 2011 01:44 PM

Barry - you're right, my mistake.

Posted by: saurabh at April 22, 2011 02:53 PM

wouldn't the chance of dying be about 5%? Although it's likely my simple binomial distribution probably is too naive for the problem.

Posted by: Bill Murray at April 22, 2011 03:46 PM

If the amount of Cs-137 released in a reactor incident is X, we are told that, with a half life of 30 years, all the Cs-137 will have decayed to a stable isotope of Cs in 10 half lives ... 300 years. So what if the amount released is 2X? Still all gone in 300 years or is it 600 years?

Posted by: Parmenides Wheelspin at April 22, 2011 05:06 PM

Parmenides Wheelspin: that's not exactly what you're being told, at least if the science reporting is halfway responsible. In ten half-lives, the proportion of remaining radioactivity is 1/2^10, or one thousandth of the original. (In an isotope whose products are radioactive, this doesn't always hold, but Cs-137's products are very short-lived and decay to stable Ba-137, as far as I can tell.)

I think they're saying that the radioactivity levels will be "less than background" or "undetectable" or somesuch at that point; the complete absence of radioactivity is unrealistic in the natural environment. Because the decay is exponential, doubling the initial count of radioactive Cs-137 atoms will mean one additional half-life required to bring it down to previously-projected safe levels.

That is, in your terminology, if X amount takes 300 years--ten half-lives--to become undetectable, then 2X would take 330 years, or eleven half-lives, since after 30 years, there would be X amount left, and you've reduced the question to one you've already answered.

Posted by: grendelkhan at April 22, 2011 06:22 PM

bobs above had the right answer. It's the equivalent of flipping four coins and returning "dead" if more than two come up heads.

So, there are sixteen possible configurations. We want to know how many of those configurations have two or fewer heads. (4 choose 0) + (4 choose 1) + (4 choose 2) = 11 (paste it into Wolfram Alpha if you're curious), so the odds of survival are 11/16, or just under 69%.

I agree with saurabh, though--I don't see the point of this thought experiment, when the quantities in question are large enough to decay(practically) deterministically for all realistic purposes.

Posted by: grendelkhan at April 22, 2011 06:28 PM

...in the environment.

Posted by: Amandasaurus at April 22, 2011 06:42 PM

Thank you, grendelkhan for the response.

So, if the level in the sea water of I-131 was 7.5 million times the 'acceptable' level a couple weeks ago and the half life of I-131 is 8 days and if the source of the the I-131 is stifled now, how long before the Iodine will be radioactively harmless? Calling it 8 million instead of 7.5 million suggests 23 half lives or 184 days.

Still one encounters the notion of '10 half lives' taking care of the problem, as here: http://www.youtube.com/BerkeleyLab#p/u/0/t3cdf51zbLU

Posted by: Parmenides Wheelspin at April 23, 2011 03:45 AM

Does this mean, that since ONLY the Iodine (NONE of the other stuff, I've noticed)is ever mentioned in the press releases, once the Iodine has decayed then the water can be dumped into the environment and its "all good"? Am I wrong or does the magic really work?

Posted by: Mike Meyer at April 23, 2011 02:31 PM