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[Phys-L] Re: Radioactive Bananas



If you want to check those bananas for K-40, be sure to take a good
background spectrum with your NaI(Tl) or solid state detector, because
there is K-40 in most cement and concrete. Then subtract the background
spectrum from the measurement that you take of the bananas.

If you want some really radioactive food try Brazil nuts. They are from
trees that grow in Monazite sands that have a lot natural radioactivity.
Typically, 1000 to 7000 pCi/kg of Ra-226 and daughters is found in
Brazil nuts.

By the way, bananas are not unusually high in K-40. White potatoes have
about the same amount of K-40.

Dr. Mark H. Shapiro
Professor of Physics, Emeritus
California State University, Fullerton
Phone: 714 278-3884
FAX: 714 278-5810
email: mshapiro@fullerton.edu
web: http://chaos.fullerton.edu/Shapiro.html
travel and family pictures:
http://community.webshots.com/user/mhshapiro


-----Original Message-----
From: Forum for Physics Educators [mailto:PHYS-L@list1.ucc.nau.edu] On
Behalf Of Thomas O'Neill
Sent: Sunday, October 30, 2005 1:07 PM
To: PHYS-L@LISTS.NAU.EDU
Subject: Re: Radioactive Bananas

Michael,
Thanks for a thorough explanation of the whole affair. I was on
the point of grabbing my Geiger counters and checking a pile of bananas
just for the giggle of it. I may do it, anyway, but you just saved me a
great deal of time in background research.

THO

Thomas O'Neill
Physics
Shenandoah Valley Governor's School

-----Original Message-----
From: Michael Edmiston [mailto:edmiston@BLUFFTON.EDU]
Sent: Sunday, October 30, 2005 1:41 PM
To: PHYS-L@LISTS.NAU.EDU
Subject: Re: Radioactive Bananas

Potassium is the most stable mass-39 nuclide, and also the most stable
mass-41 nuclide. 40-Potassium would be the most stable mass-40 nuclide
if it weren't for the fact it is odd-odd. When a nuclide is in this
situation it usually ends up that the even-even nuclides on either side
of it are more stable than it is, and that means it can decay either way
(by beta-minus or by beta-plus/electron-capture).

In the case of 40K, it beta-minus decays to the ground state of 40Ca
about 89% of the time, and there is no gamma. It electron-capture
decays to an excited state of 40Ar about 11% of the time, and there is a
1.46 MeV gamma. It can also beta-plus decay to the ground state of 40Ar
about 0.001% of the time, in which case there is no gamma but there
would be two 0.511 MeV annihilation photons when the beta-plus
encounters and electron. It can also EC to the ground state of 40Ar
about 0.2% of the time, in which case there is no radiation.

Summary...

(1) 89% of 40K decays are beta-minus to the ground state. These yield a
beta-minus emission with endpoint energy of 1.31 MeV, and no gamma.

(2) 11% of 40K decays are EC to an excited state. These emit a 1.46 MeV
gamma.

(3) 0.2% of 40K decays are EC to the ground state. These emit nothing.

(4) 0.001% of 40K decays are beta-plus to the ground state. These emit
a beta-plus which quickly annihilate to two 0.511 MeV photons.

There are is neutrino and antineutrino emission, but forget about seeing
these.

Now... note that I said the beta-minus *endpoint energy* is 1.31 MeV.
You won't see any 1.31 MeV beta-minus emissions because the endpoint
energy is shared between the beta and the antineutrino. On average, the
emitted beta is about half the endpoint energy, or about 0.65 MeV.

In the end, if there are 10^6 decays of 40K, the likely emissions would
be about...

(1) 890,000 beta-minus with energies near 0.65 Mev
(2) 110,000 gamma of 1.46 MeV
(3) 20 annihilation photons of 0.511 Mev

In a banana, the beta-minus are going to be difficult to detect because
of self-absorption. Most of the 650-keV betas will not make it out of
the banana. If you have a truck of bananas it doesn't improve things.
Radiation from the bananas in the center will never make it out, plus
now we have crates and the walls of the truck, and a lot of air that the
betas would have to go through.

Thus, we are most easily going to detect the bananas by the 1.46 MeV
gamma. Since this is high energy, the detection efficiency with a GM
tube is poor. We need to use a NaI detector. These, of course, are
much more expensive, especially the big ones (of which you need several)
to survey trucks passing through customs. My NaI detector is a two-inch
detector. It is a 2"-diameter cylinder 2" long. Mounted on a
photomultiplier tube my unit cost about $2000 in 1985. When I set a
whole bottle of KCl (500 g) on top of the detector, I can notice an
increase over background in about two or three minutes, but I need to
collect data for about an hour before a convincing 1.46-MeV peak becomes
obvious in the pulse-height (energy) spectrum.

Detecting the 40K in a banana is not easy.

Michael D. Edmiston, Ph.D.
Professor of Physics and Chemistry
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu