## Tag Archives: potassium

Waterford Crystal – A beautiful addition to any home! It shines, looks lovely, and it is radioactive!

I recently tested a piece of Waterford crystal to see why it was radioactive. For a long time I have wondered by my Geiger counter went crazy when near it. The truth was quickly revealed by Gamma Spectroscopy! The little Angel statue contained potassium 40, (element K). All potassium contains 93+ % Potassium 39, stable potassium, and a little bit of the isotope Potassium 41, perhaps 6+ %. But, all potassium also contains a tiny fraction of the radioactive isotope of potassium, Potassium 40 (K40). Potassium 40 undergoes three forms of decay, beta -, rarely beta+, and electron capture. The last step emits a gamma ray with an energy of 1461 keV. It is this gamma ray that I detected.

Amazing!

Waterford Crystal Angel

Waterford Crystal - Gamma spectrum

## Radioactive Banana! Peeling Away the Mystery

Radioactive Banana! Peeling Away the Mystery

All bananas contain potassium (element K). All potassium contains 93+ % Potassium 39, stable potassium, and a little bit of the isotope Potassium 41, perhaps 6+ %. But, all potassium also contains a tiny fraction of the radioactive isotope of potassium, Potassium 40 (K40).

Potassium 40 undergoes three forms of decay, beta -, rarely beta+, and electron capture. The last step emits a gamma ray with an energy of 1461 keV. It is this gamma ray that I detected.

My calculations for the typical radioactivity of a banana:

The number of Potassium (K) atoms per gram of potassium:
(Avogadro’s Number / Atomic Weight of K40) = 6.022 x10^23 / 39 = 1.544 × 10^22 K Atoms/gram

The amount of Potassium in a Banana (approx):
grams of Potassium in a banana = 0.442 grams

Natural abundance of K40 per normal Potassium (A): 0.000117

Half life of Potassium: 3.9357×10^16 seconds (T 1/2).

Calculation:
((Avogadro’s Number) / (Atomic weight)) x (0.442 g) x (A) x (ln 2) / (T1/2)
(((6.022*10^23 / 39)*0.442) x 0.000117) x ln2 / (3.9357×10^16)
=14.0633 decays per second per banana
= 14.0633 Bq Banana^-1
=lol

For \$350 USD – \$500 USD
http://beeresearch.com.au/ – Quality and inexpensive MCA
http://www.physics.usyd.edu.au/~marek/pra/index.html – PRA software for viewing peaks.

You can often find a good scintillation probe on eBay for a few hundred dollars (USD), but you have to shop for it.

\$4,000-\$5,000 USD

Some potassium and banana sites!!!
http://www.orau.org/ptp/collection/consumer%20products/potassiumgeneralinfo.htm
http://en.wikipedia.org/wiki/Banana_equivalent_dose
http://www.chiquitabananas.com/Worlds-Favorite-Fruit/bananas-and-potassium.aspx
http://www.ncbi.nlm.nih.gov/pubmedhealth/PMH0000089/

## Beta Radiation, Geiger Counters 101 Part Three

Filed under Rants & Misc

Beta radiation was first detected over 100 years ago. Beta radiation is the ejection of energetic electrons and anti-electrons (called positrons) from an atom. These electrons and positrons are moving with incredible force. They smash into our bodies and may cause harm. Beta radiation is one of the largest sources of radiation humans are exposed to.

You Eat This Each Day!!!

This atom is Potassium-40, a radioactive version of the Potassium you and I eat each day in our bananas and vitamins. When you eat light salt or Potassium salt, you are eating Potassium Chloride, mostly, which contains small amounts of K-40, radioactive Potassium! Yum!

Beta radiation is stopped by thin metal easily, a sheet of aluminum being the most often cited method.

Three more videos are coming!
Gamma and X-Ray Radiation, Geiger Counters 101 Part Four
Radiation Math, Geiger Counters 101 Part Five

Important Websites:
Anti-Proton.com
GeigerCounters.com
Medcom.com
Seintl.com

## Converting CPM to uSv/hr, Plus Calibration

You CANNOT use micro sieverts per hour (uSv/hr), MREM/hr, mR/hr, or any other such energy-based unit with a Geiger counter without first knowing the exact energy of the object you are testing. With this knowledge, you can calculate your equivalent dose from exposure, or even calibrate your Geiger counter to do it for you.

NOTE: The math I use was created by me, but is widely used in science, and not new. DO NOT USE the techniques or math I show you for anything you really need to be 100% sure of. This is for fun and there are many additional variables I did not calculate.

USE AT OWN RISK
This is only an approximate means.

The real method for finding the energy of a given isotope is to use a scintillator to measure the energy emitted by the sample. If you REALLY must use a Geiger counter, you might try something like this:

Sv/hr = E(joules) * A(Bq) * 3600s * Wr * H * (100-(r/(ρ cm^3/CSDA cm^2)))/kg Hr
Where:
E = energy per Bq in joules
A = activity in Bq
3600s = 60seconds x 60min
Wr = weighting factor of radiation. Gamma and Beta=1, Alpha=20
H = weighting factor of body part. Use 1 for entire body
/kg Hr = the entire effective energy is divided per kilogram, and per hour.
The last part calculates the change in energy of particle with respect to position within a medium. ρ is the density of air. dE/dx, basically.

Watch the video to see in detail.