Calculating Radiation Emitted from K-40?

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Discussion Overview

The discussion revolves around calculating the amount of radiation emitted from 3 grams of K-40, focusing on the theoretical and practical aspects of determining its specific activity and decay rate.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant seeks a method to calculate the microcuries of radiation emitted from K-40, expressing uncertainty about the complexity of their approach.
  • Another participant provides detailed calculations regarding the half-life of K-40, the number of atoms in 3 grams, and the resulting decays per second, leading to a conversion into curies.
  • A third participant introduces the concept of specific activity, suggesting that while calculations can be made from first principles, it is often more practical to reference literature tables for specific activity values.
  • A later reply shifts the topic slightly, inquiring about the radiation emission from an unrelated device, indicating a potential distraction from the main topic.

Areas of Agreement / Disagreement

Participants present varying approaches to the calculation, with no consensus on the preferred method. The discussion remains open with multiple viewpoints on how to best determine the radiation emitted from K-40.

Contextual Notes

There are limitations regarding the assumptions made in the calculations, such as the reliance on specific activity values from literature versus first-principle calculations. The inquiry about the unrelated device introduces additional uncertainty regarding the focus of the discussion.

ampzor
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How would I be able to calculate the amount of microcuries of radiation emitted from 3 grams of K-40? I think I have a way of figuring it out but it is pretty lengthy and I am sure that I am over thinking this a bit too much.
 
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The half life of K-40 is 1.277 x 109 years = 4.03 x 1016 seconds. The half life is Ln(2) = 0.693 x the 1/e lifetime. So the 1/e lifetime is about 5.81 x 1016 seconds.

3 grams of K-40 contains (3/40) x 6.02 x 1023 = 4.51 x 1022 atoms.

There are about 4.51 x 1022/5.81 x 1016 = 7.76 x 105 dps (decays per second) in 3 grams of K-40..

1 Curie = 3.7 x 1010 dps.

Bob S
 
The specific radioactivity per unit mass for a particular radionuclide, measured in Ci/g or Bq/g for example, is more commonly known as the "specific activity". Whilst you can calculate it from first principles if you know the half-life, as elucidated above, usually you just look up the specific activity of whatever radionuclide you're working with in literature tables.
 
Thank you Bob S.
 
does anyone know if the http://iphonecarmount.org" gives out radiation?
 
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