Radiotracer Decay - What's Left?

  • Thread starter Chyral
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In summary: If the daughter product were to bind strongly to the tracer, then you would see a decrease in the tracer concentration over time due to radioactive decay.
  • #1
Chyral
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Greetings all. I'm a just a chemist banging my head about some what seems to me as fundamentally a nuclear physics question, so please bear with me.

If you have a molecule bearing a radioactive atom, a PET radiotracer, say F-18 FDG or some such beast, and it decays (to O-18 correct?) what exactly happens to the entire molecule? Another way of putting it is: As the positron-neutrino pair is released, is the rest of the tracer molecule blown apart by kinetic energy from the decay event?
Or am I thinking too much in a classical sense and you are left with a molecule happily bearing O-18 rather than F-18?
And lastly, does the answer depend on what type of decay is occurring?
 
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  • #2
It depends on the electron configuration of the radioactive elements daugher.

You can look on wikipedia for example, on all decay modes. Then there are things like "Internal conversion", "electron capture"; which changes the electron configuration of the daugher atom.
 
  • #3
Thanks for the reply! So it looks like this was the wrong place to post (sorry about that). I have asked people here in radiochemistry and medical physics and I still am left without an answer, though. Oh well.
 
  • #4
I'm not a radiochemist, but I believe that most of the time the daughter product dissociates from the tagged molecule/pharmaceutical (which remains intact)
 
  • #5
Chyral said:
If you have a molecule bearing a radioactive atom, a PET radiotracer, say F-18 FDG or some such beast, and it decays (to O-18 correct?) what exactly happens to the entire molecule? Another way of putting it is: As the positron-neutrino pair is released, is the rest of the tracer molecule blown apart by kinetic energy from the decay event?
Or am I thinking too much in a classical sense and you are left with a molecule happily bearing O-18 rather than F-18?
And lastly, does the answer depend on what type of decay is occurring?

Hello Chyral,

I would say that the recoiling of the nucleus is negligible (specially for F) wrt the energy of positron-neutrino pair. In some way, that's why you can do imaging.
 

Related to Radiotracer Decay - What's Left?

1. What is radiotracer decay?

Radiotracer decay is the process by which a radioactive isotope breaks down into more stable forms over time, emitting radiation in the process.

2. How does radiotracer decay work?

Radiotracer decay occurs when the nucleus of a radioactive atom becomes unstable and releases energy in the form of radiation. This can happen through several different types of decay, including alpha, beta, and gamma decay.

3. What is the half-life of a radiotracer?

The half-life of a radiotracer is the amount of time it takes for half of the original amount of the isotope to decay into a more stable form. This can vary greatly depending on the specific isotope, ranging from fractions of a second to billions of years.

4. How is radiotracer decay used in scientific research?

Radiotracer decay is often used in scientific research as a way to track and measure various processes. For example, radioactive isotopes can be used to trace the movement of molecules in biological systems or to determine the age of archaeological artifacts.

5. Is radiotracer decay dangerous?

While exposure to radiation can be harmful, most radiotracer decay processes are relatively safe when handled properly. The amount of radiation emitted by a radiotracer is typically very small and decays quickly, so the risk of harm is minimal. However, it is important to follow proper safety protocols when working with radioactive materials.

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