Question about PET scans (Positron emission tomography)

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Why there are no free radicals formed with PET scan?
I understand that a PET scan will produce Positrons which will come into contact with an electron and produce Gamma rays in the area where there is a high uptake of sugar (assume glucose). In this process of annihilating an electron, some poor atom will lose an electron which I assume would produce a free radical(s). I assume would produce a large number of free radicals.
So what is the real story? PET scan is supposedly safe but I don't understand how!
thanks
 

.Scott

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The radionucleotide used when mimicking glucose is Florine-18. Its decay products include Oxygen and a positron. So there is not much ionization due to the electron it absorbs - because going from Florine to Oxygen yields a "spare" electron.

It is ionizing, and some free radicals can be created - but not exactly by the mechanism you describe.

According to this article: http://www.radioactivity.eu.com/site/pages/Doses_Diagnostics.htm
The positron emission tomography (PET) is still rare compared to scintigraphy. Most of these tests use as a marker 18F-FDG whose irradiating character is estimated by the ICRP – for an adult - to 19 µSv / MBq. For a standard PET (injected activity of 10 millicuries or 370 MBq), the resulting dose is 7 mSv, about the same as for a scanner. If tomography is performed in conjunction with a scanner (PET-SCAN) the dose from the scanner must be added.
That would be the same radiation damage you would expect to get in about 2 years from natural background radiation.

Whether or not it is "safe" depends on what you are comparing it to. In most cases, someone getting a PET scan has more serious conditions to worry about than 7 mSv.
 
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there is much ionization due to the electron it absorbs - because going from Florine to Oxygen yields a "spare" electron
I think you mean there is not much ionization, because the positron emitted in the nuclear reaction will most likely meet up with the electron that now no longer has an atomic orbital because oxygen has one less atomic electron than fluorine. To a first approximation this should produce zero ionization; the complete reaction is fluorine + 9 electrons -> oxygen + 8 electrons + positron + spare electron -> oxygen + 8 electrons (i.e., a neutral oxygen atom) + gamma rays.
 

.Scott

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I think you mean there is not much ionization, because the positron emitted in the nuclear reaction will most likely meet up with the electron that now no longer has an atomic orbital because oxygen has one less atomic electron than fluorine. To a first approximation this should produce zero ionization; the complete reaction is fluorine + 9 electrons -> oxygen + 8 electrons + positron + spare electron -> oxygen + 8 electrons (i.e., a neutral oxygen atom) + gamma rays.
Yes, that is what I meant. I have corrected the post.
 
Thanks for such a rapid response. So, since the reactions take place where there is more uptake there is most likely a canceling out with positron and the spare electron. I didn’t know about the decay that produces Oxygen.

The reason I asked this was that I had a PET scan a couple of years ago and found it like SIFI!

I need to do lots of reading to further understand everything. I did have Chemistry back in Junior high and then again in High School but that was back in 1965. It seems that Chemistry has become a branch of Physics.

As a Social Implications of Technology, I think that the public needs to have a better understanding and appreciation of Technology.
 

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