A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Which radioactive isotope is added to glucose for positron emission tomography (PET)? carbon-14 fluorine-18 cobalt-60 technitium-99

  • This Question is Closed
  1. Rushwr
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What do u think ?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Maybe Carbon-14?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Rushwr

  4. cuanchi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    https://en.wikipedia.org/wiki/Fludeoxyglucose_(18F)

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So it's flourine? @Cuanchi

  6. cuanchi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it looks like that. It is what I found. I would not use C14 since the half life is too long and I dont know if it can decay producing positron. C14 decay by beta decay to produce N14 that is stable and non radioactive isotope

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.