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Half life Tc-99m injected into a patient

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A patient comes into the hospital for a bone scan and is injected with a dye containing Tc-99m. The half-life of Tc-99m is 6.03 h. What fraction of the original technetium will remain the patient 36 h after the procedure if radioactive decay is the only means by which it is removed?

    2. Relevant equations




    3. The attempt at a solution

    Af = Ao (1/2)t/h
    Af = 99 (1/2)36/6.03
    Af = 1.579214746

    1.57921476 / 98 = 0.016

    0.016 x 100 = 1.6 %

    I am getting 1.6 % but the answer key states the answer as 0.016 %. What am I doing wrong. Is the m after Tc-99m play a role in this that I'm missing?
     
    Last edited: Apr 28, 2013
  2. jcsd
  3. Apr 28, 2013 #2

    SteamKing

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    No. Assume the original amount of TC-99 which is injected is 1.
     
  4. Apr 28, 2013 #3
    But if its a ratio why does it matter what value I put in for the original amount? For example, for every other question involving half life I have used the amu given, i.e. Aluminium-30 I used 30 amu as the original and it always gives the right answer. So what is so unique about this question?
     
  5. Apr 28, 2013 #4

    SteamKing

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    "What fraction of the original technetium will remain ... after 36 hours ..." So if the original amount of the sample is 100% of the total amount injected, then what fraction will remain after 36 hours have elapsed?
     
  6. Apr 28, 2013 #5

    haruspex

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    Why are you substituting 99 there? That's not the amount of radioactive material. You want Af/Ao, right?
     
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