Half life Tc-99m injected into a patient

[needingtoknow]

Homework Statement

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?

Homework Equations

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?

 

[SteamKing]

No. Assume the original amount of TC-99 which is injected is 1.

 

[needingtoknow]

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?

 

[SteamKing]

"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?

 

[haruspex]

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

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