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needingtoknow
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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?
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