To determine the percentage of technetium-99m left in the patient's body 12 hours later, we need to determine the number of half-lives that have passed in that time period.
Since the half-life of technetium-99m is 6 hours, there are 12 hours / 6 hours = 2 half-lives that have passed.
Each half-life reduces the amount of technetium-99m by half.
Therefore, after 2 half-lives, the percentage of technetium-99m left in the patient's body is (1/2)*(1/2) = 1/4 = 25%.
Thus, the answer is (d) 25%.
Technetium-99m is a radioactive isotope commonly used in medicine as a radioactive tracer. A radioactive tracer is an isotope injected into the body to help create images for diagnosis of health problems. Technetium-99m has a half-life of 6 hours. If a patient receives a dose of technetium-99m one morning, about what percentage of the technetium-99m will be left in the patient's body 12 hours later?
a
12.5%
b
6.25%
c
93.8%
d
25.0%
1 answer
4 answers