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
25.0%
b
12.5%
c
6.25%
d
93.8%
To determine the percentage of Technetium-99m that will be left in the patient's body 12 hours later, we can use the concept of half-life.
The half-life of Technetium-99m is given as 6 hours. This means that after 6 hours, half of the original amount of Technetium-99m would have decayed, leaving behind 50% of the initial amount.
Since 12 hours have passed, we can calculate how many half-lives occurred in this time by dividing 12 by 6. In this case, two half-lives have occurred.
After the first half-life of 6 hours, there would be 50% of the Technetium-99m remaining.
After the second half-life of another 6 hours, there would be 50% of the previous 50% remaining.
To find the total percentage remaining after 12 hours, we multiply the two percentages together: 50% * 50% = 25% = 0.25.
Therefore, approximately 25% of the Technetium-99m will be left in the patient's body 12 hours later. So, the correct answer is option a) 25.0%.