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?

Question

a
93.8%
b
25.0%
c
6.25%
d
12.5%

Answer 1

To calculate the remaining percentage of Technetium-99m in the patient's body after 12 hours, we need to determine how many half-lives have elapsed in that time. Given that Technetium-99m has a half-life of 6 hours:

1. Calculate the number of half-lives in 12 hours: \[ \text{Number of half-lives} = \frac{12 \text{ hours}}{6 \text{ hours/half-life}} = 2 \text{ half-lives} \]

2. Calculate the remaining amount after each half-life:

   • After 1 half-life (6 hours), 50% remains.
   • After 2 half-lives (12 hours), the remaining amount is: \[ 50% \times \frac{1}{2} = 25% \]

Therefore, after 12 hours, about 25.0% of the technetium-99m will be left in the patient's body.

The correct answer is b) 25.0%.