Question

Technetium-99m is a radioactive isotope frequently employed in medical applications as a radioactive tracer. A radioactive tracer involves introducing an isotope into the body to facilitate the generation of diagnostic images for health condition assessment. Technetium-99m possesses a half-life of 6 hours. If a patient is administered a dose of technetium-99m in the morning, what approximate percentage of the technetium-99m will remain within the patient's body after 12 hours?

Answer 1

To determine the percentage of technetium-99m remaining in the body after 12 hours, we can use the concept of half-lives.

Given that technetium-99m has a half-life of 6 hours, we can determine how many half-lives have passed in 12 hours:

\[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{12 \text{ hours}}{6 \text{ hours}} = 2 \]

After each half-life, the remaining quantity of the isotope is halved. Starting with 100% of the original dose, we can calculate the remaining percentage after 2 half-lives as follows:

\[ \text{Remaining percentage after 1 half-life} = 100% \times \frac{1}{2} = 50% \]

\[ \text{Remaining percentage after 2 half-lives} = 50% \times \frac{1}{2} = 25% \]

Thus, after 12 hours, approximately 25% of the technetium-99m administered to the patient will remain in their body.