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?(1 point)

Question

Responses

93.8%
93.8%

12.5%
12.5%

6.25%
6.25%

25.0%

Answer 1

To calculate the remaining percentage of technetium-99m after 12 hours, we can use the half-life formula.

The half-life of technetium-99m is 6 hours. Therefore, after each half-life, half of the substance remains.

After 6 hours (1 half-life), 50% remains. After 12 hours (2 half-lives), the calculation is as follows:

• Start with 100%.
• After the first half-life (6 hours): 100% / 2 = 50%.
• After the second half-life (another 6 hours): 50% / 2 = 25%.

Thus, after 12 hours, 25% of the technetium-99m will remain in the patient's body.

The correct answer is 25.0%.