Question

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 6.25% c 93.8% d 12.5%

Answer 1

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

The half-life of Technetium-99m is 6 hours, which means that every 6 hours, half of the remaining isotope decays.

We can calculate the number of half-lives that have passed in 12 hours:

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

Now we can apply the decay formula. After each half-life, the amount of the isotope left can be calculated as:

\[ \text{Remaining amount} = \left(\frac{1}{2}\right)^n \times 100% \]

where \( n \) is the number of half-lives.

Substituting \( n = 2 \):

\[ \text{Remaining amount} = \left(\frac{1}{2}\right)^2 \times 100% = \frac{1}{4} \times 100% = 25% \]

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

The answer is (a) 25.0%.