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Question 10.5: A patient is injected with a sample of technetium-99m (t 1/2...

A patient is injected with a sample of technetium-99m (t_{1 / 2} = 6.0 h), which has an activity of 32 mCi. What activity is observed after 12 h?

Analysis

First determine how many half-lives have occurred in the time interval. Then, multiply the initial activity by the number of half-lives to determine the remaining activity.

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[1] Use the half-life of technetium-99m as a conversion factor to convert the number of hours to the number of half-lives.

12  \not h \times \frac{1 \text { half-life }}{6.0  \not h}=2.0 \text { half-lives }

[2] For each half-life, multiply the initial activity by 1/2 to obtain the final activity.

10.5

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