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To determine the percentage of Technetium-99m that will be left in the patient's body 12 hours later, we first need to determine how many half-lives have passed.
12 hours is equivalent to 12/6 = 2 half-lives.
Each half-life represents a reduction to half of the initial amount. Therefore, after 2 half-lives, the amount of Technetium-99m left in the body will be 1/2 * 1/2 = 1/4 of the initial amount.
This is equivalent to 1/4 * 100% = 25%.
Therefore, the answer is (a) 25.0%.