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
12.5%
b
6.25%
c
93.8%
d
25.0%
To determine the percentage of technetium-99m left in the patient's body 12 hours later, we need to determine the number of half-lives that have passed in that time period.
Since the half-life of technetium-99m is 6 hours, there are 12 hours / 6 hours = 2 half-lives that have passed.
Each half-life reduces the amount of technetium-99m by half.
Therefore, after 2 half-lives, the percentage of technetium-99m left in the patient's body is (1/2)*(1/2) = 1/4 = 25%.
Thus, the answer is (d) 25%.
To determine the percentage of technetium-99m left in the patient's body 12 hours later, we need to calculate how many half-lives have occurred during that time.
Since technetium-99m has a half-life of 6 hours, we can divide the 12-hour interval by the half-life to find out how many half-lives have passed.
12 hours / 6 hours = 2 half-lives
Each half-life reduces the amount of technetium-99m by half. Thus, after 2 half-lives, the remaining percentage can be calculated as:
(1/2) * (1/2) = 1/4 = 0.25 = 25%
So, about 25% of the technetium-99m will be left in the patient's body 12 hours later.
Therefore, the correct option is (d) 25.0%.
To determine the percentage of technetium-99m remaining in the patient's body 12 hours later, we can use the concept of half-life.
The half-life of technetium-99m is 6 hours, which means that after 6 hours, half of the initial amount of technetium-99m will decay.
Since 12 hours is equal to two half-lives (6 hours * 2), we can use the formula: Percentage Remaining = (1/2)^n * 100%, where n is the number of half-lives.
In this case, n = 2, as we have two half-lives.
Substituting the values into the formula:
Percentage Remaining = (1/2)^2 * 100%
Percentage Remaining = (1/4) * 100%
Percentage Remaining = 25%
Therefore, about 25% of the technetium-99m will be left in the patient's body 12 hours later.
The correct answer is (d) 25.0%.