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
6.25%
b
93.8%
c
25.0%
d
12.5%
The correct answer is d) 12.5%.
To calculate the percentage of technetium-99m left in the patient's body after 12 hours, divide the time elapsed by the half-life and raise it to the power of the number of half-lives.
In this case, 12 hours is equal to 2 half-lives (12/6 = 2).
Using the formula:
Percentage remaining = (1/2)^(number of half-lives) * 100
Plug in the values:
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.
To determine the percentage of technetium-99m left in the patient's body 12 hours later, we need to calculate the number of half-lives that have occurred in that time period.
Given that technetium-99m has a half-life of 6 hours, we can divide the 12-hour time period by the half-life to find the number of half-lives that have occurred:
12 hours / 6 hours = 2 half-lives
Each half-life reduces the amount of technetium-99m by half. So, after 2 half-lives, we have:
1/2 * 1/2 = 1/4 or 25% of the original amount remaining.
Therefore, the correct answer is option c) 25%.
To determine the percentage of technetium-99m that will be left in the patient's body 12 hours later, we need to calculate the number of half-lives that have passed.
The half-life of technetium-99m is 6 hours. Therefore, in 12 hours, 2 half-lives will have passed (12 hours / 6 hours per half-life).
Each half-life represents a 50% decay of the initial amount. So, after 2 half-lives, the percentage remaining can be calculated as:
(0.5)^(number of half-lives) = (0.5)^2 = 0.25 = 25%
Therefore, about 25% of the technetium-99m will be left in the patient's body 12 hours later.
The correct answer is: c) 25.0%