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?

To determine the percentage of technetium-99m left in the patient's body after 12 hours, we need to calculate how many half-lives have passed.

Given that the half-life of technetium-99m is 6 hours, we divide 12 hours by 6 hours per half-life:

12 hours / 6 hours per half-life = 2 half-lives.

For each half-life, the amount of technetium-99m is reduced by half. Therefore, after two half-lives, the amount remaining is:

(1/2) * (1/2) = 1/4 = 0.25

So, about 25% of the technetium-99m will be left in the patient's body after 12 hours.

To determine the percentage of technetium-99m remaining in the patient's body after 12 hours, we need to calculate the number of half-lives that have elapsed.


Given that the half-life of technetium-99m is 6 hours, we can calculate the number of half-lives in 12 hours as follows:

Number of half-lives = Total time / Half-life
= 12 hours / 6 hours
= 2 half-lives

Each half-life reduces the amount of technetium-99m by half. Therefore, after 2 half-lives, we can calculate the remaining percentage as follows:

Remaining percentage = (1/2) ^ Number of half-lives * 100%
= (1/2) ^ 2 * 100%
= (1/4) * 100%
= 25%

Therefore, approximately 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 can use the concept of half-life. The half-life of technetium-99m is given as 6 hours, which means that after every 6 hours, the amount of technetium-99m will decrease by half.

First, let's calculate the number of half-lives that have passed in 12 hours. Since each half-life is 6 hours, we have:

12 hours ÷ 6 hours/half-life = 2 half-lives

So, in 12 hours, two half-lives have occurred. Now, we can calculate the percentage of technetium-99m remaining after two half-lives.

Each half-life reduces the amount of technetium-99m by half. After the first half-life, we have 50% of the original amount left. After the second half-life, we have only 50% of that remaining, which is:

50% × 50% = 25%

Thus, about 25% of the technetium-99m will be left in the patient's body 12 hours later.