Question

An important consideration of the medical use of radioisotopes is how long they're active. Technetium-99 is a short-lived isotope of technetium with a half-life of 6 hours. If a patient is treated with technetium-99 for a PET scan, when will the patient have less than 1 percent of the original technetium-99 present?

Answer 1

To determine when the patient will have less than 1 percent of the original technetium-99 present, we can use the concept of half-life.

The half-life of technetium-99 is 6 hours, which means that every 6 hours, half of the technetium-99 present will decay. To find when less than 1 percent remains, we can set up the following equation:

(1/2)^n = 1/100

Here, "n" represents the number of half-lives required for the remaining amount to be less than 1 percent (1/100).

Now, let's solve for "n":

(1/2)^n = 1/100

Take the natural logarithm (ln) of both sides to solve for "n":

ln((1/2)^n) = ln(1/100)

Using the properties of logarithms:

n * ln(1/2) = ln(1/100)

Now, divide both sides by ln(1/2):

n = ln(1/100) / ln(1/2)

Using a calculator:

n ≈ 6.64

Since "n" represents the number of half-lives, and we can't have a fraction of a half-life, we need to round up to the nearest whole number because we want the patient to have less than 1 percent remaining. So, the patient will have less than 1 percent of the original technetium-99 present after approximately 7 half-lives.

Now, we know that the half-life of technetium-99 is 6 hours, so to find the time it takes for 7 half-lives:

7 * 6 hours = 42 hours

So, the patient will have less than 1 percent of the original technetium-99 present after approximately 42 hours.