Answer:
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.