Consider a radioactive sample with a half-life of one week. How much of the original sample will be left at the end of the second week? The third week? The fourth week?

1st week: 1/2

2nd week: 1/4
3rd week: 1/8
...

Thank you, so if I am understanding this correctly, each cycle is being cut each week by half.

yes. the half-life is how long it takes for the remaining amount to go down by 1/2. Since the half-life is given as 1 week, that means that each week there is only half as much at the end of the week as there was at the beginning of that week.

Thank you for your help.

Consider a radioactive sample with a half-life of one week.

How much of the original sample will be left at the end of the second week?
The third week? 3rd week n will be equal to 3, so amount left = 123 = 0.125 = 12.5%
The fourth week?

1/16

To determine how much of the original radioactive sample will be left at the end of each week, we need to understand the concept of half-life. The half-life of a radioactive substance is the amount of time it takes for half of the substance to decay. In this case, the half-life of the radioactive sample is one week.

To calculate the remaining amount of the sample at the end of each week, we can use the formula:

Remaining Amount = Initial Amount x (1/2)^(number of half-lives)

Let's calculate the remaining amount at the end of the second, third, and fourth weeks.

1. At the end of the second week:
- The number of half-lives is 2 because two weeks have passed.
- The remaining amount can be calculated as:
Remaining Amount = Initial Amount x (1/2)^2

2. At the end of the third week:
- The number of half-lives is 3 because three weeks have passed.
- The remaining amount can be calculated as:
Remaining Amount = Initial Amount x (1/2)^3

3. At the end of the fourth week:
- The number of half-lives is 4 because four weeks have passed.
- The remaining amount can be calculated as:
Remaining Amount = Initial Amount x (1/2)^4

To find the solution, you'll need to know the initial amount of the radioactive sample.