5 mg iodine 131 its half life is 8.1 days what is the time when 1% remain

k = 0693/t1/2

Solve for k and substitute into the equation below.

ln(No/N) = kt
Use 5mg for No.
Use 1% x 5 = 0.05 for N.
k from above.
Solve for t.
t will be in days if you used 8.1 days for the half-life.

To determine the time when 1% of iodine-131 remains, we can use the concept of half-life. The half-life of iodine-131 is given as 8.1 days. This means that every 8.1 days, the amount of iodine-131 will be halved.

Let's break down the problem step by step:

1. Start with 100% (or the initial amount) of iodine-131, which is 5 mg.

2. After one half-life (8.1 days), half of the iodine-131 will remain, which is 50% or 0.5 mg.

3. After two half-lives (2 x 8.1 days), another half of the remaining iodine-131 will decay. So, from the previous step, we have 0.5 mg remaining, and half of that will be 0.25 mg.

4. Continue this process until we reach 1% of the initial amount:

- After three half-lives (3 x 8.1 days), we will have 0.5 x 0.5 = 0.25 mg remaining.
- After four half-lives (4 x 8.1 days), we will have 0.25 x 0.5 = 0.125 mg remaining.
- After five half-lives (5 x 8.1 days), we will have 0.125 x 0.5 = 0.0625 mg remaining.

5. As we can see, after five half-lives, which is equivalent to (5 x 8.1) = 40.5 days, only 0.0625 mg, or 1% of the initial amount, remains.

Therefore, the time when 1% of the iodine-131 remains is approximately 40.5 days.