Technetium-99m is an ideal radioisotope for scanning organs because it has a half-life of 6.0 hr and is a pure gamma emitter. Suppose that 200mg were prepared in the technetium generator this morning. How many milligrams would remain after the following intervals?

1- One half life.
2- two half-lives
3- 18 hr
4- 30 hr

k = 0.693/t1/2

Then
ln(No/N) = kt
No = 200 mg
N = solve for this
k from above
t time i hours.

To calculate the amount of Technetium-99m remaining after a certain time interval, we can use the formula:

Amount remaining = Initial amount × (1/2)^(elapsed time / half-life)

Let's calculate the amount remaining for each time interval:

1- One half-life:
The half-life of Technetium-99m is 6.0 hours. Therefore, after one half-life (6.0 hours), the fraction remaining is (1/2)^(6.0/6.0) = 1/2.
Thus, the amount remaining would be 200 mg × 1/2 = 100 mg.

2- Two half-lives:
After two half-lives (2 × 6.0 hours = 12.0 hours), the fraction remaining is (1/2)^(12.0/6.0) = 1/4.
Therefore, the amount remaining would be 200 mg × 1/4 = 50 mg.

3- 18 hours:
Since 18 hours is exactly three half-lives (3 × 6.0 hours), the fraction remaining is (1/2)^(18.0/6.0) = 1/8.
Hence, the amount remaining would be 200 mg × 1/8 = 25 mg.

4- 30 hours:
Since 30 hours is exactly five half-lives (5 × 6.0 hours), the fraction remaining is (1/2)^(30.0/6.0) = 1/32.
Therefore, the amount remaining would be 200 mg × 1/32 = 6.25 mg.

Therefore, the amount of Technetium-99m remaining after each interval is:
1- One half-life: 100 mg
2- Two half-lives: 50 mg
3- 18 hours: 25 mg
4- 30 hours: 6.25 mg.