Question

For technetium-99 with a half-life of 6 hours, what percent of the radioactivity would remain after 2 days (48 hours)?

a) 6.25%
b) 12.5%
c) 25%
d) 50%

Answer 1

After calculating the number of half-lives for technetium-99 in 48 hours, we use the radioactive decay expression to determine that less than 0.4% of the original radioactivity would remain, which is not an option listed in the question.

Step-by-step explanation:

The question asks us to calculate the remaining radioactivity of technetium-99 after 2 days or 48 hours. With a half-life of 6 hours, we can determine the remaining radioactivity by calculating how many half-lives have passed in the given time period and then applying the concept of radioactive decay.

First, we find out how many half-lives occur in 48 hours:

48 hours ÷ 6 hours/half-life = 8 half-lives.

Next, we use the expression for radioactive decay: after 1 half-life, 50% remains; after 2 half-lives, 25% remains; and so on. We halve the remaining amount of radioactivity with each successive half-life. So, after 8 half-lives, the math works as follows:

1. After 1 half-life: 50%
2. After 2 half-lives: 25%
3. After 3 half-lives: 12.5%
4. After 4 half-lives: 6.25%
5. After 5 half-lives: 3.125%
6. After 6 half-lives: 1.5625%
7. After 7 half-lives: 0.78125%
8. After 8 half-lives: 0.390625%

Therefore, after 48 hours, which is 8 half-lives, the remaining radioactivity of technetium-99 will be 0.390625%, which is less than any of the options provided in the question, suggesting it may contain an error or required approximation.