Max's observations suggest that the amount of radioactive iodine decreases over time following a predictable pattern. Let's analyze this step by step:
1. Initial Observation:
- At time [tex]\( t = 0 \)[/tex] days, the amount of iodine is 100 grams.
2. After 8 days:
- The amount of iodine in the container is 50 grams. This indicates that the iodine amount reduces to half its initial value in 8 days.
Given this observation, we infer that the half-life period of the radioactive iodine is 8 days.
3. After 16 days:
- We need to calculate the amount of iodine left after 16 days.
- Since the half-life period is 8 days, after one half-life (8 days), the amount is halved to 50 grams.
- After another half-life (another 8 days), the amount will halve again, so after 16 days, the amount of iodine will be:
[tex]\[
\frac{50}{2} = 25 \text{ grams}
\][/tex]
4. After 24 days:
- Similarly, after 24 days (which is three half-life periods of 8 days each), the amount of iodine is:
[tex]\[
100 \times \left( \frac{1}{2} \right)^3 = 100 \times \frac{1}{8} = 25 \text{ grams}
\][/tex]
5. After 32 days:
- We need to calculate the amount of iodine left after 32 days, which is four half-life periods.
- The amount of iodine is:
[tex]\[
100 \times \left( \frac{1}{2} \right)^4 = 100 \times \frac{1}{16} = 6.25 \text{ grams}
\][/tex]
Therefore, based on the observations and calculations, the correct inference is:
- After 32 days, the amount of iodine left in the container will be 6.25 grams.
This matches the fourth option in the question. Thus, Max's most likely inference would be:
- After 32 days, the amount of iodine left in the container will be 6.25 grams.
