Final answer:
The isotope S-35, with a half-life of 87 days, would have the greatest amount of the original isotope remaining after 100 days because more than half but less than a full half-life has passed, leaving more than half of the original amount.
The Correct Answer is Option .B.
Step-by-step explanation:
To determine which of the isotopes would have the greatest amount of the original isotope remaining after 100 days, we must consider the half-life of each isotope. The half-life indicates the time required for half of the isotope to decay. The isotope with the longest half-life would have the most original material remaining after a given timeframe.
- Fe-59 has a half-life of 44.5 days, so after 100 days, just over 2 half-lives would have passed, leaving less than a quarter of the original amount.
- S-35 has a half-life of 87 days, so after 100 days, more than half but less than a full half-life has passed, leaving more than half of the original amount.
- P-32 has a half-life of 14.3 days, so after 100 days, numerous half-lives have passed, leaving only a tiny fraction of the original. Similarly, Cu-64 with a half-life of 12.7 hours and I-131 with a half-life of 8.0 days would have very little of the original isotope remaining.
Therefore, the answer is B) 1.0 g of S-35 after 100 days (half-life = 87 days), as it has the longest half-life and hence, would retain the most original isotope after 100 days.