Final answer:
The half-life of phosphorus-32 is 14 days. This is determined by the data given that a 2.0 gram sample decayed to 0.25 grams after 42 days, which represents three half-lives of decay.
Step-by-step explanation:
The question is asking to calculate the half-life of phosphorus-32 given that a sample of phosphorus-33 decayed from 2.0 grams to 0.25 grams after 42 days. To solve this problem, we need to understand the concept of a half-life. The half-life of a radioactive isotope is the time it takes for half of the isotope in a sample to decay. From the provided information, we know that phosphorus-32 has a half-life of 14.3 days. This means that if you start with 100 mg of phosphorus-32, you will have 50 mg left after 14.3 days, and 25 mg left after 28.6 days, and so on.
When we apply this concept to the student's problem with phosphorus-33 (noting the typo in the original question, which should refer to phosphorus-32), we can determine how many half-lives have elapsed during the 42-day period. If a 2.0 gram sample decays to 0.25 grams, this represents three half-lives (2.0 grams to 1.0 gram, then to 0.5 grams, and finally to 0.25 grams). Since 42 days encompass three half-lives, we divide 42 days by 3 to find that the half-life of phosphorus-32 is 14 days.