205k views
1 vote
What is the half-life for the first-order decay of phosphorus-32?

a) 2.43 days
b) 14.22 days
c) 33.33 days
d) 1.43 days

1 Answer

5 votes

Final answer:

The half-life of phosphorus-32 is 14.3 days, during which time half of the substance will radioactively decay. The correct answer is (b) 14.22 days.

Step-by-step explanation:

The half-life for the first-order decay of phosphorus-32 is 14.3 days. If one starts with 100 mg of phosphorus-32, after 14.3 days, only 50 mg would remain, which is the definition of a half-life. This is a characteristic of first-order decay, where the quantity of a substance diminishes by half over a set period of time known as its half-life.

The half-life of an isotope is a constant that represents the time it takes for half of a sample of the radioactive substance to decay. Phosphorus-32 is known to have a half-life that you can look up in scientific references or reliable databases.

After a quick reference to a reliable source, one would find that the half-life of phosphorus-32 is approximately 14.29 days. Since the provided options don't include 14.29 days exactly, we should choose the option that is closest to this value.

Looking at the options: a) 2.43 days b) 14.22 days c) 33.33 days d) 1.43 days The option that is closest to the known half-life of phosphorus-32 (14.29 days) is: b) 14.22 days So, the correct answer is (b) 14.22 days.

User Ukliviu
by
7.8k points