Final answer:
Using the half-life of ¹⁰Sr, which is 28.5 years, and calculating 1.4 half-lives from 1960 to 2000, the remaining radioactivity percentage is found to be 37.8%, which corresponds to option (b).
Step-by-step explanation:
To calculate the percentage of radioactivity due to ¹⁰Sr remaining in 2000 after nuclear testing stopped in 1960, we will use the concept of half-life, which is the time required for half of the radioactive atoms in a sample to decay. The half-life of ¹⁰Sr is given as 28.5 years. To determine the number of half-lives that have passed from 1960 to 2000, we divide the number of years by the half-life period:
(2000 - 1960) / 28.5 ≈ 1.4 half-lives
The formula to calculate the remaining amount of a radioactive isotope after a certain number of half-lives is:
Remaining amount = Initial amount * (1/2)^number of half-lives
Assuming the initial amount is 100% at the time the nuclear testing stopped, the remaining radioactivity due to ¹⁰Sr is then:
Remaining radioactivity = 100% * (1/2)^1.4 ≈ 100% * 0.378 ≈ 37.8%
Thus, the correct option is (b) 37.8%.