Final answer:
Correct option: (a) 1/8.
The question involves calculating the initial ratio of the number of nuclei of two radioactive substances A and B, where substance A's half-life is twice that of B. After three half-lives of A, the number of nuclei are equal, leading to the conclusion that the initial ratio NB/NA is 1/8.
Step-by-step explanation:
The problem involves understanding the concept of half-life in radioactive decay.
The half-life is the duration required for half of the radioactive substance to decay. If the half-life of substance A is twice that of substance B, and after three half-lives of substance A, the number of nuclei of A and B are equal, then we can calculate the remaining nuclei of both substances to find the initial ratio NB/NA.
Since three half-lives of A have passed, the number of remaining A nuclei is given by NA*(1/2)^3 = NA/8. During this same duration, six half-lives of substance B would have occurred (since the half-life of B is half that of A), so the number of remaining B nuclei is given by NB*(1/2)^6 = NB/64.
At this point, NA/8 = NB/64, so solving for the ratio NB/NA would give us 1/8.
This allows us to conclude that the correct answer is (a) 1/8.