Final answer:
The ratio of combined activity of an equimolar mixture of two radioactive substances A and B after 12 hours to its initial activity is 9/64, which is determined by applying the concept of half-life for each substance. Option A is correct.
Step-by-step explanation:
The question concerns the half-life of two radioactive substances, A and B, and asks for the ratio of the activity after 12 hours to the initial activity of an equimolar mixture of A and B. To solve this, we apply the concept of half-life which states that after one half-life, half of the radioactive nuclei will remain.
For substance A with a half-life of 2 hours, after 12 hours, which is 6 half-lives, the fraction remaining will be ½6 or 1/64. For substance B with a half-life of 4 hours, after 12 hours, which is 3 half-lives, the fraction remaining will be ½3 or 1/8.
As the mixture is equimolar, the initial activity ratio is 1:1. After the decay, the total activity is the sum of the activities of A and B. Thus, A's activity contributes 1/64 and B's activity contributes 1/8 of their respective initial activities.
The combined activity of both substances after 12 hours will be the sum: (1/64) + (1/8). Converting 1/8 to the same denominator as 1/64 gives us 8/64. Adding these up, we get (1/64) + (8/64) = 9/64.
Therefore, the ratio of the combined activity after 12 hours to the initial activity of an equimolar mixture of A and B is 9/64, which matches option A in the multiple-choice question.