After 5 half-lives have passed for 87Rb, what percentage of daughter isotopes would be present?

• A. 87.5%
• B. 50%
• C. 75%
• D. 96.9%

Answer: (D)

Verified step by step guidance

1. Understand that after each half-life, half of the remaining parent isotope decays into the daughter isotope. The fraction of the parent isotope remaining after n half-lives is given by the formula: \(\left(\frac{1}{2}\right)^n\).
2. Identify the number of half-lives passed, which is 5 in this problem. Substitute n = 5 into the formula to find the fraction of the parent isotope remaining: \(\left(\frac{1}{2}\right)^5\).
3. Calculate the fraction of the parent isotope remaining after 5 half-lives, which represents the amount of 87Rb still present.
4. Determine the fraction of daughter isotopes formed by subtracting the fraction of the parent isotope remaining from 1: \$1 - \left(\frac{1}{2}\right)^5$.
5. Convert the fraction of daughter isotopes to a percentage by multiplying by 100 to find the percentage of daughter isotopes present after 5 half-lives.