A radioactive isotope has a half-life of 10 days. What is the time required for the activity of the isotope to fall to [math: 116]th of its original value?

Question

BlackTom AI · Accepted Answer

Each half-life reduces the activity by a factor of 1/2, so after n half-lives the remaining activity is (1/2)^n of the original.

1. Relevant knowledge points and concepts
- Half-life T is the time required for a quantity to fall to half its value.
- After n half-lives the fraction remaining = (1/2)^n.
- Exponential decay can also be written A = A0 (1/2)^(t/T) or using logarithms t = T · log(A/A0) / log(1/2).

2. Step-by-step reasoning
We are given T = 10 days and we want the activity to fall to 1/16 of the original, so set (1/2)^n = 1/16, where n is the number of half-lives elapsed.

3. Intermediate calculation steps
(1/2)^n = 1/16
Recognize 1/16 = 1/2^4, because 16 = 2^4.
Thus (1/2)^n = (1/2)^4, so n = 4.

Time t = n × T = 4 × 10 days = 40 days.

Alternatively using logarithms:
t = T · log(A/A0) / log(1/2)
= 10 · log(1/16) / log(1/2)
= 10 · (−log 16) / (−log 2) = 10 · (log 16 / log 2) = 10 · 4 = 40 days.

4. Why the correct answer is correct
Four half-lives are required to reduce a quantity to 1/16 (because 1/2^4 = 1/16). With each half-life 10 days long, 4 × 10 = 40 days is the required time.

5. Why the other options are incorrect
- a. 20 days: equals 2 half-lives → remaining fraction = (1/2)^2 = 1/4, not 1/16.
- c. 50 days: equals 5 half-lives → remaining fraction = (1/2)^5 = 1/32, not 1/16.
- d. 160 days: equals 16 half-lives → remaining fraction = (1/2)^16 = 1/65536, far smaller than 1/16.

6. Conclusion
The time required is 40 days, matching the provided correct answer (b. 40 days).

A radioactive isotope has a half-life of 10 days. What is the time required for the activity of the isotope to fall to [math: 116]th of its original value?单项选择题

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