In 168 seconds the activity of a radioisotope falls to [math: 18] th of its original value. What is its half-life?

Question

BlackTom AI · Accepted Answer

Radioactive decay follows A = A0*(1/2)^(t/T1/2), so reducing to 1/8 means a whole-number of half-lives has elapsed.

1) Relevant concepts:
- Half-life T1/2 is the time for activity to drop by a factor of 2.
- After n half-lives the remaining activity is (1/2)^n of the original.
- Exponential form: A/A0 = (1/2)^(t/T1/2).

2) Step-by-step reasoning:
Set A/A0 = 1/8 and t = 168 s. Using the half-life relation,
(1/2)^(t/T1/2) = 1/8.

Recognize 1/8 = (1/2)^3, so the exponent must equal 3:
t/T1/2 = 3.

3) Intermediate calculation:
Solve for T1/2:
T1/2 = t / 3 = 168 s / 3 = 56 s.

(Alternatively using logarithms: t/T1/2 = ln(1/8)/ln(1/2) = (-ln 8)/(-ln 2) = ln 8/ln 2 = 3, same result.)

4) Why the correct answer is correct:
If the half-life is 56 s, three half-lives occur in 168 s, and (1/2)^3 = 1/8 of the original activity remains, matching the given condition.

5) Why the other options are incorrect:
- a. 42 s: 168/42 = 4 half-lives → remaining fraction (1/2)^4 = 1/16, not 1/8.
- b. 21 s: 168/21 = 8 half-lives → (1/2)^8 = 1/256, not 1/8.
- d. 84 s: 168/84 = 2 half-lives → (1/2)^2 = 1/4, not 1/8.

Conclusion: The half-life is 56 seconds, so option c. 56 seconds is correct. This matches the provided answer.

In 168 seconds the activity of a radioisotope falls to [math: 18] th of its original value. What is its half-life?单项选择题

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