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ACS General Chemistry 2 Practice Exam

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The half-life for first-order radioactive decay of 32P is 14.3 days. How many days would be required for the activity to decrease to 20.0% of its initial value?
Correct Answer:
33.0 days
Explanation:
In first-order decay, activity falls exponentially with time, and the fraction remaining ties directly to time via N/N0 = (1/2)^{t/t1/2}. Here, we want the activity to be 20% of its initial value, so 0.20 = (1/2)^{t/14.3}. Taking natural logs gives t/14.3 = ln(0.20)/ln(0.50). Using ln(0.20) ≈ -1.6094 and ln(0.50) ≈ -0.6931, we get t ≈ 14.3 × (2.322) ≈ 33.2 days. So about 33 days. Intuitively, two half-lives (about 28.6 days) would leave 25% remaining; reaching 20% requires a bit more time, which matches a little over 33 days. The other time options correspond to significantly more or fewer half-lives and would produce fractions well below 20%.

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