Chemistry · Semester B TEKS 14A-14C
HardCalcWord
A radioactive sample decays from 100.0 g to 12.5 g over exactly 21.6 days, with no half-life value given directly. Determine the isotope's half-life, working backward from the observed decay fraction to the number of half-lives elapsed.
A7.2 days, from recognizing 12.5/100.0 = 1/8 = 3 half-lives, then dividing the 21.6-day total by 3
B2.7 days, mistaking the denominator of the remaining fraction 1/8 for a direct count of 8 half-lives
C≈10.8 days, treating the decay as a simple linear relationship between percent mass lost and elapsed time
D21.6 days, reporting the total elapsed time itself as the half-life without recognizing multiple half-lives occurred
Explanation
The fraction remaining is 12.5/100.0 = 0.125 = 1/8, which corresponds to exactly 3 half-lives, since (1/2)³ = 1/8. Since 3 half-lives took 21.6 days total, one half-life is 21.6 ÷ 3 = 7.2 days. Dividing 21.6 days by 8 instead of by 3 (mistaking the DENOMINATOR of the remaining fraction 1/8 for the number of half-lives, rather than correctly solving (1/2)ⁿ=1/8 for n=3) gives an incorrect half-life of 21.6÷8 = 2.7 days. Treating the decay as though it followed a simple linear relationship between mass lost and time (reasoning that 87.5% of the mass was lost, so the 'half-life' should be roughly half of 21.6 days ≈ 10.8 days) misapplies linear reasoning to what is fundamentally an exponential decay process. Reporting 21.6 days itself as the half-life ignores that MULTIPLE half-lives, not just one, were needed to reach the observed 1/8 remaining fraction.
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