Selenium-75, a radioactive isotope, has a half-life of 120 days. Given that radioactive decay is a first-order process, how many days does it take for 80% of the 75Se to disappear?

17B

Question

Selenium-75, a radioactive isotope, has a half-life of 120 days. Given that radioactive decay is a first-order process, how many days does it take for 80% of the 75Se to disappear?

17B

BlackTom AI · Accepted Answer

First, identify the given data and the goal: a first-order radioactive decay with half-life Th = 120 days, and we want the time t when 80% has disappeared, meaning 20% remains (N/N0 = 0.20).
Option 1 (39 days): To check this, use the formula N/N0 = (1/2)^(t/Th). Solving for t gives t = Th * log(0.20) / log(0.50). Numerically, log(0.20) ≈ -0.69897 and log(0.50) ≈ -0.30103, so t ≈ 120 * (-0.69897)/(-0.30103) ≈ 120 * 2.32 ≈ 278.4 days. If you instead plug into the 0.80 remaining scenario (since 80% decayed), you’d get t ≈ 39 days, which corresponds to only 20% decayed, not 80%. This shows why 39 days does not fit the requirement of 80% decayed.
Option 2 (279 days): As calculated above, t ≈ 278–279 days yields N/N0 ≈ 0.20, i.e., 80% decayed. This aligns exactly with the target condition for a first-order process with a 120-day half-life.
Option 3 (96 days): Plugging t = 96 into the formula gives N/N0 = (1/2)^(96/120) = (1/2)^(0.8) ≈ 0.574. That means about 42.6% has decayed, not 80%. Thus this time is far too short to reach 80% disappearance.
Option 4 (72 days): Here N/N0 = (1/2)^(72/120) = (1/2)^(0.6) ≈ 0.659, so roughly 34.1% decayed—still far from 80%—so this duration cannot produce 80% disappearance.
Option 5 (192 days): For t = 192 days, N/N0 = (1/2)^(192/120) = (1/2)^(1.6) ≈ 0.325, meaning about 67.5% decayed. While substantial, it does not reach the targeted 80% disappearance.
In summary, only option 279 days yields a remaining fraction around 0.20, consistent with 80% of the substance having decayed under a first-order decay with a 120-day half-life.

Selenium-75, a radioactive isotope, has a half-life of 120 days. Given that radioactive decay is a first-order process, how many days does it take for 80% of the 75Se to disappear? 17B

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