Radioactive decay problem: If 75% of a radioactive sample decays in 6 hours (i.e., only 25% remains), what is the half-life of the nuclide, expressed in hours?

Introduction / Context:Half-life problems are staple quantitative questions in nuclear science. They require translating a percentage of remaining material into an exponential-decay relationship to solve for the characteristic time constant—the half-life (t_1/2).

Given Data / Assumptions:

• After 6 hours, 75% has decayed, so 25% remains.
• Decay is first order: N(t) = N0 * (1/2)^(t / t_1/2).
• No growth or additional sources; closed system for the nuclide.

Concept / Approach:Set the fraction remaining equal to (1/2)^(t / t_1/2). Solve algebraically for t_1/2 by taking logarithms or matching powers of 1/2. Because 25% = 1/4 = (1/2)^2, the exponent reveals how many half-lives elapsed.

Step-by-Step Solution:Write remaining fraction: N/N0 = 0.25.Express 0.25 as power of 1/2: 0.25 = (1/2)^2.Equate exponents: (1/2)^(t / t_1/2) = (1/2)^2 ⇒ t / t_1/2 = 2.Solve for half-life: t_1/2 = t / 2 = 6 / 2 = 3 hours.

Verification / Alternative check:Two half-lives reduce a sample to 25%: After first half-life (3 h) → 50%; after second (another 3 h) → 25%. This matches the 6-hour condition, confirming t_1/2 = 3 h.

Why Other Options Are Wrong:3/4 or 1/6 hours: not consistent with two full half-lives in 6 hours.4 or 6 hours: would leave fractions of 0.3536 or 0.5 after 6 h, not 0.25.

Common Pitfalls:Using linear instead of exponential decay.Confusing “decayed” fraction (75%) with “remaining” fraction (25%).

Final Answer:3