Radioactive decay with half-lives: a sample has a half-life of 100 days. If we start with 1 g, how much remains after 400 days (i.e., after four half-lives)?

Introduction / Context:
Half-life problems test understanding of exponential decay in nuclear physics. The half-life t1/2 is the fixed interval over which a radioactive sample's quantity reduces to half its current value. This question asks how much of a 1 g sample remains after several half-lives have elapsed.

Given Data / Assumptions:

• Initial mass, m0 = 1 g.
• Half-life, t1/2 = 100 days.
• Total time elapsed, t = 400 days.
• Decay follows N(t) = N0 * (1/2)^(t / t1/2).

Concept / Approach:
Radioactive decay is a first-order process. Each half-life reduces the remaining amount by a factor of 1/2, regardless of how much remains. The number of half-lives elapsed is n = t / t1/2. The remaining mass is m = m0 * (1/2)^n. No complicated algebra is needed when t is a whole-number multiple of the half-life—simply halve successively.

Step-by-Step Solution:
Compute n: n = 400 / 100 = 4 half-lives.Apply the decay formula: m = 1 * (1/2)^4.Evaluate the power: (1/2)^4 = 1/16.Hence m = 1/16 g.

Verification / Alternative check:
Successive halving: 1 g → 1/2 g (100 d) → 1/4 g (200 d) → 1/8 g (300 d) → 1/16 g (400 d). This confirms the calculation without formulas.

Why Other Options Are Wrong:
1/2: corresponds to one half-life (100 days), not four.1/4: corresponds to two half-lives (200 days).1/8: corresponds to three half-lives (300 days).

Common Pitfalls:
Confusing total time with number of half-lives; adding halves linearly instead of multiplying by 1/2 each interval; rounding fractional half-lives when the time matches an exact multiple.

Final Answer:
1/16