Choice A is the correct answer. Find the decay factor per cycle.

1. Model: $C(n) = C_0 b^n$ where $b$ is decay factor per cycle.
2. After 1000 cycles: $0.80C_0 = C_0 b^{1000}$.
3. Simplify: $0.80 = b^{1000}$.
4. Solve: $b = (0.80)^{1/1000} = 0.80^{0.001}$.
5. Calculate: $b \approx 0.9998$ (loses about 0.02% per cycle).

💡 Strategic Tip: Small decay per cycle compounds to significant loss over many cycles.

Choice B is incorrect because this loses 1% per cycle (too much). Choice C is incorrect because this loses 0.05% per cycle. Choice D is incorrect because this loses 2% per cycle (way too much).