A radioactive substance has a half-life of 3 hours, meaning that every 3 hours, exactly half of the remaining substance...

1. TRANSLATE the problem information

• Given information:
  • Half-life = 3 hours
  • Initial amount = \(1.024 \times 10^6\) grams
  • Target amount = \(1.0 \times 10^3\) grams
  • Find: number of hours

• What this tells us: We need to find how many 3-hour periods it takes for the substance to decay from \(1.024 \times 10^6\) to \(1.0 \times 10^3\) grams.

2. TRANSLATE the decay relationship into mathematics

• The radioactive decay formula is: \(\mathrm{Amount = Initial \times (1/2)^n}\)
• Where \(\mathrm{n}\) = number of half-life periods
• Set up equation: \(1.0 \times 10^3 = 1.024 \times 10^6 \times (1/2)^n\)

3. SIMPLIFY to isolate the exponential term

• Divide both sides by the initial amount:
  \((1/2)^n = (1.0 \times 10^3) \div (1.024 \times 10^6)\)

• Calculate the fraction: \(1000 \div 1,024,000 = 1/1024\)

4. INFER the solution using powers of 2

• We need to solve: \((1/2)^n = 1/1024\)
• Recognize that \(1024 = 2^{10}\)
• So: \((1/2)^n = 1/2^{10} = (1/2)^{10}\)
• Therefore: \(\mathrm{n = 10}\) half-life periods

5. TRANSLATE back to real time

• Total time = \(10 \text{ periods} \times 3 \text{ hours per period} = 30 \text{ hours}\)

Answer: D) 30 hours

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students set up the equation incorrectly, often writing something like "\(\mathrm{Amount = Initial \times (1/2) \times n}\)" instead of "\(\mathrm{Amount = Initial \times (1/2)^n}\)", treating the decay as linear rather than exponential.

This fundamental misunderstanding of the decay process leads to incorrect calculations and may cause them to select Choice A (18 hours) or get stuck and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when calculating \(1000 \div 1,024,000\), perhaps getting \(1/1000\) instead of \(1/1024\), or they don't recognize that \(1024 = 2^{10}\).

Without this key insight, they cannot solve the exponential equation efficiently and may resort to trial-and-error with the answer choices, potentially selecting Choice B (21 hours) or Choice C (27 hours).

The Bottom Line:

This problem requires both understanding exponential decay conceptually (not linear decay) and recognizing powers of 2. Students who miss either piece will struggle to reach the correct systematic solution.