The half-life of arsenic-74 is about 18 days. If a sample initially contains 100.00 mg arsenic-74, what mass (in mg) would be left after 72 days?

• A. 0.0625 mg
• B. 6.25 mg
• C. 25 mg
• D. 12.5 mg
• E. 50 mg

Answer: (B)

Verified step by step guidance

1. Understand the concept of half-life: The half-life of a radioactive isotope is the time required for half of the isotope to decay. For arsenic-74, this is 18 days.
2. Determine how many half-lives have passed in 72 days: Divide the total time (72 days) by the half-life (18 days) to find the number of half-lives.
3. Calculate the remaining mass after each half-life: Start with the initial mass of 100.00 mg. After each half-life, the mass is halved.
4. Use the formula for exponential decay: The remaining mass can be calculated using the formula \( m = m_0 \times \left( \frac{1}{2} \right)^n \), where \( m_0 \) is the initial mass and \( n \) is the number of half-lives.
5. Substitute the values into the formula: Plug in the initial mass (100.00 mg) and the number of half-lives calculated in step 2 into the formula to find the remaining mass.