The half-life of arsenic-74 is about 18 days. If a sample initially contains 5.13 × 10⁴ mg arsenic-74, what mass (in mg) would be left after 80 days?

• A. 2.36 × 10³ mg
• B. 7.02 × 10² mg
• C. 1.43 × 10³ mg
• D. 1.14 × 10⁸ mg

Answer: (A)

Verified step by step guidance

1. Identify the given information: the half-life (t_{1/2}) of arsenic-74 is 18 days, the initial mass (m_0) is 5.13 \(\times\) 10^{4} mg, and the time elapsed (t) is 80 days.
2. Calculate the number of half-lives that have passed using the formula: \\ \(n = \frac{t}{t_{1/2}} = \frac{80}{18}\).
3. Use the half-life decay formula to find the remaining mass (m): \\ \(m = m_0 \times \left(\frac{1}{2}\right)^n\).
4. Substitute the values of \(m_0\) and \(n\) into the formula to express the remaining mass in terms of known quantities.
5. Evaluate the expression to find the mass of arsenic-74 left after 80 days (do not calculate the final number here, just set up the expression).