During World War I radium-226 was used in the manufacturing of luminous paint. If it takes 2.12 x 10⁴ days for its degradation to be 2.49% complete, what is its decay constant?

• A. 1.74 x 10^–4 days^–1
• B. 1.08 x 10^–5 days^–1
• C. 1.19 x 10^–6 days^–1
• D. 2.14 x 10^–5 days^–1

Answer: (C)

Verified step by step guidance

1. Identify the type of decay process: Radium-226 undergoes first-order radioactive decay, which is characterized by a constant half-life and a decay rate that is proportional to the amount of substance present.
2. Use the first-order decay formula: The formula for first-order decay is \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the remaining quantity of the substance, \( N_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is the time elapsed.
3. Determine the fraction of radium-226 remaining: Since the degradation is 2.49% complete, 97.51% of the radium-226 remains. Therefore, \( \frac{N(t)}{N_0} = 0.9751 \).
4. Rearrange the decay formula to solve for the decay constant \( k \): Take the natural logarithm of both sides of the equation \( 0.9751 = e^{-kt} \) to get \( \ln(0.9751) = -kt \).
5. Calculate the decay constant \( k \): Substitute \( t = 2.12 \times 10^4 \) days into the equation \( \ln(0.9751) = -k(2.12 \times 10^4) \) and solve for \( k \) to find the decay constant.