Carbon-14, which is present in all living tissue, radioactively decays via a first-order process. A one-gram sample of wood taken from a living tree gives a rate for carbon-14 decay of 13.6 counts per minute (CPM). If the half-life for carbon-14 is 5720 years, calculate the time it would take for the count rate to decrease to 6.8 CPM.

• A. 1430 years
• B. 5720 years
• C. 11440 years
• D. 2860 years

Answer: (B)

Verified step by step guidance

1. Identify that the decay of Carbon-14 is a first-order process, which means the rate of decay is proportional to the amount of Carbon-14 present.
2. Use the first-order decay formula: \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the count rate at time \( t \), \( N_0 \) is the initial count rate, \( k \) is the decay constant, and \( t \) is the time.
3. Calculate the decay constant \( k \) using the half-life formula for first-order reactions: \( k = \frac{\ln(2)}{\text{half-life}} \). Substitute the given half-life of 5720 years into the formula.
4. Set up the equation using the initial count rate \( N_0 = 13.6 \) CPM and the final count rate \( N(t) = 6.8 \) CPM. Substitute these values into the first-order decay formula to solve for \( t \).
5. Rearrange the equation to solve for \( t \): \( t = \frac{\ln(N_0/N(t))}{k} \). Substitute the values for \( N_0 \), \( N(t) \), and \( k \) to find the time it takes for the count rate to decrease to 6.8 CPM.