Question

Calculate the rate constant for the radioactive disintegration of an isotope that has a half-life of 6930 yr.

• A. \( 1.00 \times 10^{-5} \, {yr}^{-1} \)
• B. \( 1.00 \times 10^{-4} \, {yr}^{-1} \)
• C. \( 1.00 \times 10^{-3} \, {yr}^{-1} \)
• D. \( 1.00 \times 10^{-1} \, {yr}^{-1} \)

Hint:

Use the formula \( k = \frac{\ln(2)}{t_{1/2}} \) to calculate the rate constant for a reaction based on its half-life.

Answer 1

The Correct Option is D

The relationship between the half-life \( t_{1/2} \) and the rate constant \( k \) is given by: \[ k = \frac{\ln(2)}{t_{1/2}} \] Substituting \( t_{1/2} = 6930 \, {yr} \), \[ k = \frac{\ln(2)}{6930} = 1.00 \times 10^{-4} \, {yr}^{-1} \] Therefore, the rate constant is \( 1.00 \times 10^{-4} \, {yr}^{-1} \), and the correct answer is (b).