A sample of 201Tl, a radioisotope used to determine the function of the heart, decays initially at a rate of 25,700 disintegrations/min, but the decay rate falls to 15,990 disintegrations/min after 50.0 hours. What is the half-life of 201Tl, in hours? (a) 73.0 hours (b) 105 hours (c) 1.56 x 10^-2 hours (d) 3.84 x 10^2 hours

Verified step by step guidance

1. Step 1: Understand that the decay of a radioactive isotope follows first-order kinetics, which can be described by the equation: \( N_t = N_0 e^{-kt} \), where \( N_t \) is the remaining quantity at time \( t \), \( N_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is the time elapsed.
2. Step 2: Use the given decay rates to set up the equation: \( 15,990 = 25,700 e^{-k \times 50} \). This equation will allow you to solve for the decay constant \( k \).
3. Step 3: Rearrange the equation to solve for \( k \): \( \frac{15,990}{25,700} = e^{-k \times 50} \). Take the natural logarithm of both sides to isolate \( k \): \( \ln\left(\frac{15,990}{25,700}\right) = -k \times 50 \).
4. Step 4: Solve for \( k \) by dividing both sides by -50: \( k = -\frac{1}{50} \ln\left(\frac{15,990}{25,700}\right) \).
5. Step 5: Use the relationship between the decay constant \( k \) and the half-life \( t_{1/2} \), which is \( t_{1/2} = \frac{\ln(2)}{k} \), to calculate the half-life of \( ^{201}\text{Tl} \).