\begin{tabular}{||c||}\hline Berkelium -246 has a half-life of \\ I.8 days. A sample used to contain \\ 73.6 mg of the isotope but now \\ only contains 2.3 mg . How \\ many days have passed? \end{tabular}

To find the number of days that have passed, we can use the formula for half-lives, which is: \[ N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \] Where: - \( N \) is the remaining quantity of the isotope. - \( N_0 \) is the initial quantity of the isotope. - \( t \) is the time that has passed. - \( T_{1/2} \) is the half-life of the isotope. Given: - \( N_0 = 73.6 \, \text{mg} \) - \( N = 2.3 \, \text{mg} \) - \( T_{1/2} = 1.8 \, \text{days} \) We first find how many half-lives have passed: \[ 2.3 = 73.6 \left(\frac{1}{2}\right)^{t/(1.8)} \] To simplify: \[ \frac{2.3}{73.6} = \left(\frac{1}{2}\right)^{t/(1.8)} \] \[ 0.0312 = \left(\frac{1}{2}\right)^{t/(1.8)} \] Next, we can take the logarithm of both sides: \[ \log(0.0312) = \frac{t}{1.8} \log(0.5) \] Now, solving for \( t \): \[ t = \frac{\log(0.0312)}{\log(0.5)} \times 1.8 \] Using a calculator: \[ \log(0.0312) \approx -1.504 \] \[ \log(0.5) \approx -0.301 \] So: \[ t = \frac{-1.504}{-0.301} \times 1.8 \approx 9.0 \, \text{days} \] Therefore, 9 days have passed.