5. \( [-/ 1 \) Points \( ] \) DETAILS MY NOTES SPRECALC8 4.6.025. This exercise uses the radioactive decay model. If 250 mg of a radloactive element decays to 230 mg in 12 hours, find the half-life (in hr) of the element. (Round your answer to the nearest whole number. \( \square \mathrm{hr} \)

To find the half-life of the radioactive element using the decay information given, we first note that the decay formula can be expressed as: \[ N(t) = N_0 e^{-kt} \] Where: - \( N(t) \) is the remaining quantity after time \( t \), - \( N_0 \) is the initial quantity, - \( k \) is the decay constant, - \( t \) is time. From the problem, we have \( N_0 = 250 \, \text{mg} \) and \( N(t) = 230 \, \text{mg} \) after \( t = 12 \, \text{hr} \). 1. We can set up the equation: \[ 230 = 250 e^{-12k} \] 2. Dividing both sides by 250 gives: \[ \frac{230}{250} = e^{-12k} \] 3. Taking the natural logarithm on both sides: \[ \ln\left(\frac{230}{250}\right) = -12k \] 4. Solving for \( k \): \[ k = -\frac{1}{12} \ln\left(\frac{230}{250}\right) \] Next, to find the half-life \( T_{1/2} \), we use the following relationship: \[ T_{1/2} = \frac{\ln(2)}{k} \] 5. Substitute \( k \) into the half-life equation: \[ T_{1/2} = \frac{\ln(2)}{-\frac{1}{12} \ln\left(\frac{230}{250}\right)} \] Now calculating this step-by-step: 1. Calculate \( k \): \[ k \approx -\frac{1}{12} \ln\left(0.92\right) \approx 0.0065 \] 2. Therefore, \[ T_{1/2} = \frac{0.693}{0.0065} \approx 106.46 \, \text{hr} \] After rounding to the nearest whole number, the half-life is approximately **106 hr**.