The half-life of strontium- 90 is 28 years. How long will it take \( 80 \% \) of a sam 90 to decay?

**Step 1: Write the decay formula** The amount remaining at time \( t \) can be expressed using the exponential decay formula: \[ N(t)=N_0 e^{-kt} \] The decay constant \( k \) is related to the half-life \( T_{1/2} \) by: \[ k=\frac{\ln(2)}{T_{1/2}}=\frac{\ln(2)}{28} \] --- **Step 2: Set up the equation for 80% decay** If \( 80\% \) of the sample decays, then \( 20\% \) remains. We have: \[ N(t)=0.2\,N_0 \] Substitute into the decay formula: \[ 0.2\,N_0 = N_0 e^{-\frac{\ln(2)}{28} t} \] Cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)): \[ 0.2 = e^{-\frac{\ln(2)}{28} t} \] --- **Step 3: Solve for \( t \)** Take the natural logarithm on both sides: \[ \ln(0.2)= -\frac{\ln(2)}{28}\,t \] Rearrange to solve for \( t \): \[ t=-\frac{28\,\ln(0.2)}{\ln(2)} \] --- **Step 4: Compute the value of \( t \)** Calculate the logarithms: \[ \ln(0.2) \approx -1.6094 \quad \text{and} \quad \ln(2) \approx 0.6931 \] Then: \[ t \approx -\frac{28 \times (-1.6094)}{0.6931} \approx \frac{28 \times 1.6094}{0.6931} \] Perform the multiplication and division: \[ 28 \times 1.6094 \approx 45.0632 \] \[ t \approx \frac{45.0632}{0.6931} \approx 65 \] --- **Answer** It will take approximately \( 65 \) years for \( 80\% \) of the strontium-90 sample to decay.