4 A 2088 sample of sodium-24 decays to 13.0 g of sodium- 24 within 60.0 h . What is the half-life of this radioactive isotope?

To find the half-life of sodium-24, we can use the formula for radioactive decay, which relates the initial amount, the remaining amount, the half-life, and the time elapsed. The formula is: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] Where: - \( N(t) \) is the remaining quantity of the substance after time \( t \), - \( N_0 \) is the initial quantity of the substance, - \( T_{1/2} \) is the half-life, - \( t \) is the elapsed time. Given: - \( N_0 = 2088 \) g (initial amount), - \( N(t) = 13.0 \) g (remaining amount after 60.0 h), - \( t = 60.0 \) h. We can rearrange the formula to solve for the half-life \( T_{1/2} \): \[ \frac{N(t)}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \] Taking the logarithm of both sides: \[ \log\left(\frac{N(t)}{N_0}\right) = \frac{t}{T_{1/2}} \log\left(\frac{1}{2}\right) \] Now, substituting the values: \[ \frac{N(t)}{N_0} = \frac{13.0}{2088} \] Calculating this ratio: \[ \frac{N(t)}{N_0} \approx 0.00622 \] Now, substituting this into the logarithmic equation: \[ \log(0.00622) = \frac{60.0}{T_{1/2}} \log\left(\frac{1}{2}\right) \] Calculating \( \log(0.00622) \) and \( \log(0.5) \): \[ \log(0.00622) \approx -2.206 \] \[ \log(0.5) \approx -0.301 \] Now substituting these values into the equation: \[ -2.206 = \frac{60.0}{T_{1/2}} \times -0.301 \] Solving for \( T_{1/2} \): \[ -2.206 = \frac{-18.06}{T_{1/2}} \] \[ T_{1/2} = \frac{60.0 \times 0.301}{2.206} \] Calculating \( T_{1/2} \): \[ T_{1/2} \approx \frac{18.06}{2.206} \approx 8.19 \text{ hours} \] Thus, the half-life of sodium-24 is approximately **8.19 hours**.