Question

A radioactive sample has equal numbers of \(^{15} \mathrm{O}\) and $^{19} \mathrm{O}$ nuclei. Use the half-lives found in Appendix B to determine how long it will take before there are twice as many \(^{15} \mathrm{O}\) nuclei as \(^{19} \mathrm{O} .\) What percent of the \(^{19} \mathrm{O}\) nuclei have decayed during this time?

Short answer

Based on the given information about the radioactive sample containing equal numbers of \(^{15}\mathrm{O}\) and \(^{19}\mathrm{O}\) nuclei, it will take approximately 34.52 seconds for there to be twice as many \(^{15}\mathrm{O}\) nuclei as \(^{19}\mathrm{O}\) nuclei. During this time, approximately 60.7% of the \(^{19}\mathrm{O}\) nuclei will have decayed.

Step-by-step solution

Find the decay constants of the isotopes

The decay constants (\(\lambda\)) can be found from the half-lives (\(t_{1/2}\)) using the formula: \(\lambda = \frac{ln(2)}{t_{1/2}}\) We can look up the half-lives of the isotopes from the Appendix B of the textbook, which gives: - \(t_{1/2} = 122.24\,\text{s}\) for \(^{15} \mathrm{O}\) - \(t_{1/2} = 26.91\,\text{s}\) for \(^{19} \mathrm{O}\) Now, we can find the decay constants using the half-life values: \(\lambda_{15} = \frac{ln(2)}{122.24\,\text{s}}\) and \(\lambda_{19} = \frac{ln(2)}{26.91\,\text{s}}\)

Set up the radioactive decay equations with the required ratio

We need to find the time duration when there are twice as many \(^{15}\mathrm{O}\) nuclei as \(^{19}\mathrm{O}\) nuclei. Let's write the radioactive decay equations for the two isotopes: \(N_{15}(t) = N_0 e^{-\lambda_{15} t}\) \(N_{19}(t) = N_0 e^{-\lambda_{19} t}\) Since we want the number of \(^{15}\mathrm{O}\) nuclei to be twice the number of \(^{19}\mathrm{O}\) nuclei at a certain time duration, we can write: \(N_{15}(t) = 2 N_{19}(t)\) Now, substitute \(N_{15}(t)\) and \(N_{19}(t)\) from the radioactive decay equations: \(N_0 e^{-\lambda_{15} t} = 2 N_0 e^{-\lambda_{19} t}\)

Solve for the time duration

Now, we need to solve this equation for the time duration, \(t\). \(\frac{N_0 e^{-\lambda_{15} t}}{N_0 e^{-\lambda_{19} t}} = 2\) Divide by \(N_0\): \(e^{-\lambda_{15} t} = 2 e^{-\lambda_{19} t}\) Take the natural logarithm of both sides: \(-\lambda_{15} t = ln(2) - \lambda_{19} t\) Now, solve for \(t\): \(t = \frac{ln(2)}{\lambda_{19} - \lambda_{15}}\) Substitute the calculated decay constants: \(t = \frac{ln(2)}{\frac{ln(2)}{26.91\,\text{s}} - \frac{ln(2)}{122.24\,\text{s}}}\) Calculate the value of \(t\): \(t \approx 34.52\,\text{s}\) This means it will take approximately 34.52 seconds before there are twice as many \(^{15}\mathrm{O}\) nuclei as \(^{19}\mathrm{O}\) nuclei.

Calculate the percentage of \(^{19}\mathrm{O}\) nuclei decayed

To find the percentage of \(^{19}\mathrm{O}\) nuclei decayed during this time, we will calculate how many nuclei remain after 34.52 seconds and compare it to the initial amount. Since we set \(N_0 = 1\) for simplicity, this will not affect the calculation. Using the radioactive decay equation for \(^{19}\mathrm{O}\) nuclei: \(N_{19}(t) = N_0 e^{-\lambda_{19} t}\) Substitute \(t \approx 34.52\,\text{s}\): \(N_{19}(34.52\,\text{s}) = N_0 e^{-\lambda_{19} \cdot 34.52\,\text{s}}\) Calculate the remaining \(^{19}\mathrm{O}\) nuclei: \(N_{19}(34.52\,\text{s}) = N_0 e^{-\frac{ln(2)}{26.91\,\text{s}} \cdot 34.52\,\text{s}} \approx 0.393 N_0\) To find the percentage of \(^{19}\mathrm{O}\) nuclei decayed, we compute: Percentage decayed = \(\frac{N_0 - N_{19}(34.52\,\text{s})}{N_0} \times 100\%\) Percentage decayed = \(\frac{N_0 - 0.393 N_0}{N_0} \times 100\% \approx 60.7\%\) This means that approximately 60.7% of the \(^{19}\mathrm{O}\) nuclei have decayed during this time.