The bromine- 82 nucleus has a half-life of \(1.0 \times 10^{3}\) min. If you wanted 1.0 g \(^{82}\mathrm{Br}\) and the delivery time was 3.0 days, what mass of NaBr should you order (assuming all of the Br in the NaBr was $^{82} \mathrm{Br}$ )?

We are given that the half-life of Bromine-82 is \(1.0 \times 10^3\) minutes. We can use this information to calculate the decay constant 'λ' using the following formula: \[t_{1/2} = \frac{\ln 2}{\lambda}\] Rearranging for λ, we get: \[\lambda = \frac{\ln 2}{t_{1/2}}\] Plug in the values: \[\lambda = \frac{\ln 2}{1.0 \times 10^3 \, \text{min}}\]

The delivery time is given as 3.0 days, we need to convert this into minutes. \[\text{Delivery time} = 3.0 \, \text{days} \times \frac{24 \, \text{hours}}{1 \, \text{day}} \times \frac{60 \, \text{min}}{1 \, \text{hour}} = 4320 \, \text{min}\]

Using the decay equation, we can find the initial mass of Bromine-82 by considering its final mass after 3 days. \[N_t = N_0 e^{-\lambda t}\] We want to find \(N_0\). Rearranging for \(N_0\), we get: \[N_0 = \frac{N_t}{e^{-\lambda t}}\] Now, plug in the values for \(N_t\), \(t\), and the decay constant λ that we calculated in Step 1: \[N_0 = \frac{1.0 \, \text{g}}{e^{-\frac{\ln 2}{1.0 \times 10^3 \, \text{min}} \times 4320 \, \text{min}}}\]

Now that we have the initial mass of \(^{82}\mathrm{Br}\), we can calculate the mass of NaBr required. We will use the molar masses of Na (22.99 g/mol) and \(^{82}\mathrm{Br}\) (81.92 g/mol). The mass ratio of Na and \(^{82}\mathrm{Br}\) in NaBr can be determined as follows: \[\text{Mass ratio} = \frac{\text{Molar mass of Na}}{\text{Molar mass of } ^{82}\mathrm{Br}} = \frac{22.99 \, \text{g/mol}}{81.92 \, \text{g/mol}}\] So, the mass of NaBr needed can be calculated by multiplying the initial mass of \(^{82}\mathrm{Br}\) with the mass ratio: \[\text{Mass of NaBr} = N_0 \times \text{Mass ratio} = \frac{1.0 \, \text{g}}{e^{-\frac{\ln 2}{1.0 \times 10^3 \,\text{min}} \times 4320 \, \text{min}}} \times \frac{22.99 \, \text{g/mol}}{81.92 \, \text{g/mol}}\] Finally, compute the mass of NaBr needed to obtain 1.0 g \(^{82}\mathrm{Br}\).