Radioactive copper-64 decays with a half-life of 12.8 days. a. What is the value of \(k\) in \(\mathrm{s}^{-1} ?\) b. A sample contains 28.0 \(\mathrm{mg}^{64} \mathrm{Cu}\) . How many decay events will be produced in the first second? Assume the atomic mass of $^{64} \mathrm{Cu}\( is 64.0 \)\mathrm{u} .$ c. A chemist obtains a fresh sample of \(^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below 25\(\%\) of the initial measured value. How long does she have to do the experiment?

To find the decay constant, we can use the half-life formula: \[t_\frac{1}{2} = \frac{\ln 2}{k}\], where \(t_\frac{1}{2}\) is the half-life. We know that the half-life of copper-64 is 12.8 days. Let's convert it to seconds by multiplying by 24 hours/day, 60 minutes/hour, and 60 seconds/minute: \[12.8 \, \text{days} = 12.8 \times 24 \times 60 \times 60 \, \text{s} = 1105920 \, \text{s}\]. Now we can solve for \(k\): \[k = \frac{\ln 2}{t_\frac{1}{2}} = \frac{\ln 2}{1105920 \, \text{s}}\].

The given sample has a mass of 28.0 mg and an atomic mass of 64.0 u. To find the number of copper-64 atoms in the sample, we first need to convert the mass of the sample to moles, using the atomic mass as a conversion factor: \[\text{moles of }^{64}\mathrm{Cu} = \frac{28.0 \, \text{mg}}{64.0 \, \text{u/mol}} = 0.4375 \, \text{mol}\]. Next, we need to find the number of atoms in 0.4375 moles using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol): \[N_0 = 0.4375 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 2.63 \times 10^{23} \, \text{atoms}\]. Now we can find the number of decay events, which is equal to the rate of change of radioactive atoms with respect to time. First, find the derivative of the radioactive decay equation with respect to time: \[\frac{dN(t)}{dt} = -k \cdot N_0 \cdot e^{-k \cdot t}\]. At \(t = 0\), the rate of decay events is: \[\frac{dN(0)}{dt} = -k \cdot N_0 = -\frac{\ln 2}{1105920 \, \text{s}} \times 2.63 \times 10^{23} \, \text{atoms}\].

To find the time at which the radioactivity falls below 25% of the initial radioactivity value, we can set \(N(t) = 0.25 \cdot N_0\) and solve for \(t\): \[0.25 \cdot N_0 = N_0 \cdot e^{-k \cdot t}\]. Taking the natural logarithm of both sides and simplifying: \[\ln 0.25 = -k \cdot t\]. Now, divide both sides by \(-k\) and substitute the value of \(k\) from Step 1: \[t = \frac{\ln 0.25}{-k} = \frac{\ln 0.25 \cdot 1105920 \, \text{s}}{\ln 2}\]. Now you can calculate each value to obtain the results: a. The value of \(k\) in \(\mathrm{s}^{-1}\) is \(\frac{\ln 2}{1105920 \, \text{s}}\). b. In the first second, \(-\frac{\ln 2}{1105920 \, \text{s}} \times 2.63 \times 10^{23} \, \text{atoms}\) decay events will occur. c. The radioactivity of the sample needs to remain above 25% of the initial value for \(\frac{\ln 0.25 \cdot 1105920 \, \text{s}}{\ln 2}\) seconds.