(a) Calculate the number of free electrons per cubic meter for silver, assuming that there are 1.3 free electrons per silver atom. The electrical conductivity and density for Ag are \(6.8 \times 10^{7}(\Omega-\mathrm{m})^{-1}\) and \(10.5 \mathrm{g} / \mathrm{cm}^{3}, \mathrm{re}\) spectively. (b) Now compute the electron mobility for Ag.

First, we need to find the number of atoms per cubic meter using the given density. The density of silver is given as 10.5 g/cm³. To convert this to the number of atoms per cubic meter, we need to know the molar mass of silver (Ag), which is 107.8682 g/mol, and Avogadro's number which is \(6.022\times10^{23}\) atoms/mol. First, convert the density from g/cm³ to kg/m³: $$ 10.5 \frac{\text{g}}{\text{cm}^{3}} = 10.5 \times 1000 \frac{\text{kg}}{\text{m}^{3}} = 10,500 \frac{\text{kg}}{\text{m}^{3}} $$ Next, we'll find the number of moles of silver per cubic meter: $$ \frac{10,500 \frac{\text{kg}}{\text{m}^{3}}}{107.8682 \frac{\text{g}}{\text{mol}}} = 97.32 \, \text{moles/m}^{3} $$ Then, we'll multiply the number of moles by Avogadro's number to get the number of silver atoms per cubic meter: $$ 97.32 \, \text{moles/m}^{3} \times 6.022\times10^{23} \frac{\text{atoms}}{\text{mol}} = 5.86\times10^{28} \frac{\text{atoms}}{\text{m}^{3}} $$

Now we have to find the number of free electrons per cubic meter using the given number of free electrons per silver atom (1.3 free electrons/atom). We'll multiply the number of atoms per cubic meter by the number of free electrons per atom: $$ 5.86\times10^{28} \frac{\text{atoms}}{\text{m}^{3}} \times 1.3 \frac{\text{free electrons}}{\text{atom}} = 7.61\times10^{28} \frac{\text{free electrons}}{\text{m}^{3}} $$

Finally, we need to calculate the electron mobility using the given electrical conductivity (\(6.8\times10^{7}(\Omega\text{-m})^{-1}\)) and the previously calculated number of free electrons per cubic meter. We'll use the formula for electrical conductivity: $$ \sigma = nq\mu $$ Where \(\sigma\) is electrical conductivity, \(n\) is the number of free electrons per cubic meter, \(q\) is the charge of an electron (\(1.6\times10^{-19}\) C), and \(\mu\) is electron mobility. We'll solve for \(\mu\): $$ \mu = \frac{\sigma}{nq} = \frac{6.8\times10^{7} (\Omega\text{-m})^{-1}}{7.61\times10^{28}\frac{\text{free electrons}}{\text{m}^{3}} \times 1.6\times10^{-19} \text{C}} = 5.62 \times 10^{-3} \frac{\text{m}^{2}}{\text{V}\cdot\text{s}} $$ The electron mobility for silver is \(5.62\times10^{-3}\frac{\text{m}^{2}}{\text{V}\cdot\text{s}}\).