(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 \(\mathrm{Ag}\) are \(6.8 \times 10^{7}(\Omega \cdot \mathrm{m})^{-1}\) and \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. (b) Now, compute the electron mobility for \(\mathrm{Ag}\).

Question: Calculate the number of free electrons per cubic meter for silver and compute the electron mobility for silver. Solution: Step 1: Convert the density of silver into atoms per cubic meter. The density of silver converted to atoms/m³ is: $$ \frac{10.5 \times 10^6 \times 6.022 \times 10^{23}}{107.87} \ \mathrm{atoms/m^{3}} $$ Step 2: Calculate the number of free electrons per cubic meter. The number of free electrons per cubic meter for silver (n) is: $$ n = 1.3 \times \frac{10.5 \times 10^6 \times 6.022 \times 10^{23}}{107.87} $$ Step 3: Calculate the electron mobility. Electron mobility (μ) is calculated using the electrical conductivity (σ) and the free electron density (n) as follows: $$ μ = \frac{σ}{ne} $$ Plugging in our values, we get: $$ μ = \frac{6.8 \times 10^{7}}{1.3 \times \frac{10.5 \times 10^6 \times 6.022 \times 10^{23}}{107.87} \times 1.6 \times 10^{-19}} $$ After simplifying the expression, the electron mobility for silver can be calculated.