1. **Problem statement:** Calculate the equilibrium electron concentration ($n$), hole concentration ($p$), mobilities of electrons ($\mu_n$) and holes ($\mu_p$), and electrical resistivity ($\rho$) for a silicon semiconductor doped with $N_A = 1.4 \times 10^{16}$ cm$^{-3}$ boron atoms (acceptors) and $N_D = 1.0 \times 10^{16}$ cm$^{-3}$ phosphorus atoms (donors) at 27 °C. Intrinsic carrier concentration $n_i = 1.5 \times 10^{10}$ cm$^{-3}$. 2. **Key formulas and rules:** - Charge neutrality: $p + N_D = n + N_A$ - Mass action law: $np = n_i^2$ - For p-type or n-type, majority carrier concentration approximates to net doping. - Electron mobility $\mu_n$ and hole mobility $\mu_p$ depend on doping concentration. - Resistivity formula: $$\rho = \frac{1}{q(n\mu_n + p\mu_p)}$$ where $q = 1.6 \times 10^{-19}$ C. 3. **Calculate net doping:** $$N_A = 1.4 \times 10^{16}, \quad N_D = 1.0 \times 10^{16}$$ Net acceptor concentration: $$N_A - N_D = 1.4 \times 10^{16} - 1.0 \times 10^{16} = 0.4 \times 10^{16} = 4.0 \times 10^{15}$$ Since $N_A > N_D$, the semiconductor is p-type. 4. **Calculate majority and minority carrier concentrations:** Majority holes: $$p \approx N_A - N_D = 4.0 \times 10^{15}$$ Minority electrons from mass action law: $$n = \frac{n_i^2}{p} = \frac{(1.5 \times 10^{10})^2}{4.0 \times 10^{15}} = \frac{2.25 \times 10^{20}}{4.0 \times 10^{15}} = 5.625 \times 10^{4}$$ 5. **Estimate mobilities:** Using typical values for silicon at room temperature: - Electron mobility $\mu_n$ for doping $\approx 10^{16}$ cm$^{-3}$ is about 1350 cm$^2$/V·s - Hole mobility $\mu_p$ for doping $\approx 10^{16}$ cm$^{-3}$ is about 480 cm$^2$/V·s 6. **Calculate electrical resistivity:** Charge of electron: $$q = 1.6 \times 10^{-19} \text{ C}$$ Calculate conductivity $\sigma$: $$\sigma = q(n\mu_n + p\mu_p) = 1.6 \times 10^{-19} \times (5.625 \times 10^{4} \times 1350 + 4.0 \times 10^{15} \times 480)$$ Calculate terms: $$5.625 \times 10^{4} \times 1350 = 7.59375 \times 10^{7}$$ $$4.0 \times 10^{15} \times 480 = 1.92 \times 10^{18}$$ Sum: $$7.59375 \times 10^{7} + 1.92 \times 10^{18} \approx 1.92 \times 10^{18}$$ So, $$\sigma \approx 1.6 \times 10^{-19} \times 1.92 \times 10^{18} = 0.3072 \text{ S/cm}$$ Resistivity: $$\rho = \frac{1}{\sigma} = \frac{1}{0.3072} \approx 3.25 \ \Omega \cdot \text{cm}$$ **Final answers:** - Electron concentration $n = 5.63 \times 10^{4}$ cm$^{-3}$ - Hole concentration $p = 4.0 \times 10^{15}$ cm$^{-3}$ - Electron mobility $\mu_n = 1350$ cm$^2$/V·s - Hole mobility $\mu_p = 480$ cm$^2$/V·s - Electrical resistivity $\rho = 3.25 \ \Omega \cdot \text{cm}$