An intrinsic silicon sample at room temperature has $n_i \approx 1.5\times 10^{16}\,\text{m}^{-3}$. After doping with phosphorus to a concentration $N_d = 10^{22}\,\text{m}^{-3}$, the equilibrium electron concentration is essentially $N_d$, and the hole concentration $n_h$ is given by the mass-action law $n_e n_h = n_i^2$. Compute $n_h$.

1. The law of mass action for an extrinsic semiconductor: $n_e\,n_h = n_i^2$, where $n_i$ is the intrinsic carrier concentration at that temperature. 2. After doping with phosphorus (n-type), essentially every donor contributes one electron, so $n_e \approx N_d = 10^{22}\,\text{m}^{-3}$. 3. Solve for the hole concentration: $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5\times 10^{16})^2}{10^{22}}$. 4. $(1.5)^2 = 2.25$, and $(10^{16})^2 = 10^{32}$. So $n_i^2 = 2.25\times 10^{32}$. 5. Divide by $10^{22}$: $n_h = 2.25\times 10^{10}\,\text{m}^{-3}$. 6. Notice how dramatic the asymmetry is: holes drop from $1.5\times 10^{16}$ (intrinsic) to $2.25\times 10^{10}$ (after doping) — a million-fold reduction. Electrons rise correspondingly. This is the defining feature of extrinsic semiconductors: majority carrier concentration ≫ minority. 7. Other options reflect common slips: A is the intrinsic value (forgot doping), B is just $N_d$, D is $n_i^2$ without dividing. _Source: NCERT Class 12 Physics Part 2, Ch 14, §14.4 (Extrinsic semiconductor, $n_e n_h = n_i^2$), p. 8–9._