A silicon crystal is doped with 5 10¹⁷ cm ⁻³ atoms of boron. The electron concentration (n_e) at room temperature is (n_i = 1.5 10¹⁰ cm ⁻³)

Step-by-step solution

Boron is a trivalent impurity, so doping silicon with boron creates a p-type semiconductor. The hole concentration is approximately equal to the acceptor concentration: n_h N_A = 5 10¹⁷ cm ⁻³ According to the mass action law: n_e n_h = n_i^2 Substituting the given values: n_e = n_i^2 n_h n_e = (1.5 10¹⁰)^2 5 10¹⁷ n_e = 2.25 10²⁰ 5 10¹⁷ n_e = 0.45 10^3 cm ⁻³ n_e = 4.5 10^2 cm ⁻³ Answer: 4.5 10^2 cm ⁻³