A silicon sample is doped simultaneously with donor impurity phosphorus at a concentration of N_D = 3 10²² m⁻³ and acceptor impurity boron at a concentration of N_D = 2.8 10²² m⁻³ . The intrinsic carrier concentration of silicon at room temperature is n_i = 1.5 10¹⁶ m⁻³ . Assuming complete ionization, the hole concentration is: A. 1.125 10¹⁰ m⁻³ B. 1.125 10¹¹ m⁻³ C. 2.125 10¹⁰ m⁻³ D. 2.125 10¹¹ m⁻³

Step-by-step solution

Given donor concentration N_D = 3 10²² m⁻³ and acceptor concentration N_A = 2.8 10²² m⁻³ . Since N_D > N_A , the semiconductor is n-type. The effective donor concentration is N_D - N_A = 3 10²² - 2.8 10²² = 0.2 10²² = 2 10²¹ m⁻³ . Assuming complete ionization, the electron concentration n is approximately equal to the effective donor concentration: n N_D - N_A = 2 10²¹ m⁻³ Using the mass action law, n p = n_i^2 , where n_i = 1.5 10¹⁶ m⁻³ . The hole concentration p is: p = n_i^2 n = (1.5 10¹⁶)^2 2 10²¹ p = 2.25 10³²