MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Mass action law states that in a semiconductor at thermal equilibrium, the product of electron and hole concentrations equals the square of the intrinsic concentration: $n_e n_h = n_i^2$.Explanation:Pure silicon at 300 K has equal electron and hole concentrations: $n_i = 10^{16} \, \text{m}^{-3}$.Phosphorus is a pentavalent donor impurity, so doping with $10^{21}$ atoms per m³ converts silicon into an n-type semiconductor.In an n-type material, electrons are the majority carriers and holes are the minority carriers.Since the donor concentration is much larger than the intrinsic concentration, the new electron concentration is approximately $n_e \approx 10^{21} \, \text{m}^{-3}$.At thermal equilibrium, the mass action law holds:$$n_e n_h = n_i^2$$Rearranging for the new hole concentration:$$n_h = \frac{n_i^2}{n_e}$$Substituting the given values:$$n_h = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{11} \, \text{m}^{-3}$$Answer:The new hole concentration in the doped silicon crystal is $10^{11}$ per m³.Correct option: D.