MHTCET 22 Apr 2025 Shift 1 Paper

Concept:In a doped semiconductor, the product of electron and hole concentrations remains constant as per the law of mass action.Explanation:Given, intrinsic carrier concentration at 301 K is $n_i = 10^{16} \mathrm{~m}^{-3}$.When $10^{21}$ phosphorus atoms are added per cubic metre, the crystal becomes n-type.For an n-type semiconductor, the electron concentration $n$ equals the donor concentration $N_D$.Thus, $n = N_D = 10^{21} \mathrm{~m}^{-3}$.Using the law of mass action, $n \cdot p = n_i^2$.Substituting the known values, we get $p = \frac{n_i^2}{n}$.Therefore, $p = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{11} \mathrm{~m}^{-3}$.This drastic increase in electrons suppresses the hole concentration to maintain thermal equilibrium.Answer:The new hole concentration in silicon is $10^{11} \mathrm{~m}^{-3}$.Hence, the correct option is B.