Question

A semiconductor has equal electron and hole concentration of 2 × 10⁸ m^-3. On doping with a certain impurity, the electron concentration increases to 4 × 10¹⁰ m^-3 , then calculate the new hole concentration of the semiconductor.

Answer 1

Given: n_i = 2 × 10⁸ m^-3, n = 4 × 10¹⁰ m^-3

After doping n_h = 10²¹ m^-3

To find: Number density of holes (n_h)

Formulae: n_i² = n_en_h

Calculation: From formula.