If $n_i$ is the concentration of intrinsic charge carriers in a pure semiconductor and $n_e$ and $n_h$ be the concentrations of electrons and holes, respectively after the semiconductor is doped, then at thermal equilibrium.

$n_e\, n_h = {n_i}^2$

The correct answer is Option (2) → $n_e\, n_h = {n_i}^2$

For a doped semiconductor at thermal equilibrium, the product of electron and hole concentrations remains constant and equal to the square of the intrinsic carrier concentration:

$n_e \cdot n_h = n_i^2$

This is known as the mass-action law for semiconductors. Here:

- $n_i$ = intrinsic carrier concentration

- $n_e$ = electron concentration after doping

- $n_h$ = hole concentration after doping

Final Answer: $n_e \cdot n_h = n_i^2$