A Ge specimen is doped with Al. The concentration of acceptor atoms is ≈ $10^{21}$ atoms/m³. Given that the intrinsic concentration of electrons in the specimen is $10^{19} / m^3$. The new electron concentration is

$10^{17} /m^3$

When Ge specimen is doped with $Al$, then concentration of acceptor atoms is also called concentration of holes.

Using formula

$n_i^2=n_en_h$

$n_i$ = concentration of electron hole pair = $10^{19}/ m^3$

$n_e$ = concentration of electrons

$n_h$ = concentration of holes = $10^{21} atoms/m^3$

$∴ (10^{19})^2 = 1021 × n_e$

$n_e = 10^{17} /m^3$