What is the relation between the number density of holes $(n_h)$ and free electrons $(n_e)$ in an intrinsic semiconductor at room temperature?

In a semiconductor,the number densities of electrons and holes are $8 \times 10^{18} \ m^{-3}$ and $5 \times 10^{18} \ m^{-3}$ respectively. If the mobilities of electrons and holes are $2.3 \ m^2/V \cdot s$ and $0.01 \ m^2/V \cdot s$ respectively,then the semiconductor is:

What is the number density of donor atoms which must be added to a pure germanium semiconductor to make an $n$-type semiconductor of conductivity $6.4 \ \Omega^{-1} \ cm^{-1}$? The mobility of electrons in $n$-type germanium is $4 \times 10^3 \ cm^2 \ V^{-1} \ s^{-1}$. Neglect the contribution of holes to conductivity.

What will be the conductivity of a pure silicon crystal at $300 \ K$? Given that the intrinsic carrier concentration is $1.072 \times 10^{10} \ cm^{-3}$ and at this temperature,the electron mobility $\mu_n = 1350 \ cm^2 / (V \cdot s)$ and hole mobility $\mu_p = 480 \ cm^2 / (V \cdot s)$.

In a pure silicon crystal,the electron-hole concentration is $10^{16} \ m^{-3}$ at $301 \ K$. Now,$10^{21}$ atoms of phosphorus are added per cubic metre. The new hole concentration in silicon is (in per $m^3$):

For a pure $Si$ crystal at $300 \ K$,the electron $(n_e)$ and hole $(n_h)$ concentrations are equal,being $1.5 \times 10^{16} \ m^{-3}$. On doping with Indium,the hole concentration increases to $4.5 \times 10^{22} \ m^{-3}$. Calculate the new electron concentration $(n_e)$ in the doped $Si$.