A pure silicon crystal at temperature $300$ K has electron and hole concentration ($n_i$) $10^{16}$ per m$^3$ each and $10^{21}$ phosphorus atoms ($n_e$) are added per cubic metre. The new hole concentration in silicon crystal is

According to the mass action law for semiconductors:

$n_e n_h = n_i^2$

Given the intrinsic carrier concentration $n_i = 10^{16}$ m$^{-3}$.

Phosphorus is a pentavalent impurity, which acts as a donor. Since the number of donor atoms added is much greater than the intrinsic concentration, the new electron concentration is approximately equal to the donor concentration:

$n_e \approx N_D = 10^{21}$ m$^{-3}$

Substituting the values into the mass action law to find the new hole concentration $n_h$:

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

$n_h = \dfrac{(10^{16})^2}{10^{21}}$

$n_h = \dfrac{10^{32}}{10^{21}} = 10^{11}$ m$^{-3}$

Answer: $10^{11}$ per m$^3$