Answer
$n_P = 5.0 \times 10^{21}m^{-3} $
Work Step by Step
The added charge carrier density is
$n_P = 10^{-7} n_{Si} $
$n_{Si} = \frac{\rho si N_A}{M_{Si}}$
Where
$\rho_Si = 2330 kg/m^3$
$M_{Si} = 0.0281 kg/mol$
$N_A$ is $6.022 \times 10^{23}/mol$
Substitute all the values into equation
$n_{Si} = \frac{(2330 kg/m^3)(6.022 \times 10^{23}/mol)}{(0.0281 kg/mol)}$
$n_{Si} = 5.0 \times 10^{28} m^{-3}$
Now find $n_P$
$n_P = 10^{-7} (5.0 \times 10^{28} m^{-3}) $
$n_P = 5.0 \times 10^{21}m^{-3} $
