The number of silicon atoms per  $m^3$ is  $5\times {10}^{28}$. This is doped simultaneously with $5\times {10}^{22}$ atoms per $m^3$ of arsenic and  $5\times {10}^{20}$ atoms per $m^3$ of Indium .Given that $n_i= 1.5 \times {10}^{16} m^{-3}$. If the number of electrons is $z\times {10}^{21}{m}^{-3}$ then z is

$Donor Density \left(Arsenic\right) = N_D=5\times {10}^{22}{m}^{-3}$

$Acceptor Density \left(Indium\right) = N_A=5\times {10}^{20}{m}^{-3}$

$n_i=1.5\times {10}^{16}{m}^{-3}$

 

$$\begin{gathered} If the semi conductor is electrically neutral , then \\ {N}_D-N_A=n_e-n_h.......\left(1\right) \end{gathered}$$

$here N_D-N_A = 5\times {10}^{22}-5\times {10}^{20}$

$$\begin{gathered} =5\times {10}^{20}({10}^2-1) \\ =5\times {10}^{20}\times 99 = 495\times {10}^{20} \end{gathered}$$

$Also, n_e{n}_h={n_i}^2...............\left(2\right)$

$Now, we can write {\left(n_e+n_h\right)}^2={(n_e-n_h)}^2+4n_e{n}_h.....................\left(3\right)$

$$\begin{gathered} Put \left(1\right) and \left(2\right) in \left(3\right) \\ \Rightarrow (n_e+n_h)=\sqrt{{(N_D-N_A)}^2+4{n_i}^2} \end{gathered}$$

$\Rightarrow n_e+n_h+n_e-n_h=N_D-N_A+\sqrt{{(N_D-N_A)}^2+4{n_i}^2}$

$\Rightarrow 2n_e=N_D-N_A+\sqrt{{(N_D-N_A)}^2+4{n_i}^2}$

$\Rightarrow 2n_e=495\times {10}^{20}+\sqrt{{(495\times {10}^{20})}^2+4\times {(1.5\times {10}^{16})}^2}$

$\Rightarrow 2n_e=495\times {10}^{20}+\sqrt{(495\times 495\times {10}^{40})+4\times (2.25\times {10}^{32})}$

$\Rightarrow 2n_e\approx 495\times {10}^{20}+\sqrt{{495}^2\times {10}^{40}}$

$\Rightarrow 2n_e\approx 495\times {10}^{20}+495\times {10}^{20}$

$\Rightarrow 2n_e\approx 495\times {10}^{20}\times 2$

$\Rightarrow n_e\approx 495\times {10}^{20}=49.5\times {10}^{21}$