pure Si crystal has $$6×10^{28}$$​ atoms $$\text{m}^3$$​. It is doped by 1 ppm concentration of pentavalent As. Then the number of electrons and holes are:

(Given that $$\text{n}_\text i=1.5×10^{16}\space\text{m} ^{-3}$$​)

Correct option is B

​$$\begin{aligned}&\textbf{ Given Data:} \\&\quad \circ~\text{Silicon atomic density: } 6 \times 10^{28}\,\text{atoms/m}^3 \\&\quad \circ~\text{Doping concentration: 1 part per million (ppm) of As} \\&\quad \circ~\text{Intrinsic carrier concentration: } n_i = 1.5 \times 10^{16}\,\text{m}^{-3} \\\\&\quad n_e = 6 \times 10^{28} \times 10^{-6} = 6 \times 10^{22}\,\text{m}^{-3} \\&\quad \text{\textit{Each As atom donates one free electron}} \\\\&\quad \text{Using the mass-action law for semiconductors:} \\&\quad n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{6 \times 10^{22}} \\&\quad = \frac{2.25 \times 10^{32}}{6 \times 10^{22}} = 3.75 \times 10^9\,\text{m}^{-3} \\\\&\textbf{Conclusion:} \\&\quad \circ~\text{Electron concentration: } 6 \times 10^{22}\,\text{m}^{-3} \\&\quad \circ~\text{Hole concentration: } 3.75 \times 10^9\,\text{m}^{-3}\end{aligned}$$​​