42. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 1016 m−3 . Doping by indium increases nh to 4.5 × 1022 m−3 . The doped semiconductor is of ?

\(42.\ \text{Pure Si at } 500\,\mathrm{K} \text{ has equal number of electron } (n_e) \text{ and hole } (n_h) \text{ concentrations of } 1.5 \times 10^{16}\,\mathrm{m^{-3}} .\ \text{Doping by indium increases } n_h \text{ to } 4.5 \times 10^{22}\,\mathrm{m^{-3}} .\ \text{The doped semiconductor is of ?}\)

\((a)\ \text{n-type with electron concentration } n_e = 5 \times 10^{22}\,\mathrm{m^{-3}}\)

\((b)\ \text{p-type with electron concentration } n_e = 2.5 \times 10^{10}\,\mathrm{m^{-3}}\)

\((c)\ \text{n-type with electron concentration } n_c = 2.5 \times 10^{23}\,\mathrm{m^{-3}}\)

\((d)\ \text{p-type having electron concentration } n_e = 5 \times 10^{9}\,\mathrm{m^{-3}}\)

Answer :  \((d)\ \text{p-type having electron concentration } n_e = 5 \times 10^{9}\,\mathrm{m^{-3}}\)

\( n_e = n_h = 1.5 \times 10^{16}\ \mathrm{m^{-3}} \ \text{(initial)} \)

\( n_h = 4.5 \times 10^{22}\ \mathrm{m^{-3}} \)

\( n_e \cdot n_h = n_i^2 \quad ( \text{Law of mass action} ) \)

\( n_e \,(4.5 \times 10^{22}) = (1.5 \times 10^{16})^2 \)

\( n_e = \frac{(1.5)^2}{4.5} \times 10^{10} \)

\( n_e = 0.5 \times 10^{10} \)

\( n_e = 5 \times 10^{9} \)