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Pure \mathrm{Si} at \mathrm{500\ K} has equal number of electron \mathrm{(n_{e})} and hole \mathrm{(n_{h})} concentrations of \mathrm{1.5\times 10^{16}m^{-3}} . Doping by indium increases nh to \mathrm{4.5\times 10^{22}m^{-3}} . The doped semiconductor is of

Option: 1

\mathrm{n}−type with electron concentration \mathrm{n_e=2.5 \times 10^{23} \mathrm{~m}^{-3}}


Option: 2

\mathrm{p}−type having electron concentration \mathrm{n_e=5 \times 10^9 \mathrm{~m}^{-3}}


Option: 3

\mathrm{n}−type with electron concentration \mathrm{n_e=5 \times 10^{22} \mathrm{~m}^{-3}}


Option: 4

\mathrm{p}−type with electron concentration \mathrm{n_e=2.5 \times 10^{10} \mathrm{~m}^{-3}}


Answers (1)

best_answer

\begin{aligned} &\mathrm{ n_i^2=n_e n_h} \\ &\mathrm{ \left(1.5 \times 10^{16}\right)^2=n_e\left(4.5 \times 10^{22}\right)} \\ &\mathrm{ n_e=0.5 \times 10^{10}=5 \times 10^9} \\ &\mathrm{ n_h=4.5 \times 10^{22}} \\ &\mathrm{ n_h \gg n_e} \end{aligned}
Semiconductor is \mathrm{p}- type and \mathrm{n_e=5 \times 10^9 \mathrm{~m}^{-3}}

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Anam Khan

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