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

Option: 1

n–type with electron concentration \mathrm{\mathrm{n}_{\mathrm{c}}=5 \times 10^{22} \mathrm{~m}^{-3}}


Option: 2

p–type with electron concentration \mathrm{\mathrm{n}_{\mathrm{c}}=2.5 \times 10^{10} \mathrm{~m}^{-3}}


Option: 3

n–type with electron concentration \mathrm{\mathrm{n}_{\mathrm{e}}=2.5 \times 10^{23} \mathrm{~m}^{-3}}


Option: 4

p–type having electron concentration \mathrm{\mathrm{n}_{\mathrm{e}}=5 \times 10^9 \mathrm{~m}^{-3}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{n}_{\mathrm{i}}^2=\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}} \\ & \left(1.5 \times 10^{16}\right)^2=\mathrm{n}_3\left(4.5 \times 10^{22}\right) \\ & \Rightarrow \mathrm{n}_{\mathrm{e}}=0.5 \times 10^{10} \\ & \text { or } \quad \mathrm{n}_{\mathrm{e}}=5 \times 10^9 \\ & \text { Given } \mathrm{n}_{\mathrm{h}}=4.5 \times 10^{22} \end{aligned}}

\mathrm{\Rightarrow \mathrm{n}_{\mathrm{h}}=4.5 \times 10^{22}}

Semiconductor is p-type and \mathrm{\mathrm{n}_{\mathrm{e}}=5 \times 10^9 \mathrm{~m}^{-3}}

Posted by

avinash.dongre

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