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The resistivity of pure silicon is \mathrm{2300 \Omega -m} and the mobilities of electrons and holes in it are 0.135 and 0\mathrm{0.048 \; m^{2}/V^{-S}}
respectively. The resistivity of a specimen of silicon doped with \mathrm{10}^{19} atoms of phosphorus per meter. Is:

Option: 1

\mathrm{1.6 \Omega-\mathrm{m} }


Option: 2

2.6 \Omega-\mathrm{m}


Option: 3

3.6 \Omega-m


Option: 4

\mathrm{4.6 \Omega-m}


Answers (1)

best_answer

The resistivity of pure silicon is \mathrm{2300\Omega} m

and \mathrm{\mu_{\mathrm{e}}=0.135 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}, \mu \mathrm{h}=0.048 \mathrm{~m}^2 / \mathrm{N}-\mathrm{s} .}

using, \mathrm{\sigma=1 / \rho=\left(n_1 \mu_e+n_1 \mu_n\right) e}

\mathrm{\begin{aligned} & (2300)^{-1}=n_I(0.135+0.048) \times 1.6 \times 10^{-19} \\ & n_I=1.5 \times 10^{16} / \mathrm{m}^3 . \end{aligned}}

Is the intrinsic electron & hole concentration. The resistivity of a specimen doped with \mathrm{10^{19}P-atoms/m^{3}} can be found from:

\mathrm{\begin{aligned} \sigma(\text { conductivity }) & =n_e \text { e. } \mu_e \quad\left(\because n_e=10^{19} / \mathrm{m}^3>\mathrm{n_l} \right) \ =10^{19} \times 1.6 \times 10^{-19} \times 0.135 \\ & =0.216 \mathrm{mho} / \mathrm{m} . \\ \rho & =\frac{1}{\sigma}=4.6 \Omega-\mathrm{m} \end{aligned}}

Posted by

Ajit Kumar Dubey

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