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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 \mathrm{0.048 \mathrm{~m}^{2} / \mathrm{V}}-s respectively. The resistivity of a specimen of silicon doped with \mathrm{10^{19}} atoms of phosphorus per meter is:  

Option: 1

\mathrm{4.6 \Omega-m}


Option: 2

\mathrm{5.2 \Omega-m}


Option: 3

\mathrm{5.6 \Omega-m}


Option: 4

6.4 \Omega-\mathrm{m}


Answers (1)

best_answer

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

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

\text{Using,}\: \sigma=1 / \rho=\left(n_{1} \mu_{e}+n_{\mid} \mu_{h}\right) e

(2300)^{-1}=n_{\perp}(0.135+0.048) \times 1.6 \times 10^{-19}
\mathrm{n_{I}=1.5 \times 10^{16} / \mathrm{m}^{3}}

Is the intrinsic electron & hole concentration. The resistivity of a specimen doped with \mathrm{10^{19} \mathrm{P}}- atoms/m3 can be found from :

\mathrm{\sigma \text { (conductivity) } =\mathrm{n}_{\mathrm{e}} \text { e. } \mu_{\mathrm{e}} \quad\left(\because \mathrm{n}_{\mathrm{e}}=10^{19} / \mathrm{m}^{3}>>\mathrm{n}_{\mathrm{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}

Posted by

Rishi

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