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The length of germanium rod is \mathrm{0.928\ cm} and its area of cross-section is \mathrm{1\ mm^{2}} . If for germanium \mathrm{n_{i}=2.5 \times 10^{19}m^{-3}}  \mathrm{\mu_h=0.19 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}, \mu_e=0.39 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}}

Option: 1

\mathrm{2.5\ k \Omega}


Option: 2

\mathrm{4.0\ k \Omega}

 


Option: 3

\mathrm{5.0\ k \Omega}


Option: 4

\mathrm{10.0\ k \Omega}


Answers (1)

best_answer

\begin{aligned} & R=\frac{\rho l}{A}=\frac{L}{n_i e\left(\mu_{\mathrm{e}}+\mu_{\mathrm{h}}\right) \mathrm{A}} \\ & =\frac{0.928 \times 10^{-2}}{2.5 \times 10^{19} \times 1.6 \times 10^{-19}(0.39+0.19) \times 10^{-6}} \\ & =4000 \Omega \end{aligned}

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Ritika Harsh

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