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  • A Ge specimen is doped with Al. The concentration of acceptor atoms is<img alt="\mathrm{\approx 10^{21}atoms/m^{3}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Capprox%2010%

A Ge specimen is doped with Al. The concentration of acceptor atoms is\mathrm{\approx 10^{21}atoms/m^{3}} . Given that the intrinsic concentration of electrons in the specimen is 10^{19}/m^{3}. The new electron concentration is

Option: 1

\mathrm{10^{17} / \mathrm{m}^3 }


Option: 2

\mathrm{ 10^{15} / \mathrm{m}^3}


Option: 3

\mathrm{10^{4} / \mathrm{m}^3}


Option: 4

\mathrm{10^2 / \mathrm{m}^3}


Answers (1)

best_answer

When Ge specimen is doped with Al, then concentration of acceptor atoms is also called concentration of holes. Using formula

\mathrm{n_1^2=n_{\varepsilon} n_n}

\mathrm{\mathrm{n}_{\mathrm{i}}}= concentration of electron hole pair \mathrm{=10^{19} / \mathrm{m}^3}

\mathrm{\mathrm{n}_{\mathrm{e}}} = concentration of electrons

\mathrm{\mathrm{n}_{\mathrm{h}}} = concentration of holes \mathrm{=10^{21} \text { atoms } / \mathrm{m}^3}

\mathrm{\begin{aligned} \therefore \quad & \left(10^{19}\right)^2=10^{21} \times n_e \\ & n_e=10^{17} / \mathrm{m}^3 \end{aligned}}

Posted by

Gautam harsolia

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