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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 \mathrm{\approx 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\mathrm{Al}, then concentration of acceptor atoms is also called concentration of holes.

Using formula

\mathrm{n^2=n_e n_n}

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

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

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

\mathrm{\begin{gathered} \therefore\left(10^{19}\right)^2=10^{21} \times \mathrm{n}_{\mathrm{e}} \\ \mathrm{n}_{\mathrm{e}}=10^{17} / \mathrm{m}^3 \end{gathered}}

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HARSH KANKARIA

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