Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in
anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

Latest Question