Question

A Ge specimen is doped with $Al$. The concentration of acceptor atoms is $\sim {10}^{21}atoms/m^3$. Given that the intrinsic concentration of electron-hole pairs is $\sim {10}^{19}/m^3$, the concentration of electrons in the specimen is

• A. ${10}^{17}/m^3$
• B. ${10}^{15}/m^3$
• C. ${10}^4/m^3$
• D. ${10}^2/m^3$

Answer 1

Answer: (A)

${10}^{17}/m^3$

$n_e{n}_h=n_i^2$
$n_e$ is concentration of electron, $n_h$ is concentration of holes and $n_i$ is concentration of electron pairs in intrinsic semi-conductor.
Here $n_h={10}^{21},n_e=$?, $n_i={10}^{19}$
${10}^{21}\times n_e={10}^{19}\times {10}^{19}$
$n_e=\frac{{10}^{38}}{{10}^{21}}={10}^{17}{m}^{-3}$