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

We need to relate the concentration of holes to the concentration of electron pairs and concentration of electron-hole pairs. Doping is done in a semiconductor to increase the conductivity of the semiconductor. We are given with Ge being doped with Al.

Formula Used: The formulae used in the solution are given here.
We know the relationship between them is given as-
${n_e}{n_h} = {n_i}^2$, where ${n_e}$ concentration of electron is, ${n_h}$ is concentration of holes and ${n_i}$ is concentration of electron pairs in intrinsic semiconductor.

Complete step by step answer
Doping is done in a semiconductor to increase the conductivity of the semiconductor. Doping means the addition of impurities into a semiconductor to define modification of conductivity. Two of the most important materials with which silicon can be doped with, are boron which has 3 electrons in its outermost shell and phosphorus which have 5 electrons in its outermost shell. Other materials that are commonly used are aluminium, arsenic, antimony etc. For p-doped semiconductors, the acceptor energy level is close to the valence band. And for n- doped semiconductors the donor energy level is close to the conduction band.
It has been given that, Ge being doped with Al. We need to find the concentration of the electron specimen in the sample.
Let ${n_e}$ is concentration of electrons, ${n_h}$ is concentration of holes and ${n_i}$ is concentration of electron pairs in intrinsic semiconductor. We know the relationship between them is given as-
${n_e}{n_h} = {n_i}^2$.
Given in the question, ${n_h} = {10^{21}}atoms/{m^3}$ and ${n_i} = {10^{19}}atoms/{m^3}$. We are required to find the concentration of electrons, that is, the value of, ${n_e}$
Now, it can also be written, ${n_e} = \dfrac{{{n_i}^2}}{{{n_h}}}$
Thus the value of ${n_e}$ is:
$\Rightarrow {n_e} = \dfrac{{{{\left( {{{10}^{19}}} \right)}^2}}}{{{{10}^{21}}}} = \dfrac{{{{10}^{38}}}}{{{{10}^{21}}}} = {10^{17}}atoms/{m^3}$

Thus, the correct answer is Option A.

Note
Elements with 3 valence electrons are used for p-type doping and elements with 5 valence electrons are used for n-type doping. P-type semiconductors have holes as their majority charge carriers and n-type semiconductors have electrons as majority charge carriers.