Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The density of an electron - hole pair in a pure germanium is $3 \times {10^{16}}{m^{ - 3}}$ at room temperature. On doping with aluminium, the hole density increases to $4.5 \times {10^{22}}{m^{ - 3}}$ . Now the electron density ( in ${m^{ - 3}}$ ) in doped germanium will be?
(A) $1 \times {10^{10}}$
(B) $2 \times {10^{10}}$
(C) $0.5 \times {10^{10}}$
(D) $4 \times {10^{10}}$

seo-qna
Last updated date: 23rd May 2024
Total views: 45.9k
Views today: 1.45k
Answer
VerifiedVerified
45.9k+ views
Hint: Use the formula : ${n_h}{n_e} = {n_i}^2$
where, ${n_h}$ is the extrinsic number density of holes after doping in the semiconductor
${n_e}$ is the extrinsic number density of electrons after doping in the semiconductor
and, ${n_i}$ is the intrinsic number density of electron - hole pairs in pure semiconductor

Complete step by step solution
We are given the density of electron hole pairs in pure germanium to be $3 \times {10^{16}}{m^{ - 3}}$. This is the intrinsic density or the density of electron hole pairs, before doping.
\[ \Rightarrow {n_i} = 3 \times {10^{16}}{m^{ - 3}}\]
We are also given the density of holes in extrinsic germanium to be $4.5 \times {10^{22}}{m^{ - 3}}$, i.e. after it has been doped with aluminium.
\[ \Rightarrow {n_h} = 4.5 \times {10^{22}}{m^{ - 3}}\]
We have to find the density of electrons in extrinsic germanium, i.e. after it has been doped with aluminium. This means we have to find ${n_e}$.
We know the formula ${n_h}{n_e} = {n_i}^2$, we will find an expression for ${n_e}$ from this,
\[
   \Rightarrow {n_h}{n_e} = {n_i}^2 \\
   \Rightarrow {n_e} = \dfrac{{{n_i}^2}}{{{n_h}}} \\
 \]
Now substituting all the known values to the right hand side of above equation,
\[ \Rightarrow {n_e} = \dfrac{{{{\left( {3 \times {{10}^{16}}{m^{ - 3}}} \right)}^2}}}{{4.5 \times {{10}^{22}}{m^{ - 3}}}}\]
\[ \Rightarrow {n_e} = \dfrac{{9 \times {{10}^{32}}{m^{ - 6}}}}{{4.5 \times {{10}^{22}}{m^{ - 3}}}}\]
\[ \Rightarrow {n_e} = 2 \times {10^{10}}{m^{ - 3}}\]

Therefore, option (B) is correct.

Note: Doping means the introduction of impurities into a pure semiconductor crystal on purpose, to alter the concentration of electrons or holes in it. Here, aluminium is added to a pure germanium semiconductor crystal. Aluminium has 3 valence electrons, i.e. it is a trivalent impurity. Addition of a trivalent impurity causes the concentration of holes to increase, and the semiconductor to become a p -type extrinsic semiconductor. Hence, the density of holes should be more than the density of electrons for such a case, which is in agreement with our values of \[{n_e}\] and \[{n_h}\].