Question

What is the conductivity of a semiconductor sample having electron concentration of $5\times {{10}^{18}}{{m}^{-3}}$, hole concentration of $5\times {{10}^{19}}{{m}^{-3}}$, electron mobility of $2.0{{m}^{2}}{{v}^{-1}}{{s}^{-1}}$ and hole mobility of $0.01{{m}^{2}}{{v}^{-1}}{{s}^{-1}}$?
(take charge of the electron as $1.6\times {{10}^{-19}}C$).
A) $1.68{{\left( \Omega -m \right)}^{-1}}$
B) $0.59{{\left( \Omega -m \right)}^{-1}}$
C) $1.83{{\left( \Omega -m \right)}^{-1}}$
D) $1.20{{\left( \Omega -m \right)}^{-1}}$

Answer 1

Hint: The conductivity of a semiconductor depends upon the sum of the products of the hole concentration and hole mobility, and the electron concentration and electron mobility. This problem can be solved by plugging in the values given in the question in the formula for the conductivity of a semiconductor based on the hole concentration and mobility, and electron concentration and mobility.

Formula used:
The conductivity $\sigma $ of a semiconductor is given by
$\sigma =e\left( {{n}_{e}}{{\mu }_{e}}+{{n}_{h}}{{\mu }_{h}} \right)$
where ${{n}_{e}},{{n}_{h}}$ are the electron and hole concentrations in the semiconductor respectively, ${{\mu }_{e}},{{\mu }_{h}}$ are the mobilities of the electrons and holes respectively and $e=1.6\times {{10}^{-19}}C$ is the magnitude of the charge of an electron.

Complete step by step answer:
As explained in the hint, we can solve this question by plugging in the values of the electron and hole concentrations and their mobilities into the formula for the conductivity. Hence, let us proceed to do that.
The conductivity $\sigma $ of a semiconductor is given by
$\sigma =e\left( {{n}_{e}}{{\mu }_{e}}+{{n}_{h}}{{\mu }_{h}} \right)$ --(1)
where ${{n}_{e}},{{n}_{h}}$ are the electron and hole concentrations in the semiconductor respectively, ${{\mu }_{e}},{{\mu }_{h}}$ are the mobilities of the electrons and holes respectively and $e=1.6\times {{10}^{-19}}C$ is the magnitude of the charge of an electron.
Therefore, let us analyze the question.
The electron concentration of the semiconductor is ${{n}_{e}}=5\times {{10}^{18}}{{m}^{-3}}$.
The hole concentration of the semiconductor is ${{n}_{h}}=5\times {{10}^{19}}{{m}^{-3}}$.

The mobility of the electron in the semiconductor is ${{\mu }_{e}}=2.0{{m}^{2}}{{v}^{-1}}{{s}^{-1}}$.
The mobility of the hole in the semiconductor is ${{\mu }_{h}}=0.01{{m}^{2}}{{v}^{-1}}{{s}^{-1}}$.
Let the conductivity of the semiconductor be $\sigma $.
Given the magnitude of the charge on the electron is $e=1.6\times {{10}^{-19}}C$.
Hence, using (1), we get,
$\sigma =e\left( {{n}_{e}}{{\mu }_{e}}+{{n}_{h}}{{\mu }_{h}} \right)$
$\therefore \sigma =\left( 1.6\times {{10}^{-19}} \right)\times \left[ \left( 5\times {{10}^{18}}\times 2 \right)+\left( 5\times {{10}^{19}} \right)\times 0.01 \right]=\left( 1.6\times {{10}^{-19}} \right)\times \left[ 1\times {{10}^{19}}+\left( 0.05\times {{10}^{19}} \right) \right]$
$\therefore \sigma =\left( 1.6\times {{10}^{-19}} \right)\times {{10}^{19}}\left[ 1+0.05 \right]=1.6\times {{10}^{-19}}\times 1.05\times {{10}^{19}}=1.68{{\left( \Omega -m \right)}^{-1}}$
Hence, the conductivity of the semiconductor is $1.68{{\left( \Omega -m \right)}^{-1}}$.

Hence, the correct answer is option A. i.e., $1.68{{\left( \Omega -m \right)}^{-1}}$.

Note: Equation (1) can give the students a feel and understanding why trivalent or pentavalent impurities can increase the conductivity of a semiconductor. Trivalent and pentavalent impurities add free holes and electrons to the semiconductor respectively and increase the respective concentration. This increase in the concentration leads to an increase in the conductivity of the semiconductor as is evident from equation (1).