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The electrical conductivity of semiconductors is
A)${10^8}{\Omega ^{ - 1}}c{m^{ - 1}}$
B) ${10^{ - 23}}{\Omega ^{ - 1}}c{m^{ - 1}}$
C) ${10^{ - 9}}{\text{ to }}{10^2}{\text{ }}{\Omega ^{ - 1}}c{m^{ - 1}}$
D) None of the above.

Answer
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Hint: Electrical conductivity is defined as the measurement of electric flow in the given material. It has differed from one material to the other material and it is reciprocal to the resistivity of the material.

Complete Step by Step Solution:
As we know that electrical conductivity is the measurement of the flow of charge or flow of current in different- different materials. It is written as;
\[\sigma = \dfrac{1}{\rho }\]
Here, $\sigma $ is known as the electrical conductivity
            $\rho $ is known as resistivity.
The unit of electrical conductivity is ${\Omega ^{ - 1}}c{m^{ - 1}}$
According to the above equation we can say that the electrical conductivity is the inverse of the resistivity.
Now, Semiconductors are that material which come in between conductor and insulator. In a conductor, the flow of the current is large, therefore its electrical conductivity is also having a large value, whereas in the insulator the flow of the charge very low even though we can say that it is negligible to flow the charge in this type of material. Therefore, its electrical conductivity is very low as compared to the conductor.
Now, we know that the semiconductor is between conductor and insulator. Therefore, its electrical conductivity is also in between the conductor and insulator.
The values of electrical conductivity are shown below,
1. For conductor - $\sigma > {10^3}{\Omega ^{ - 1}}c{m^{ - 1}}$
2. For semiconductor - $\sigma = {10^{ - 9}} < \sigma < {10^2}\,{\Omega ^{ - 1}}c{m^{ - 1}}$
3. Insulator - $\sigma = {10^{ - 8}}{\Omega ^{ - 1}}c{m^{ - 1}}$
Hence, option (C) is the correct answer.

Note: Semiconductors are smaller in size and they are possessing very less weight as compared to the other types of conductors. Its resistance decreases when we increase the temperature and vice versa.