Question

39. If the number of electrons (majority carrier) in a semiconductor is 5 X 1020 m and pe is 0.135 mho, find the
resistivity of the semiconductor.
A.
00.0926 m
В.
00.0945 Om
C.
00.09120m
00.0978 Om​

Answer 1

Explanation:

If the number of electrons (majority carrier) in a semiconductor is 5 X 1020 m and pe is 0.135 mho, find the resistivity of the semiconductor. A. 00.0926 m.6

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Answer 2

The resistivity of semiconductor is 0.0926om option (A) is the correct.

Given,

Number of electrons in a semiconductor =

$5 \times {10 }^{20} {m}^{ - 3}$

ue = 0.135mho

To Find,

The resistivity of the semiconductor.

Solution,

we can solve the given problem in the following manner.

Now,

we know the conductivity = no of electron × ue

= 5×1.6× 0.135×10 mho/m

= 10.8mho/m

so resistivity = 1/ conductivity

$= \frac{1}{10.8}$

= 0.0926 om

Hence, the resistivity of the conductor is 0.0926 om so option a is the correct answer.