JEE Main 2017PhysicsSemiconductorsEasyMCQ

JEE Main 2017 — Semiconductors Question with Solution

JEE Main 2017 (08 Apr Online)

Question

The conductivity of a semiconductor sample having electron concentration of 5×1018 electrons m-3, hole concentration of 5×1019 holes m-3, electron mobility of 2.0 m2 V-1 s-1 and hole mobility of 0.01 m2 V-1 s-1 is

(Take charge of an electron as 1.6×10-19 C )
 

Choose an option

Show full solutionCorrect option: B
Correct answer
B1.65 Ω m-1

Step-by-step explanation

The conductivity of a semiconductor is given by

σ=eneμe+nhμh

=1.6×10-195×1018×2+5×1019×0.01

=1.6×10-191019+0.05×1019

=1.6+1.05

=1.65 (Ω m)−1

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About this question

This is a previous-year question from JEE Main 2017, covering the Semiconductors chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.