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Question:

What is the conductivity of a semiconductor sample having electron concentration of 5x1018m-3, hole concentration of 5x1019m-3, electron mobility of 2.0m2V-1s-1 and hole mobility of 0.01m2V-1s-1?

1.68(Ω⁻¹m⁻¹)

0.59(Ω⁻¹m⁻¹)

1.20(Ω⁻¹m⁻¹)

1.83(Ω⁻¹m⁻¹)

Solution:

Given : Hole concentration nh=5x1019 m-3
Hole mobility μh=0.01 m2V-1s-1
Electron concentration ne=5x1018 m-3
Electron concentration μe=2 m2V-1s-1
Conductivity of semiconductor σ=e(neμe+nhμh)
∴σ=(1.6x10-19)[(5x1018)(2)+(5x1019)(0.01)]
⇒ σ=1.68 (Ω⁻¹m⁻¹)