An n-type semiconductor is known to have an electron concentration of 3 ' 1018 m-3. If the electron drift velocity is 100 m/s in an electric field of 500 V/m. Calculate the mobility and conductivity of this material

Question

Engineering

Nathen Moses

23 January, 11:10

An n-type semiconductor is known to have an electron concentration of 3 ' 1018 m-3. If the electron drift velocity is 100 m/s in an electric field of 500 V/m. Calculate the mobility and conductivity of this material

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Answers (1)

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Dawson Arnold · Answer 1

Electron Mobility of the material = 0.20 m2/V-s

Conductivity of the material = 0.096 (Ω-m) - 1

Explanation:

Electron mobility = Vd/E

= 100/500

= 0.20 m2/V-s

Conductivity = σ = n |e|μ

= 3 x 1018 x 1.602 x 10 - 19 x 0.20

= 0.096 (Ω-m) - 1