SemiconductorMedium
Question
Mobilities of electrons and holes in a sample of intrinsic Ge at room temperature are 0.35m2 /V-s and 0.18m2 /V-s respectively. If the electron and hole densities are each equal to 2.5 × 1019 / m3, the Ge conductivity will be
Options
A.3.12 S/m
B.2.12 S/m
C.1.12 S/m
D.4.12 S/m
Solution
Conductivity of Ge
σ = e(neμe + nhμh)
Here ne = nh = 25 × 1019 / m3
e = 1.6 × 10-19 C
μe = 0.35m2/V - s, μh = 0.18m2 /V - s
∴ σ = 1.6 × 10-19
(205 × 1019 × 0.35 + 2.5 × 1019 × 0.18)
= 1.6 × 10-19 × 2.5 × 1019 × 0.53
= 2.12 S/m
σ = e(neμe + nhμh)
Here ne = nh = 25 × 1019 / m3
e = 1.6 × 10-19 C
μe = 0.35m2/V - s, μh = 0.18m2 /V - s
∴ σ = 1.6 × 10-19
(205 × 1019 × 0.35 + 2.5 × 1019 × 0.18)
= 1.6 × 10-19 × 2.5 × 1019 × 0.53
= 2.12 S/m
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