Numerical
Mobility of electron and hole in a sample of Ge at room temperature are 0.36 m2 /V-sec and 0.17m2 /V-sec respectively. If electron and holes densities are equal and it is 2.5 ×1013 /cm3 , calculate its conductivity.
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Solution
Data :- μe = 0.36 m2 /V-sec μh = 0.17m2 /V-sec T =300°K
`n_i = 2.5×10^13/(cm^3 ) = 2.5×10^19m^3`
Formula :- σ = n(μe- μh). e
Calculations :- σ = 2.5×1019(0.36 + 0.17)×1.6×10-19
= 2.12 mho/metre
Answer :- σ = conductivity = 2.12 mho/metre.
Concept: Conductivity, mobility, current density (drift & diffusion) in semiconductors(n type and p type)
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