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Numerical

Mobility of electron and hole in a sample of Ge at room temperature are 0.36 m^{2 }/V-sec and 0.17m^{2 }/V-sec respectively. If electron and holes densities are equal and it is 2.5 ×10^{13 }/cm^{3 }, calculate its conductivity.

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#### Solution

Data :- μ_{e } = 0.36 m^{2 }/V-sec μ_{h } = 0.17m^{2 }/V-sec T =300°K

`n_i = 2.5×10^13/(cm^3 ) = 2.5×10^19m^3`

Formula :- σ = n(μ_{e}- μ_{h}). e

Calculations :- σ = 2.5×10^{19}(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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