Question

Calculate conductivity of a germanium sample if a donar impurity atoms are added to the extent to one part in 10⁶  germanium atoms at room temperature.
Assume that only one electron of each atom takes part in conduction process.

Given:- Avogadro’s number = 6.023 × 10²³ atom/gm-mol . 
Atomic weight of Ge = 72.6

Mobility of electrons = 3800` (cm^2) / ( "volts") `sec

Density of Ge = 5.32 gm/ cm³

 

Answer 1

Given data ` N = 6.023 xx 10^23 ` atoms / gm - mole , atomic weight of Ge , A = 72.6 ,

`μ_e = 0.38 m^2/V - sec ,                      ρ = 5320  kg ⁄ m^3`

Formula :- `σ=n_e  e μ_e`
Solution :-no. of atoms / unit volume =` (Nρ)/A = (6.023 xx 10^26 xx 5320)/72.6`

        = 441.35×10²⁶ 

No .of electrons added /unit vol  =`n_e = (441.35×10^26)/
10^6`
= 441.35×10²⁰ 

Conductivity, `σ=n_e eμ_e`
=441.35×10²⁰ ×1.6×10^-19  ×0.38
=2683
Conductivity =2683 mho/m.