Calculate conductivity of a germanium sample if a donar impurity atoms are added to the extent to one part in 10^{6 } germanium atoms at room temperature.

Assume that only one electron of each atom takes part in conduction process.

Given:- Avogadro’s number = 6.023 × 10^{23 }atom/gm-mol .

Atomic weight of Ge = 72.6

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

Density of Ge = 5.32 gm/ cm^{3}

#### Solution

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`

= 441.35×10^{26 }

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

10^6`

= 441.35×10^{20 }

Conductivity, `σ=n_e eμ_e`

=441.35×10^{20 }×1.6×10^{-19 } ×0.38

=2683

Conductivity =2683 mho/m.