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 $ .

Given data :- $N = 6.023 × 10^{23}atoms/ gm-mole ,\hspace{1cm} atomic weight of Ge, A = 72.6,$

$ μ_e=0.38 m^2/V-sec, \hspace{1cm} ρ=5320 kg/m^3 $

$ Formula :- σ=n_e eμ_e$

$ Solution :- no .of atoms / unit volume = Nρ/A = (6.023 × 10^{26}×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.