# 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. - Applied Physics 1

Calculate conductivity of a germanium sample if a donar impurity atoms are added to the extent to one part in 10 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

#### 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

= 441.35×1026

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.

Concept: Type I superconductors
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