A pure Ge crystal has \( 4 \times 10^{28} \, \text{atoms} \, \text{m}^{-3} \) and is doped with 2 ppm of trivalent impur
**Number of carriers** from doping: Ge crystal 4×10²⁸ atoms/m³ doped 2 ppm trivalent gives acceptor atoms 8×10²² m⁻³, holes ≈ that, for 1.5 ppm 6×10²² m⁻³, for 0.5 ppm pentavalent Si 5×10²⁸ atoms/m³ gives 2.5×10²² donors/m³. Acceptor atom effectively negative when accepts electron, donor positive when donates, but crystal neutral. 2 ppm = 2 × 10⁻⁶ . Number of acceptor atoms = 2 × 10⁻⁶ × 4 × 10²⁸ = 8 × 10²² m⁻³ , contributing holes for p-type conductivity. Substituting values gives 8 × 10²² m⁻³, which matches expected behaviour for this semiconductor device configuration, confirming doping, dep
Ref: NCERT > Physics Book > Electronic Devices > Doping, Charge Carriers and Conductivity