A Si crystal with \( 5 \times 10^{28} \, \text{atoms} \, \text{m}^{-3} \) is doped with 0.5 ppm of pentavalent impurity.
**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. 0.5 ppm = 0.5 × 10⁻⁶ . Number of donor atoms = 0.5 × 10⁻⁶ × 5 × 10²⁸ = 2.5 × 10²² m⁻³ , each contributing one electron. Substituting values gives 2.5 × 10²² m⁻³, which matches expected behaviour for this semiconductor device configuration, confirming doping, depletio
Ref: NCERT > Physics Book > Electronic Devices > Doping, Charge Carriers and Conductivity