A phosphorus‑doped silicon wafer has donor concentration $N_D=5\times 10^{16}{\text{cm}}^{-3}$ and electron mobility ${\mu }_n=1350{\text{cm}}^2{\text{V}}^{-1}{\text{s}}^{-1}$. Assuming holes are negligible, calculate its electrical conductivity $\sigma$ in ${\Omega }^{-1}{\text{cm}}^{-1}$. Take $q=1.6\times 10^{-19}\text{C}$.

Step-by-step solution

1. Write the conductivity formula for an n‑type semiconductor: $\sigma =qn{\mu }_n$, where $n\approx N_D$ because donors are fully ionized at 300 K.
2. Substitute the values: $\sigma =(1.6\times 10^{-19})(5\times 10^{16})(1350)$.
3. Perform the multiplication: $1.6\times 5=8.0$, $8.0\times 10^{-19+16}=8.0\times 10^{-3}$, then $8.0\times 10^{-3}\times 1350=10.8\times 10^1=1.08\times 10^2$.
4. Thus $\sigma =1.08\times 10^2{\Omega }^{-1}{\text{cm}}^{-1}$.

Concept tested: Conductivity of extrinsic semiconductor

Published 14 September 2026 · How this solution was prepared and checked