Discussion
The electron concentration in an n-type semiconductor is \(5 \times 10^{18}\)/m3. If the drift velocity of electrons is 100 m/s in an electric field of 500 V/m, calculate the conductivity of the semiconductor.
Answer: B
The conductivity is \(1.60 \times 10^{-1}\) S/m.
The electron mobility is the drift velocity per unit field, and conductivity follows from \(\sigma = ne\mu\):
\[ \begin{aligned} \mu &= \frac{v_{d}}{E} = \frac{100}{500} = 0.2\ \text{m}^{2}\text{/V s} \\ \sigma &= (5 \times 10^{18})(1.602 \times 10^{-19})(0.2) \\ &= 0.16\ \text{S/m} \end{aligned} \]
That is \(1.60 \times 10^{-1}\) S/m. \(0.16 \times 10^{-1}\) is ten times too small, and 2.50 and 30.05 come from using the drift velocity or field in place of the mobility.
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