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Diffusion coefficient of electrons at 300 K

Short answer

Using Einstein relation De = μe kT/e, with kT/e = 0.02586 V at 300 K. De = 1300 cm²/V·s × 0.02586 V = 33.6 cm² s⁻¹.

Calculate the diffusion coefficient of electrons at 300 K in n-type silicon semiconductor with 10¹⁵ arsenic atoms per cm³. (μ = 1300 cm² V⁻¹ s⁻¹)

The diffusion coefficient is related to mobility by the Einstein relation. With Nd=1015N_d = 10^{15} cm⁻³ the semiconductor is non-degenerate, so the relation applies, and electrons are the majority carriers.

Deμe=kTe=1.381×10−23×3001.602×10−19=0.02586 VDe=μekTe=1300×0.02586=33.6 cm2 s−1\begin{aligned} \frac{D_e}{\mu_e} &= \frac{kT}{e} = \frac{1.381\times10^{-23} \times 300}{1.602\times10^{-19}} = 0.02586\ \text{V} \\ D_e &= \mu_e \frac{kT}{e} = 1300 \times 0.02586 \\ &= 33.6\ \text{cm}^2\,\text{s}^{-1} \end{aligned}

Answer: De≈33.6D_e \approx 33.6 cm² s⁻¹ =3.36×10−3= 3.36\times10^{-3} m² s⁻¹.

✓ Answer checked by Claude Opus 5.5 (high effort)

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