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Drift mobility in potassium (molar mass 39.95)

Short answer

Drift mobility mu_d = sigma/(e n). Electron concentration n = (rho N_A)/M = 1.37×10^28 m^-3, giving mu_d = 6.33×10^-3 m^2/(V·s) (or 63.3 cm^2/(V·s)).

If electrical conductivity of potassium is 1.39×10⁵ Sm/cm, calculate the drift mobility of electron at room temperature. Molar mass and density of potassium are 39.95 and 0.91 gm/cc.

Conductivity σ=enμd\sigma = e n \mu_d, so μd=σ/(en)\mu_d = \sigma/(en). Potassium is monovalent, so the free electron concentration equals the atomic concentration.

Given: σ=1.39×105 S/cm=1.39×107 S/m\sigma = 1.39 \times 10^{5}\ \text{S/cm} = 1.39 \times 10^{7}\ \text{S/m} (the unit "Sm/cm" is read as S/cm, the standard value for potassium), M=39.95 g/molM = 39.95\ \text{g/mol}, ρ=0.91 g/cm3=910 kg/m3\rho = 0.91\ \text{g/cm}^3 = 910\ \text{kg/m}^3.

Electron concentration

n=ρNAM=910×6.022×10230.03995=1.372×1028 m−3n = \frac{\rho N_A}{M} = \frac{910 \times 6.022 \times 10^{23}}{0.03995} = 1.372 \times 10^{28}\ \text{m}^{-3}

Drift mobility

μd=σen=1.39×1071.602×10−19×1.372×1028=6.33×10−3 m2V−1s−1\mu_d = \frac{\sigma}{e n} = \frac{1.39 \times 10^{7}}{1.602 \times 10^{-19} \times 1.372 \times 10^{28}} = 6.33 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} μd=63.3 cm2V−1s−1\mu_d = 63.3\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}

The corresponding mean free time is τ=μdme/e=3.6×10−14\tau = \mu_d m_e / e = 3.6 \times 10^{-14} s.

Answer: μd≈6.33×10−3 m2V−1s−1=63.3 cm2V−1s−1\mu_d \approx 6.33 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 63.3\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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