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Drift mobility of electrons in potassium

Short answer

Electron concentration n = (910 kg/m^3 × 6.022e23) / 0.0395 kg/mol = 1.387e28 m^-3. With sigma = 1.39e7 S/m, drift mobility mu_d = sigma / (e n) = 1.39e7 / (1.602e-19 × 1.387e28) = 6.25e-3 m^2/Vs (62.5 cm^2/Vs).

If electric 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 39.5 and 0.91 gm/cc.

Conductivity of a metal is σ=enμd\sigma = e n \mu_d, so

μd=σen\mu_d = \frac{\sigma}{e n}

Potassium is monovalent (one valence electron per atom), so nn 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=39.5 g/mol=0.0395 kg/molM = 39.5\ \text{g/mol} = 0.0395\ \text{kg/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=910×6.022×10230.0395=1.387×1028 m−3n = \frac{\rho N_A}{M} = \frac{910 \times 6.022 \times 10^{23}}{0.0395} = 1.387 \times 10^{28}\ \text{m}^{-3}

Drift mobility

μd=1.39×1071.602×10−19×1.387×1028=6.25×10−3 m2V−1s−1\mu_d = \frac{1.39 \times 10^{7}}{1.602 \times 10^{-19} \times 1.387 \times 10^{28}} = 6.25 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} μd=62.5 cm2V−1s−1\mu_d = 62.5\ \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≈6.25×10−3 m2V−1s−1=62.5 cm2V−1s−1\mu_d \approx 6.25 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 62.5\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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