Einstein relation between mobility and diffusion: Exercises

Constants used throughout: $k_{B} = 1.380649 \times 1 0^{- 23} J/K$ , $q = 1.602 \times 1 0^{- 19} C$ .

Level 1 — Recognition

Goal: can you spot the relation and plug in?

Recall Solution L1·Q1

WHAT: use $D = V_{T} μ$ . WHY: the Einstein relation says $D / μ = V_{T}$ , so $D$ is just mobility scaled by the thermal voltage. $D_{n} = V_{T} μ_{n} = 0.02585 V \times 1350 \frac{cm ^{2}}{V\cdots} = 34.9 cm^{2} / s .$ The volts cancel, leaving $cm^{2} / s$ — the correct units for $D$ . ✓ (Textbook value $\approx 35$ .)

Recall Solution L1·Q2

WHAT: invert the relation, $μ = D / V_{T}$ . WHY: we know $D$ and want $μ$ , so divide by the thermal voltage. $μ_{p} = \frac{D _{p}}{V _{T}} = \frac{12}{0.02585} = 464 cm^{2} / (V\cdots) .$

Recall Solution L1·Q3

$[D] = cm^{2} / s$ and $[μ] = cm^{2} / (V\cdots)$ . Dividing: $\frac{cm ^{2} / s}{cm ^{2} / ( V\cdots )} = \frac{cm ^{2}}{s} \cdot \frac{V\cdots}{cm ^{2}} = V . ✓$ This is why the right side must be $k_{B} T / q$ (energy per charge = volts), not $k_{B} T$ (energy).

Level 2 — Application

Goal: chain the relation with one extra idea (temperature, a second carrier, a gradient).

Recall Solution L2·Q1

WHAT: since $D = V_{T} μ = (k_{B} T / q) μ$ and $μ$ is fixed, $D \propto T$ . WHY: the only temperature-dependent factor left is $T$ inside $V_{T}$ . $D_{n} (350) = 35 \times \frac{350}{300} = 40.8 cm^{2} / s .$

Recall Solution L2·Q2

Each carrier obeys $D = V_{T} μ$ with the same $V_{T}$ : $D_{n} = 0.02585 \times 1350 = 34.9, D_{p} = 0.02585 \times 480 = 12.4 cm^{2} / s .$ The ratio: $\frac{D _{n}}{D _{p}} = \frac{V _{T} μ _{n}}{V _{T} μ _{p}} = \frac{μ _{n}}{μ _{p}} = \frac{1350}{480} = 2.81.$ WHY the ratio simplifies: $V_{T}$ is identical for both, so it cancels — the diffusion ratio equals the mobility ratio.

Recall Solution L2·Q3

WHAT: differentiate the profile, then apply Fick's law (see Diffusion current and Fick's law). $\frac{d n}{d x} = - \frac{n _{0}}{L} e^{- x / L}, \frac{d n}{d x}_{0} = - \frac{n _{0}}{L} .$ Convert $L = 2 μ m = 2 \times 1 0^{- 4} cm$ : $F = - D_{n} (- \frac{n _{0}}{L}) = 35 \times \frac{1 0 ^{16}}{2 \times 1 0 ^{- 4}} = 1.75 \times 1 0^{21} cm^{- 2} s^{- 1} .$ Positive flux → carriers stream toward $+ x$ (from crowded to empty), exactly what diffusion should do.

Level 3 — Analysis

Goal: reason about the derivation, signs, and structure — not just plug numbers.

Recall Solution L3·Q1

WHAT: differentiate the Boltzmann profile (see Boltzmann distribution in semiconductors). $\frac{d n}{d x} = n_{0} exp (\frac{q ψ}{k _{B} T}) \cdot \frac{q}{k _{B} T} \frac{d ψ}{d x} = n \cdot \frac{q}{k _{B} T} \frac{d ψ}{d x} .$ Divide by $n$ : $\frac{1}{n} \frac{d n}{d x} = \frac{q}{k _{B} T} \frac{d ψ}{d x} .$ ✓ WHY it's the whole physics: it ties the concentration gradient (what diffusion feeds on) to the field (what drift feeds on) through the single constant $k_{B} T / q$ . Once you demand zero net current, this identity is what makes $D / μ$ collapse to $k_{B} T / q$ .

Recall Solution L3·Q2

Step A — particle flux (Fick): $F = - D_{n} \frac{d n}{d x}$ (particles go from high to low $n$ ). Step B — current = charge × flux: electrons carry charge $- q$ , so $J_{n, diff} = (- q) F = (- q) (- D_{n} \frac{d n}{d x}) = + q D_{n} \frac{d n}{d x} .$ WHY two minus signs: one from Fick's law (flux opposes the gradient), one from the electron's negative charge. They multiply to a plus.

Figure — Diffusion flux vs. electron current for a decaying electron cloud. The blue curve is a decaying electron cloud (crowded on the left, empty on the right). The green arrow is the particle flux $F$ : it points in $+ x$ , from crowded to empty — that is diffusion doing its natural spreading. The red arrow is the resulting electron current $J_{n}$ : it points in $- x$ , exactly opposite the green arrow, precisely because the electron's charge is negative. So the two arrows being anti-parallel is the visual proof of the "two minus signs make a plus" bookkeeping above.

Figure — Einstein relation between mobility and diffusion

Figure L3·Q2: green = particle flux $F$ (toward $+ x$ ); red = electron current $J_{n}$ (toward $- x$ ); blue = electron density $n (x)$ .

Recall Solution L3·Q3

Definition recap. $v_{t h} \equiv k_{B} T / m^{*}$ , so $v_{t h}^{2} = k_{B} T / m^{*}$ — this is exactly what equipartition $\frac{1}{2} m^{*} v_{t h}^{2} = \frac{1}{2} k_{B} T$ gives along one axis. $τ$ is the mean time between collisions and $m^{*}$ the effective mass.

1-D estimate. With $v_{t h}^{2} = k_{B} T / m^{*}$ , $D = \frac{1}{2} v_{t h}^{2} τ = \frac{1}{2} \frac{k _{B} T}{m ^{*}} τ, μ = \frac{q τ}{m ^{*}} .$ $\frac{D}{μ} = \frac{\frac{1}{2} ( k _{B} T / m ^{*} ) τ}{q τ / m ^{*}} = \frac{k _{B} T}{2 q} .$ The main point: $τ$ and $m^{*}$ appear in both $D$ and $μ$ identically, so they cancel in the ratio. That is why $D / μ$ is universal — independent of how dirty or clean the crystal is.

WHY the $\frac{1}{2}$ is spurious. The crude estimate $D = \frac{1}{2} v_{t h}^{2} τ$ pretends a carrier moves at a single fixed speed $v_{t h}$ that reverses every $τ$ — a one-dimensional caricature. In reality the carrier has a distribution of velocities in three dimensions, and the correct random-walk result is $D = ⟨ v_{x}^{2} ⟩ τ$ with $⟨ v_{x}^{2} ⟩ = k_{B} T / m^{*}$ from equipartition (no factor $\frac{1}{2}$ , because averaging $v_{x}^{2}$ over the Maxwell distribution already supplies the right numerical weight). Substituting this exact $D = (k_{B} T / m^{*}) τ$ gives $\frac{D}{μ} = \frac{( k _{B} T / m ^{*} ) τ}{q τ / m ^{*}} = \frac{k _{B} T}{q} .$ So the $\frac{1}{2}$ was never physics — it was the price of replacing a proper velocity average with a single hand-picked speed. The cancellation of $τ$ and $m^{*}$ (the deep result) survives either way.

Level 4 — Synthesis

Goal: combine the Einstein relation with a second law (built-in potential, continuity, or a full profile).

Recall Solution L4·Q1

WHAT: invert the Boltzmann profile from L3·Q1. WHY: zero net current forces the equilibrium distribution $n = n_{0} e^{q ψ / k_{B} T}$ , so a density ratio is a potential difference. $\frac{n _{2}}{n _{1}} = exp (\frac{q ( ψ _{2} - ψ _{1} )}{k _{B} T}) \Rightarrow Δ ψ = V_{T} ln \frac{n _{2}}{n _{1}} .$ $Δ ψ = 0.02585 \times ln \frac{1 0 ^{15}}{1 0 ^{17}} = 0.02585 \times (- 4.605) = - 0.119 V .$ The low-density side sits at lower potential by $119 mV$ . Notice $V_{T} = k_{B} T / q$ — the very Einstein-relation constant — is what converts a concentration ratio into a voltage.

Recall Solution L4·Q2

WHAT: apply $L_{n} = D_{n} τ_{r}$ (from Continuity equation and carrier transport). $L_{n} = 35 \frac{cm ^{2}}{s} \times 1 0^{- 6} s = 3.5 \times 1 0^{- 5} cm^{2} = 5.92 \times 1 0^{- 3} cm = 59.2 μ m .$ Using $D_{n} = V_{T} μ_{n}$ : $L_{n} = V_{T} μ_{n} τ_{r} .$ WHY this matters: the Einstein relation lets you predict how far carriers diffuse before recombining knowing only mobility, temperature, and lifetime — the mobility measurement doubles as a diffusion measurement.

Recall Solution L4·Q3

Zero net current requires $q n μ_{n} E_{0} = - q D_{n} d n / d x$ everywhere, where $W$ is the fixed width of the linear region. Compute: $\frac{d n}{d x} = - \frac{n _{0}}{W} (constant), q n μ_{n} E_{0} = q D_{n} \frac{n _{0}}{W} .$ Left side is $q μ_{n} E_{0} n_{0} (1 - x / W)$ — it varies with $x$ . Right side $q D_{n} n_{0} / W$ is constant. They can only match if the $x$ -dependence disappears, i.e. only at one specific $x$ , not everywhere. Conclusion: impossible in equilibrium with a constant field. A genuine equilibrium demands the exponential Boltzmann profile, whose gradient scales with $n$ itself so both sides carry the same $n (x)$ factor. This is exactly why Step 4 uses an exponential, not a line.

Level 5 — Mastery

Goal: push to limits, degeneracy, and full derivations where the classical form breaks.

Definition Symbols used in this level

$E_{C}$ — the conduction-band edge: the lowest energy an electron can have while still being free to conduct. Think of it as the "floor" of the band that mobile electrons sit above.
$E_{F}$ — the Fermi level: the energy at which a state has a $50%$ chance of being occupied; it marks how "full" the electron sea is. When $E_{F}$ climbs up toward and past $E_{C}$ , states near the band edge are already crowded — that is the degenerate regime.
$η \equiv (E_{F} - E_{C}) / k_{B} T$ — the Fermi level measured from the band edge in units of thermal energy. Large negative $η$ means $E_{F}$ sits far below the band (dilute, Boltzmann-like); $η ≳ 0$ means degenerate. See Fermi-Dirac vs Maxwell-Boltzmann statistics and Degenerate semiconductors & generalized Einstein relation.
$F_{j} (η)$ — the Fermi–Dirac integral of order $j$ : the proper carrier-counting integral when Fermi–Dirac (not Boltzmann) statistics apply. All you need below is the ratio $F_{1/2} / F_{- 1/2}$ , which equals 1 in the dilute limit and grows above 1 as the material becomes degenerate.

Recall Solution L5·Q1

WHAT: multiply the classical $V_{T}$ by the Fermi–Dirac correction factor (see Degenerate semiconductors & generalized Einstein relation). $\frac{D}{μ} = 0.02585 \times 1.15 = 0.02973 V = 29.7 mV .$ WHY it exceeds $k_{B} T / q$ : in a degenerate gas the Fermi–Dirac filling means adding a carrier costs more than $k_{B} T$ (states are already occupied — the Pauli exclusion raises the effective spreading pressure). The correction factor is $\geq 1$ , so degeneracy always raises $D / μ$ above the non-degenerate value.

Recall Solution L5·Q2

In the Boltzmann tail, $F_{1/2} (η) \to e^{η}$ and $F_{- 1/2} (η) \to e^{η}$ . Their ratio: $\frac{F _{1/2} ( η )}{F _{- 1/2} ( η )} \to \frac{e ^{η}}{e ^{η}} = 1.$ Therefore $D / μ \to (k_{B} T / q) \times 1 = k_{B} T / q$ . ✓ WHY: far from degeneracy, Fermi–Dirac statistics become Maxwell–Boltzmann, so the correction factor is exactly 1 and the classical Einstein relation is recovered. The classical form is the low-density limit of the general one — which is exactly the caveat stated at the top of this page.

Recall Solution L5·Q3

Step 1 — set net current to zero. Equilibrium means no wires, no net flow: $q n μ_{n} E + q D_{n} \frac{d n}{d x} = 0 \Rightarrow q n μ_{n} E = - q D_{n} \frac{d n}{d x} .$ Step 2 — substitute $E = - d ψ / d x$ . The field is the negative slope of potential: $q n μ_{n} (- \frac{d ψ}{d x}) = - q D_{n} \frac{d n}{d x} \Rightarrow n μ_{n} \frac{d ψ}{d x} = D_{n} \frac{d n}{d x} .$ Rearrange into a separable form (all $n$ on one side, all $ψ$ on the other): $\frac{d n}{n} = \frac{μ _{n}}{D _{n}} d ψ .$ Step 3 — insert the empirical ratio $μ_{n} / D_{n} = q / k_{B} T$ and integrate. $\int \frac{d n}{n} = \frac{q}{k _{B} T} \int d ψ \Rightarrow ln n = \frac{q ψ}{k _{B} T} + C .$ Exponentiating and writing $n_{0} = e^{C}$ : $n (ψ) = n_{0} exp (\frac{q ψ}{k _{B} T}) .$ WHY this closes the loop: we started from current balance without assuming any distribution, and feeding in only the Einstein ratio forced the Boltzmann profile to appear. So the Einstein relation and the equilibrium Boltzmann distribution are two faces of the same fact — assume either one and the other follows. That is the deepest statement on this page: equilibrium, the Einstein ratio, and Boltzmann statistics are logically interlocked. The figure shows the payoff — at every point the drift and diffusion currents are equal and opposite, so the net current is zero everywhere.

Figure L5·Q3: blue = equilibrium density $n (x) = n_{0} exp (q ψ / k_{B} T)$ ; at each sample point the yellow drift arrow and red diffusion arrow are equal and opposite, so net current is zero everywhere.

Connections

Einstein relation between mobility and diffusion (parent)
Drift current and mobility
Diffusion current and Fick's law
Boltzmann distribution in semiconductors
Thermal voltage V_T
Continuity equation and carrier transport
Degenerate semiconductors & generalized Einstein relation
Fermi-Dirac vs Maxwell-Boltzmann statistics