Free Electron Theory of Conduction in Metal

In the free-electron (Drude) model the drift mobility is related to the mean free time $\tau$ between collisions by

$$\mu_d = \frac{e\tau}{m_e} \quad\Rightarrow\quad \tau = \frac{\mu_d m_e}{e}$$

and the mean free path is the distance travelled at the mean speed $u$ in time $\tau$: $\lambda = u\tau$.

Given: $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 43 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $u = 1.2 \times 10^{6}\ \text{m/s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$, $e = 1.602 \times 10^{-19}\ \text{C}$.

Mean free time

$$\tau = \frac{43 \times 10^{-4} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.44 \times 10^{-14}\ \text{s}$$

Mean free path

$$\lambda = u\tau = 1.2 \times 10^{6} \times 2.44 \times 10^{-14} = 2.93 \times 10^{-8}\ \text{m} \approx 29.3\ \text{nm}$$

This is roughly 100 interatomic spacings (copper spacing is about 0.26 nm), showing that conduction electrons are not scattered by every ion but only by lattice vibrations and defects.

Answer: $\lambda \approx 2.93 \times 10^{-8}\ \text{m} = 29.3\ \text{nm}$ (with $\tau = 2.44 \times 10^{-14}\ \text{s}$).

The Fermi energy at 0 K for a free-electron metal is

$$E_{F0} = \frac{h^2}{8 m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

where $n$ is the free electron concentration.

Origin of the formula: at 0 K all states up to $E_{F0}$ are filled. Integrating the density of states $g(E) = \frac{8\pi\sqrt{2}m_e^{3/2}}{h^3}E^{1/2}$ from 0 to $E_{F0}$ gives $n = \frac{8\pi}{3}\left(\frac{2m_e E_{F0}}{h^2}\right)^{3/2}$; solving for $E_{F0}$ gives the expression above.

Given: $\rho = 8.96\ \text{g/cm}^3 = 8960\ \text{kg/m}^3$, $M = 63.5\ \text{g/mol} = 0.0635\ \text{kg/mol}$, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$. Copper is monovalent, so each atom gives one free electron.

Electron concentration

$$n = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.0635} = 8.50 \times 10^{28}\ \text{m}^{-3}$$

Fermi energy

$$\begin{aligned} \left(\frac{3n}{\pi}\right)^{2/3} &= \left(\frac{3 \times 8.50 \times 10^{28}}{\pi}\right)^{2/3} = 1.874 \times 10^{19}\ \text{m}^{-2} \\ \frac{h^2}{8m_e} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} = 6.025 \times 10^{-38}\ \text{J m}^2 \\ E_{F0} &= 6.025 \times 10^{-38} \times 1.874 \times 10^{19} = 1.13 \times 10^{-18}\ \text{J} \\ &= \frac{1.13 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.05\ \text{eV} \end{aligned}$$

The corresponding Fermi speed is $v_F = \sqrt{2E_{F0}/m_e} \approx 1.57 \times 10^{6}\ \text{m/s}$, showing that even at 0 K the top electrons move very fast.

Answer: $n = 8.50 \times 10^{28}\ \text{m}^{-3}$, $E_{F0} \approx 1.13 \times 10^{-18}\ \text{J} = 7.05\ \text{eV}$.

The saturation current of a thermionic cathode is given by the Richardson-Dushman equation:

$$J = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad I = J \times A$$

where $A$ is the emitting (curved) surface area of the cylinder, $A = \pi d l$.

Given: $T = 1500 + 273.15 = 1773.15\ \text{K}$, $l = 5\ \text{cm} = 0.05\ \text{m}$, $d = 1.5\ \text{mm} = 1.5 \times 10^{-3}\ \text{m}$, $B_0 = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\Phi = 2.6\ \text{eV}$.

Emitting area

$$A = \pi d l = \pi \times 1.5 \times 10^{-3} \times 0.05 = 2.356 \times 10^{-4}\ \text{m}^2$$

Exponent

$$\frac{\Phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 1773.15} = 17.01, \qquad e^{-17.01} = 4.10 \times 10^{-8}$$

Current density

$$J = 3 \times 10^{4} \times (1773.15)^2 \times 4.10 \times 10^{-8} = 3.87 \times 10^{3}\ \text{A/m}^2$$

Saturation current

$$I = J A = 3.87 \times 10^{3} \times 2.356 \times 10^{-4} = 0.911\ \text{A}$$

Answer: Saturation current $I \approx 0.91\ \text{A}$ (current density $\approx 3.87 \times 10^{3}\ \text{A/m}^2$).

Thermionic emission and work function

Work function $\Phi$ is the minimum energy needed to remove an electron from the Fermi level of a metal to the vacuum just outside it ($\Phi = E_{vac} - E_F$; e.g. W 4.5 eV, thoriated W 2.6 eV).

Thermionic emission is the emission of electrons from a heated metal. At high $T$ the Fermi-Dirac tail extends above $E_F + \Phi$, so some electrons have enough energy, in the direction normal to the surface, to escape. It is used in cathodes of vacuum tubes, CRTs and X-ray tubes.

 Energy
  ^  vacuum level -------------
  |                 ^
  |                 | Phi
  |  EF   ----------v----  <- top filled states
  |       /////////////////
  |  metal  (filled band)

Derivation of Richardson-Dushman equation

Take the surface normal to $x$. Electrons escape if $\tfrac{1}{2}m v_x^2 \ge E_F + \Phi$. For these high-energy electrons $E - E_F \gg kT$, so $f(E) \approx \exp[-(E - E_F)/kT]$. The number of electrons per unit volume with velocity in $dv_x\,dv_y\,dv_z$ (including spin) is

$$dn = 2\left(\frac{m}{h}\right)^3 \exp\left[-\frac{\tfrac12 m(v_x^2 + v_y^2 + v_z^2) - E_F}{kT}\right] dv_x\,dv_y\,dv_z$$

The current density is $J = e\int v_x\, dn$ over all $v_y, v_z$ and over $v_x \ge v_{x0}$, where $\tfrac12 m v_{x0}^2 = E_F + \Phi$. Using $\int_{-\infty}^{\infty} e^{-mv^2/2kT}dv = \sqrt{2\pi kT/m}$ for $v_y$ and $v_z$:

$$J = 2e\left(\frac{m}{h}\right)^3 e^{E_F/kT} \cdot \frac{2\pi kT}{m} \int_{v_{x0}}^{\infty} v_x e^{-mv_x^2/2kT}dv_x$$

The last integral equals $\frac{kT}{m}e^{-(E_F+\Phi)/kT}$. Therefore

$$J = \frac{4\pi m e k^2}{h^3} T^2 \exp\left(-\frac{\Phi}{kT}\right) = B_0 T^2 e^{-\Phi/kT}$$

with $B_0 = \dfrac{4\pi m e k^2}{h^3} = 1.20 \times 10^{6}\ \text{A m}^{-2}\text{K}^{-2}$. This is the Richardson-Dushman equation. Real metals have smaller $B_0$ (surface reflection).

Schottky effect (field-assisted emission)

With an accelerating field $\mathcal{E}$ at the cathode, the electron's potential energy outside the surface is (image force + applied field):

$$PE(x) = \Phi - \frac{e^2}{16\pi\varepsilon_0 x} - e\mathcal{E}x$$

(measured from $E_F$). Setting $d(PE)/dx = 0$ gives the maximum at $x_m = \sqrt{e/(16\pi\varepsilon_0\mathcal{E})}$, and the barrier is lowered by

$$\Delta\Phi = \sqrt{\frac{e^3\mathcal{E}}{4\pi\varepsilon_0}} = \beta_S\sqrt{\mathcal{E}}, \qquad \beta_S = \sqrt{\frac{e^3}{4\pi\varepsilon_0}}$$

So the effective work function is $\Phi_{eff} = \Phi - \beta_S\sqrt{\mathcal{E}}$ and

$$J = B_0 T^2 \exp\left[-\frac{\Phi - \beta_S\sqrt{\mathcal{E}}}{kT}\right]$$

 PE
  ^       ___ max (lowered)
  |  Phi /   \___
  |     /        \___  with field
  |    /             \___
  | EF|____________________> x
     surface   xm

Thus the emission current rises with field even at fixed temperature. Example: $\mathcal{E} = 10^8$ V/m lowers $\Phi$ by about 0.38 eV.

Thermionic emission in metals

In a metal, conduction electrons fill states up to $E_F$ at 0 K. To leave the metal an electron needs extra energy equal to the work function $\Phi$. When the metal is heated, the Fermi-Dirac tail spreads above $E_F$; the few electrons with kinetic energy normal to the surface greater than $E_F + \Phi$ escape. This is thermionic emission. The emitted current density is given by the Richardson-Dushman equation

$$J = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad B_0 = \frac{4\pi m e k^2}{h^3} \approx 1.2 \times 10^{6}\ \text{A m}^{-2}\text{K}^{-2}$$

$J$ rises very fast with $T$ and is larger for a small $\Phi$ (hence oxide- or thorium-coated cathodes).

Image charge method

When an electron is at distance $x$ outside a flat conducting surface, the induced charges on the metal act like a positive image charge $+e$ at distance $x$ inside the surface, i.e. $2x$ away.

   metal  |  vacuum
          |
   (+e) - | - (-e)
    <--x-->|<--x-->
   image  |  electron

Attractive image force:

$$F_{im} = -\frac{e^2}{4\pi\varepsilon_0 (2x)^2} = -\frac{e^2}{16\pi\varepsilon_0 x^2}$$

The work done against it from $x$ to infinity gives the potential energy (zero at infinity):

$$PE_{im}(x) = -\frac{e^2}{16\pi\varepsilon_0 x}$$

Measured from $E_F$, the electron's PE without field is $\Phi - \dfrac{e^2}{16\pi\varepsilon_0 x}$, which rises smoothly to $\Phi$ far from the surface.

Effect of an applied field

An applied field $\mathcal{E}$ pulling electrons away adds $-e\mathcal{E}x$:

$$PE(x) = \Phi - \frac{e^2}{16\pi\varepsilon_0 x} - e\mathcal{E}x$$

Maximum of the barrier:

$$\frac{d(PE)}{dx} = \frac{e^2}{16\pi\varepsilon_0 x^2} - e\mathcal{E} = 0 \quad\Rightarrow\quad x_m = \sqrt{\frac{e}{16\pi\varepsilon_0\mathcal{E}}}$$

Substituting $x_m$:

$$\begin{aligned} PE_{max} &= \Phi - \frac{e^2}{16\pi\varepsilon_0}\sqrt{\frac{16\pi\varepsilon_0\mathcal{E}}{e}} - e\mathcal{E}\sqrt{\frac{e}{16\pi\varepsilon_0\mathcal{E}}} \\ &= \Phi - 2\sqrt{\frac{e^3\mathcal{E}}{16\pi\varepsilon_0}} = \Phi - \sqrt{\frac{e^3\mathcal{E}}{4\pi\varepsilon_0}} \end{aligned}$$

So the effective work function is $\Phi_{eff} = \Phi - \beta_S\sqrt{\mathcal{E}}$, where $\beta_S = \sqrt{e^3/(4\pi\varepsilon_0)}$.

Emission current density (Schottky effect)

Replacing $\Phi$ by $\Phi_{eff}$ in the Richardson-Dushman equation:

$$J = B_0 T^2 \exp\left[-\frac{\Phi - \beta_S\sqrt{\mathcal{E}}}{kT}\right] = J_0 \exp\left(\frac{\beta_S\sqrt{\mathcal{E}}}{kT}\right)$$

where $J_0$ is the zero-field current. Thus $\ln J$ increases linearly with $\sqrt{\mathcal{E}}$. Example: at $\mathcal{E} = 10^8$ V/m, $\Delta\Phi \approx 0.38$ eV, $x_m \approx 1.9$ nm.

Electrons can be removed from silver once they reach the vacuum level, which lies at $E = E_{F0} + \Phi$ above the bottom of the band. The total number of states per unit volume from the band bottom up to energy $E$ is found by integrating the free-electron density of states:

$$S(E) = \int_0^{E} g(E')\,dE' = \frac{16\pi\sqrt{2}}{3}\,\frac{m_e^{3/2}}{h^3}\, E^{3/2}$$

Given: $E_{F0} = 5.5\ \text{eV}$, $\Phi = 4.5\ \text{eV}$, $\rho = 10.5\ \text{g/cm}^3 = 10\,500\ \text{kg/m}^3$, $M = 107.9\ \text{g/mol}$.

Total states up to the vacuum level

$E = E_{F0} + \Phi = 10\ \text{eV} = 1.602 \times 10^{-18}\ \text{J}$

$$\begin{aligned} S &= \frac{16\pi\sqrt{2}}{3} \times \frac{(9.109 \times 10^{-31})^{3/2}}{(6.626 \times 10^{-34})^3} \times (1.602 \times 10^{-18})^{3/2} \\ &= 1.44 \times 10^{29}\ \text{m}^{-3} \end{aligned}$$

Atomic concentration of silver

$$n_{at} = \frac{\rho N_A}{M} = \frac{10\,500 \times 6.022 \times 10^{23}}{0.1079} = 5.86 \times 10^{28}\ \text{m}^{-3}$$

Comparison

$$\frac{S}{n_{at}} = \frac{1.44 \times 10^{29}}{5.86 \times 10^{28}} = 2.45$$

There are about 2.5 states per atom (states counted with spin) up to the vacuum level. Silver is monovalent, so it has $5.86 \times 10^{28}$ free electrons per m³. At 0 K these fill states only up to $E_{F0}$: $S(5.5\ \text{eV}) = 5.86 \times 10^{28}\ \text{m}^{-3}$, exactly equal to the electron concentration, which confirms $E_{F0} = 5.5$ eV. The remaining states between $E_{F0}$ and the vacuum level (about $8.5 \times 10^{28}\ \text{m}^{-3}$) are empty at 0 K and become partly filled only by thermal excitation, which is why emission needs high temperature.

Answer: total states $\approx 1.44 \times 10^{29}\ \text{m}^{-3}$, atomic concentration $= 5.86 \times 10^{28}\ \text{m}^{-3}$; the states are about 2.45 times the number of atoms.

Schottky effect is the lowering of the work function (potential barrier) of a metal surface when a strong accelerating electric field is applied to it. Because the barrier is lowered, the thermionic emission current increases with field even at constant temperature.

Barrier without field (image force)

An electron at distance $x$ outside a metal surface induces an image charge $+e$ at distance $x$ inside. The attractive image force and its potential energy are

$$F = -\frac{e^2}{4\pi\varepsilon_0(2x)^2} = -\frac{e^2}{16\pi\varepsilon_0 x^2}, \qquad PE_{im} = -\frac{e^2}{16\pi\varepsilon_0 x}$$

Taking energies from the Fermi level, the electron's PE outside is

$$PE(x) = \Phi - \frac{e^2}{16\pi\varepsilon_0 x}$$

which approaches $\Phi$ at large $x$. So the full barrier to be crossed is $\Phi$.

Barrier with applied field

An applied field $\mathcal{E}$ that pulls electrons out adds PE $= -e\mathcal{E}x$:

$$PE(x) = \Phi - \frac{e^2}{16\pi\varepsilon_0 x} - e\mathcal{E}x$$

 PE
  ^   no field: -> Phi
  | Phi .......___________
  |  dPhi{  .-''-.
  |       /        '-._  with E
  |      /             '-._
  | EF  /___________________> x
     surface   xm

Finding the lowered barrier

The peak is where $d(PE)/dx = 0$:

$$\frac{e^2}{16\pi\varepsilon_0 x_m^2} = e\mathcal{E} \quad\Rightarrow\quad x_m = \left(\frac{e}{16\pi\varepsilon_0\mathcal{E}}\right)^{1/2}$$

Putting $x_m$ back:

$$\begin{aligned} PE_{max} &= \Phi - \frac{e^2}{16\pi\varepsilon_0 x_m} - e\mathcal{E}x_m \\ &= \Phi - \left(\frac{e^3\mathcal{E}}{16\pi\varepsilon_0}\right)^{1/2} - \left(\frac{e^3\mathcal{E}}{16\pi\varepsilon_0}\right)^{1/2} \\ &= \Phi - \left(\frac{e^3\mathcal{E}}{4\pi\varepsilon_0}\right)^{1/2} \end{aligned}$$

So the barrier is reduced by

$$\Delta\Phi = \beta_S\sqrt{\mathcal{E}}, \qquad \beta_S = \left(\frac{e^3}{4\pi\varepsilon_0}\right)^{1/2} = 6.08 \times 10^{-24}\ \text{J}\,(\text{V/m})^{-1/2}$$

and the effective work function is $\Phi_{eff} = \Phi - \beta_S\sqrt{\mathcal{E}}$.

Modified Richardson equation

The Richardson-Dushman equation $J = B_0T^2\exp(-\Phi/kT)$ now uses $\Phi_{eff}$:

$$J = B_0 T^2 \exp\left[-\frac{\Phi - \beta_S\sqrt{\mathcal{E}}}{kT}\right] = J_0 \exp\left(\frac{\beta_S\sqrt{\mathcal{E}}}{kT}\right)$$

where $J_0 = B_0T^2e^{-\Phi/kT}$ is the zero-field emission and $B_0 = 4\pi mek^2/h^3 \approx 1.2 \times 10^{6}\ \text{A m}^{-2}\text{K}^{-2}$.

Remarks: $\ln J$ is linear in $\sqrt{\mathcal{E}}$ (Schottky plot). For $\mathcal{E} = 10^8\ \text{V/m}$, $\Delta\Phi \approx 0.38\ \text{eV}$ and $x_m \approx 1.9\ \text{nm}$. At much higher fields ($>10^9$ V/m) the barrier becomes thin enough for tunnelling, i.e. field emission.

Drift velocity

Without a field, conduction electrons in a metal move randomly at high speed (about $10^6$ m/s) and their average velocity is zero, so there is no net current. When a field $\mathcal{E}$ is applied, each electron is accelerated opposite to the field between collisions with vibrating ions and defects. The small average velocity gained in the direction of the force is the drift velocity $v_d$.

  E field  -------->
  e- path:  \/\/\/\_/\/\/\_/\  (random + slow drift)
            <--- net drift v_d

Between collisions the acceleration is $a = e\mathcal{E}/m_e$. If $\tau$ is the mean free time (relaxation time) between collisions, and each collision randomises the velocity, the average gained velocity is

$$v_d = \frac{e\tau}{m_e}\mathcal{E}$$

Typical $v_d$ is only mm/s to cm/s, far less than the thermal speed.

Drift mobility

The drift mobility $\mu_d$ is the drift velocity per unit electric field:

$$\mu_d = \frac{v_d}{\mathcal{E}} = \frac{e\tau}{m_e} \qquad (\text{m}^2\text{V}^{-1}\text{s}^{-1})$$

It measures how easily electrons move through the lattice. It is large when $\tau$ is long (few collisions). For copper, $\mu_d \approx 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$.

Conductivity

With $n$ free electrons per unit volume moving at $v_d$, the current density is

$$J = e n v_d = e n \mu_d \mathcal{E}$$

Comparing with Ohm's law $J = \sigma\mathcal{E}$:

$$\sigma = e n \mu_d = \frac{n e^2 \tau}{m_e}, \qquad \rho = \frac{1}{\sigma}$$

Example (copper): $n = 8.5 \times 10^{28}\ \text{m}^{-3}$, $\mu_d = 43 \times 10^{-4}\ \text{m}^2/\text{V s}$ gives $\sigma = 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 43 \times 10^{-4} \approx 5.9 \times 10^{7}\ \text{S/m}$.

Factors

In metals $n$ does not change with temperature, so the fall in $\sigma$ (rise in resistivity, $\rho \propto T$) comes entirely from the fall in mobility due to increased lattice scattering.

The drift mobility and the mean free (relaxation) time $\tau$ are related by

$$\mu_d = \frac{e\tau}{m_e} \quad\Rightarrow\quad \tau = \frac{\mu_d m_e}{e}, \qquad \lambda = u\tau$$

where $u$ is the mean speed and $\lambda$ the mean free path.

Given: $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 4.3 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $u = 2 \times 10^{7}\ \text{m/s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$, $e = 1.602 \times 10^{-19}\ \text{C}$.

Relaxation time

$$\tau = \frac{4.3 \times 10^{-3} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 2.44 \times 10^{-14}\ \text{s}$$

Mean free path

$$\lambda = u\tau = 2 \times 10^{7} \times 2.44 \times 10^{-14} = 4.89 \times 10^{-7}\ \text{m} \approx 489\ \text{nm}$$

(Using the speed as given. The usual Fermi speed in copper is about $1.6 \times 10^{6}$ m/s, which would give $\lambda \approx 40$ nm; the mean free path scales directly with the speed used.)

Answer: $\tau = 2.44 \times 10^{-14}\ \text{s}$, $\lambda = 4.89 \times 10^{-7}\ \text{m}$.

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

$$\mu_d = \frac{\sigma}{e n}$$

Potassium is monovalent (one valence electron per atom), so $n$ equals the atomic concentration.

Given: $\sigma = 1.39 \times 10^{5}\ \text{S/cm} = 1.39 \times 10^{7}\ \text{S/m}$ (the unit "Sm/cm" is read as S/cm, the standard value for potassium), $M = 39.5\ \text{g/mol} = 0.0395\ \text{kg/mol}$, $\rho = 0.91\ \text{g/cm}^3 = 910\ \text{kg/m}^3$.

Electron concentration

$$n = \frac{\rho N_A}{M} = \frac{910 \times 6.022 \times 10^{23}}{0.0395} = 1.387 \times 10^{28}\ \text{m}^{-3}$$

Drift mobility

$$\mu_d = \frac{1.39 \times 10^{7}}{1.602 \times 10^{-19} \times 1.387 \times 10^{28}} = 6.25 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$$ $$\mu_d = 62.5\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$$

(The corresponding mean free time is $\tau = \mu_d m_e/e = 3.6 \times 10^{-14}$ s.)

Answer: $\mu_d \approx 6.25 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 62.5\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$.

The saturation (thermionic) current is given by the Richardson-Dushman equation

$$J = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad I = J \times A$$

where $A = \pi d l$ is the curved emitting surface of the cylindrical cathode.

Given: $l = 5\ \text{cm} = 0.05\ \text{m}$, $d = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$, $T = 1400 + 273.15 = 1673.15\ \text{K}$, $B_0 = 3\ \text{A cm}^{-2}\text{K}^{-2} = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\Phi = 2.6\ \text{eV}$.

Emitting area

$$A = \pi d l = \pi \times 3 \times 10^{-3} \times 0.05 = 4.712 \times 10^{-4}\ \text{m}^2$$

Exponential term

$$\frac{\Phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 1673.15} = 18.03, \qquad e^{-18.03} = 1.483 \times 10^{-8}$$

Current density

$$J = 3 \times 10^{4} \times (1673.15)^2 \times 1.483 \times 10^{-8} = 1.246 \times 10^{3}\ \text{A/m}^2$$

Saturation current

$$I = J A = 1.246 \times 10^{3} \times 4.712 \times 10^{-4} = 0.587\ \text{A}$$

Answer: Saturation current $I \approx 0.59\ \text{A}$ ($J \approx 1.25 \times 10^{3}\ \text{A/m}^2$).

Given: $\rho = 8.96\ \text{g/cm}^3 = 8960\ \text{kg/m}^3$, $M = 63.5\ \text{g/mol}$; copper gives one free electron per atom.

Fermi energy at 0 K

Electron concentration:

$$n = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.0635} = 8.50 \times 10^{28}\ \text{m}^{-3}$$

Free-electron Fermi energy:

$$\begin{aligned} E_{F0} &= \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} \\ &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \times \left(\frac{3 \times 8.50 \times 10^{28}}{\pi}\right)^{2/3} \\ &= 6.025 \times 10^{-38} \times 1.874 \times 10^{19} = 1.129 \times 10^{-18}\ \text{J} = 7.05\ \text{eV} \end{aligned}$$

Fermi energy at 15°C

The temperature dependence of the Fermi energy in a metal is

$$E_F(T) = E_{F0}\left[1 - \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2\right]$$

At $T = 15 + 273.15 = 288.15\ \text{K}$: $kT = \frac{1.381 \times 10^{-23} \times 288.15}{1.602 \times 10^{-19}} = 0.02484\ \text{eV}$.

$$\begin{aligned} \frac{kT}{E_{F0}} &= \frac{0.02484}{7.0487} = 3.524 \times 10^{-3} \\ \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2 &= 0.8225 \times 1.242 \times 10^{-5} = 1.02 \times 10^{-5} \\ E_F(288\ \text{K}) &= 7.0487 \times (1 - 1.02 \times 10^{-5}) = 7.0486\ \text{eV} \end{aligned}$$

The decrease is only about $7.2 \times 10^{-5}$ eV (0.001%), so the Fermi energy of a metal is practically independent of temperature.

Answer: $E_{F0} = 7.05\ \text{eV}$; at 15°C, $E_F \approx 7.05\ \text{eV}$ (lower by only $\approx 7.2 \times 10^{-5}$ eV).

For free electrons in a metal at 0 K, all states up to $E_{F0}$ are filled, giving

$$E_{F0} = \frac{h^2}{8 m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

Given: $n = 8.43 \times 10^{28}\ \text{m}^{-3}$, $h = 6.626 \times 10^{-34}\ \text{J s}$, $m_e = 9.109 \times 10^{-31}\ \text{kg}$.

Calculation

$$\begin{aligned} \left(\frac{3n}{\pi}\right)^{2/3} &= \left(\frac{3 \times 8.43 \times 10^{28}}{\pi}\right)^{2/3} = (8.050 \times 10^{28})^{2/3} = 1.864 \times 10^{19}\ \text{m}^{-2} \\ \frac{h^2}{8m_e} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} = 6.025 \times 10^{-38}\ \text{J m}^2 \\ E_{F0} &= 6.025 \times 10^{-38} \times 1.864 \times 10^{19} = 1.123 \times 10^{-18}\ \text{J} \end{aligned}$$

In electron-volts:

$$E_{F0} = \frac{1.123 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.01\ \text{eV}$$

This agrees with the measured value for copper (about 7 eV). The Fermi temperature $T_F = E_{F0}/k \approx 8.1 \times 10^{4}$ K, far above any working temperature, so $E_F$ stays almost the same at room temperature.

Answer: $E_{F0} \approx 1.12 \times 10^{-18}\ \text{J} = 7.01\ \text{eV}$.

The conductivity gives the free-electron concentration, and the concentration gives the Fermi energy:

$$\sigma = e n \mu_d \;\Rightarrow\; n = \frac{\sigma}{e\mu_d}, \qquad E_{F0} = \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

Given: $\sigma = 63.5 \times 10^{6}\ \text{S/m}$, $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 43 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$.

Electron concentration

$$n = \frac{63.5 \times 10^{6}}{1.602 \times 10^{-19} \times 43 \times 10^{-4}} = 9.22 \times 10^{28}\ \text{m}^{-3}$$

Fermi level

$$\begin{aligned} E_{F0} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \left(\frac{3 \times 9.22 \times 10^{28}}{\pi}\right)^{2/3} \\ &= 6.025 \times 10^{-38} \times 1.979 \times 10^{19} \\ &= 1.19 \times 10^{-18}\ \text{J} = \frac{1.19 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.44\ \text{eV} \end{aligned}$$

The Fermi level lies 7.44 eV above the bottom of the conduction band (close to the accepted 7.0 eV for copper; the difference comes from the slightly high conductivity value given).

Answer: $n = 9.22 \times 10^{28}\ \text{m}^{-3}$, $E_F \approx 1.19 \times 10^{-18}\ \text{J} = 7.44\ \text{eV}$.

In the free electron (Drude) model, an electron moving at mean speed $u$ is scattered when it comes within the cross-sectional area $S$ of a scattering centre. With $N_s$ scatterers per unit volume:

$$\lambda = \frac{1}{S N_s}, \qquad \tau = \frac{\lambda}{u} = \frac{1}{S N_s u}, \qquad \mu_d = \frac{e\tau}{m_e}, \qquad \sigma = e n \mu_d$$

Each copper atom (ion) is taken as a scatterer, and each gives one conduction electron, so $N_s = n$.

Given: $u = 1.5 \times 10^{6}\ \text{m/s}$, $S = 3.9 \times 10^{-22}\ \text{m}^2$, $\rho = 8960\ \text{kg/m}^3$, $M = 63.56\ \text{g/mol}$.

Atomic (and electron) concentration

$$n = N_s = \frac{\rho N_A}{M} = \frac{8960 \times 6.022 \times 10^{23}}{0.06356} = 8.49 \times 10^{28}\ \text{m}^{-3}$$

Mean free path and mean free time

$$\begin{aligned} \lambda &= \frac{1}{S N_s} = \frac{1}{3.9 \times 10^{-22} \times 8.49 \times 10^{28}} = 3.02 \times 10^{-8}\ \text{m} \\ \tau &= \frac{\lambda}{u} = \frac{3.02 \times 10^{-8}}{1.5 \times 10^{6}} = 2.01 \times 10^{-14}\ \text{s} \end{aligned}$$

Drift mobility

$$\mu_d = \frac{e\tau}{m_e} = \frac{1.602 \times 10^{-19} \times 2.01 \times 10^{-14}}{9.109 \times 10^{-31}} = 3.54 \times 10^{-3}\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 35.4\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$$

Conductivity

$$\sigma = e n \mu_d = 1.602 \times 10^{-19} \times 8.49 \times 10^{28} \times 3.54 \times 10^{-3} = 4.82 \times 10^{7}\ \text{S/m}$$

This is close to the measured value for copper ($5.9 \times 10^{7}$ S/m), showing that the simple model gives the right order of magnitude.

Answer: $\mu_d \approx 35.4\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $\sigma \approx 4.82 \times 10^{7}\ \text{S/m}$ (with $\lambda = 30.2$ nm, $\tau = 2.01 \times 10^{-14}$ s).

The saturation current is given by the Richardson-Dushman equation:

$$J = A_0 T^2 \exp\left(-\frac{\phi}{kT}\right), \qquad I = J \times (\pi d l)$$

Given (as stated): $l = 4\ \text{m}$, $d = 2\ \text{mm}$, $T = 160 + 273.15 = 433.15\ \text{K}$, $A_0 = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\phi = 2.6\ \text{eV}$.

Emitting area

$$A = \pi d l = \pi \times 2 \times 10^{-3} \times 4 = 2.51 \times 10^{-2}\ \text{m}^2$$

Exponential term

$$\frac{\phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 433.15} = 69.63, \qquad e^{-69.63} = 5.75 \times 10^{-31}$$

Current density and current

$$\begin{aligned} J &= 3 \times 10^{4} \times (433.15)^2 \times 5.75 \times 10^{-31} = 3.24 \times 10^{-21}\ \text{A/m}^2 \\ I &= J A = 3.24 \times 10^{-21} \times 2.51 \times 10^{-2} = 8.1 \times 10^{-23}\ \text{A} \end{aligned}$$

So at 160°C there is practically no thermionic emission: $kT$ (0.037 eV) is far too small compared with the 2.6 eV work function.

Note: a transmitter tube cathode normally runs near 1600°C and is a few cm long. With $T = 1873.15$ K and $l = 4$ cm: $\phi/kT = 16.10$, $J = 1.07 \times 10^{4}\ \text{A/m}^2$, $A = 2.51 \times 10^{-4}\ \text{m}^2$, giving $I \approx 2.69\ \text{A}$.

Answer (data as given): $I \approx 8.1 \times 10^{-23}\ \text{A}$, i.e. negligible. (If 1600°C and 4 cm are meant, $I \approx 2.69$ A.)

The drift mobility is related to the mean free time $\tau$ by $\mu_d = e\tau/m_e$, and the mean free path is $\lambda = u\tau$, where $u$ is the mean speed.

Given: $\mu_d = 53\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 53 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $u = 2.2 \times 10^{6}\ \text{m/s}$.

Mean free time

$$\tau = \frac{\mu_d m_e}{e} = \frac{53 \times 10^{-4} \times 9.109 \times 10^{-31}}{1.602 \times 10^{-19}} = 3.01 \times 10^{-14}\ \text{s}$$

Mean free path

$$\lambda = u\tau = 2.2 \times 10^{6} \times 3.01 \times 10^{-14} = 6.63 \times 10^{-8}\ \text{m} \approx 66.3\ \text{nm}$$

This is a few hundred atomic spacings, so electrons pass many ions before being scattered.

Answer: $\lambda \approx 6.63 \times 10^{-8}\ \text{m} = 66.3\ \text{nm}$ ($\tau = 3.01 \times 10^{-14}$ s).

Population density

The population density $n_E(E)$ is the number of electrons per unit volume per unit energy at energy $E$. It is the product of the number of available states and the probability that they are occupied:

$$n_E(E) = g(E)\, f(E), \qquad g(E) = \frac{8\pi\sqrt{2}\, m_e^{3/2}}{h^3} E^{1/2}, \quad f(E) = \frac{1}{1 + e^{(E-E_F)/kT}}$$

The area under the $n_E$ versus $E$ curve gives the total electron concentration $n$.

 nE(E)
  ^        T=0: sharp cut at EF
  |      ..---.
  |   .-'      |\  <- T>0: small tail
  |  /         | \
  | /          |  '..
  |/___________|______> E
  0            EF

Fermi energy depends only on n

At $T = 0$ K, $f(E) = 1$ for $E < E_{F0}$ and $0$ above it. Hence

$$n = \int_0^{E_{F0}} g(E)\, dE = \frac{8\pi\sqrt{2}\, m_e^{3/2}}{h^3} \cdot \frac{2}{3} E_{F0}^{3/2} = \frac{8\pi}{3}\left(\frac{2m_e E_{F0}}{h^2}\right)^{3/2}$$

Solving,

$$E_{F0} = \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

Thus $E_{F0}$ contains only constants and $n$, the free electron concentration. A metal with more free electrons per volume has a higher Fermi energy (Cu, $n = 8.5 \times 10^{28}$ m⁻³: 7.0 eV; Na, $2.5 \times 10^{28}$: 3.1 eV).

Temperature has almost no effect

At $T > 0$, the electron number must stay the same:

$$n = \int_0^{\infty} g(E) f(E)\, dE$$

Evaluating this integral (Sommerfeld expansion) gives

$$E_F(T) = E_{F0}\left[1 - \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2\right]$$

For copper at 300 K: $kT = 0.0259$ eV, $E_{F0} = 7.0$ eV, so $kT/E_{F0} = 3.7 \times 10^{-3}$ and the correction is $\frac{\pi^2}{12}(3.7 \times 10^{-3})^2 \approx 1.1 \times 10^{-5}$, a change of only about 0.001%.

Physical reason

Only electrons within a few $kT$ of $E_F$ can be thermally excited, because states deeper down have all their neighbouring states already filled (Pauli principle). Since $kT \ll E_F$, only a tiny fraction of electrons are disturbed, and the electrons moved above $E_F$ match the holes left below $E_F$ (the Fermi function is symmetric about $E_F$). Also $n$ in a metal does not change with $T$. Hence the Fermi energy of a metal is practically independent of temperature and depends only on its electron concentration.