= Richardson-Dushman law
{c}
{title2=$J=A T^2\exp[-W/(k_BT)]$}
= Richardson law
{c}
{synonym}
The <Richardson-Dushman law> models the magnitude of emitted <electric current density> as
$$
J=A T^2\exp\!\left(-\frac{W}{k_BT}\right).
$$
Here $T$ is absolute <thermodynamic temperature>, $W$ is the <work function>, $k_B$ is the <Boltzmann constant>, and $A$ is the <Thermionic Richardson constant>. The signed <electric current density> points opposite to the outgoing <Electron> motion; $J$ here is positive. For uniform emission over area $S$, the emitted <electric current> magnitude is $I=SJ$.
The <Richardson-Dushman law> assumes an emitter close to <thermal equilibrium>, a surface with a well-defined and approximately constant <work function> over the fitted <temperature> range, and predominantly thermal escape over the barrier. Interpreting a measured collector <electric current> with this law additionally requires an <emission-limited thermionic current>, known emitting area, and negligible return emission from the collector. Strong <electric fields> can lower the surface barrier or enable <quantum tunnelling>; then this zero-field expression is insufficient.
For a deliberately chosen teaching model, take $W=2.0\,\mathrm{eV}$ and $A=120\,\mathrm{A\,cm^{-2}\,K^{-2}}$, expressing the barrier in <electronvolts>. These are illustrative parameters, not measured properties of a named material. Using $k_B=8.617333262\times10^{-5}\,\mathrm{eV\,K^{-1}}$, at $1000\,\mathrm K$ the <Boltzmann factor> is about $8.33\times10^{-11}$, yet
$$
J(1000\,\mathrm K)\simeq0.00999\,\mathrm{A\,cm^{-2}},
\qquad
J(1200\,\mathrm K)\simeq0.6885\,\mathrm{A\,cm^{-2}}.
$$
A 20 percent increase in <temperature> raises the predicted <electric current density> by about 69 times. For a chosen emitting area $S=0.010\,\mathrm{cm^2}$, these correspond to approximately $0.100\,\mathrm{mA}$ and $6.89\,\mathrm{mA}$ if collection is complete. The small escaping fraction is replenished by the external <electrical circuit>.
The sensitivity is especially clear from the <logarithmic derivative> at fixed $A,W$:
$$
\frac{d\ln J}{dT}=\frac{2}{T}+\frac{W}{k_BT^2}.
$$
In the chosen model at $1000\,\mathrm K$, a $1\,\mathrm K$ rise changes $J$ by about 2.5 percent to first order. Accurate <temperature> measurement matters even when <electric current> measurement is easy.
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