Richardson's Law Example Question

A tungsten filament is heated to 2800K. Calculate:

1. The thermionic emission current density (J) using Richardson’s Law
2. The spread in energy ($\Delta E$) of the emitted electrons due to thermal broadening ($\varphi$) = $4.5eV$ ($A$) = $1.2\ast 10^{6}A{m}^{-2}{K}^{-2}$ ($k$) = $8.617\ast 10^{-5}eV{K}^{-1}$

\ Solution:

1. Calculation of Current Density ( $J$ ) Richardson’s Law is given by: $J=A{T}^2\exp \left(-\frac{\varphi }{kT}\right)$ Plugging in the given values: $\begin{array}{rl}& J=\left(1.2\times 10^6\mathrm{A}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}\right)\times (2800\mathrm{K}{)}^2\times \exp \left(-\frac{4.5\mathrm{eV}}{\left(8.617\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}\right)\times (2800\mathrm{K})}\right)\\ & J=\left(1.2\times 10^6{\mathrm{Am}}^{-2}{\mathrm{K}}^{-2}\right)\times \left(7.84\times 10^6{\mathrm{K}}^2\right)\times \exp \left(-\frac{4.5}{0.241276}\right)\\ & J=\left(1.2\times 10^6{\mathrm{Am}}^{-2}{\mathrm{K}}^{-2}\right)\times \left(7.84\times 10^6{\mathrm{K}}^2\right)\times \exp (-18.65)\\ & J\approx 7.47\times 10^4{\mathrm{Am}}^{-2}\end{array}$
2. Calculation of Spread in Energy ( $\Delta E$ ) The spread in energy due to thermal broadening is given by: $\Delta E\approx 3kT$ Plugging in the given values: $\begin{array}{rl}& \Delta E=3\times \left(8.617\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}\right)\times (2800\mathrm{K})\\ & \Delta E\approx 0.72\mathrm{eV}\end{array}$

Results

1. The thermionic emission current density ( $J$ ) is approximately $7.47\times 10^4{\mathrm{Am}}^{-2}$.
2. The spread in energy ( $\Delta E$ ) of the emitted electrons is approximately 0.72 eV .