Question
A tungsten cathode and a thoriated-tungsten cathode have the same geometric dimensions and are operated at the same temperature. The thoriated-tungsten cathode gives 5000 times more current than the other cathode. Find the operating temperature. Take relevant data from the previous problem.

Answer

Pure tungsten:
$\phi=4.5\text{eV}$
$\text{A}=60\times10^{4}\text{A/m}^2-\text{k}^2$
$\text{i}=\text{AST}^2\text{e}^{\frac{\phi}{\text{KT}}}$
Thoriated tungsten:
$\phi=2.6\text{eV}$
$\text{A}=3\times10^{4}\text{A/m}^2-\text{k}^2$
$\text{i}_{\text{Thoriated Tungsten}}=5000\text{i}_{\text{Tungsten}}$
So, $5000\times\text{S}\times60\times10^{4}\times\text{T}_2\times\text{e}^{\frac{-4.5\times1.6\times10^{-19}}{1.38\times\text{t}\times10^{-23}}}$
$\Rightarrow\text{S}\times3\times10^{4}\times\text{T}^2\times\text{e}^{\frac{-2.56\times1.6\times10^{-19}}{1.38\times\text{T}\times10^{-23}}}$
$\Rightarrow3\times10^{8}\times\text{e}^{\frac{-4.5\times1.6\times10^{-19}}{1.38\times\text{t}\times10^{-23}}}=\text{e}^{\frac{-2.56\times1.6\times10^{-19}}{1.38\times\text{T}\times10^{-23}}}\times3\times10^4$
Taking 'in'
$\Rightarrow9.21\text{T}=220.29$
$\Rightarrow\text{T}=\frac{22029}{9.21}=2391.856\text{K}$

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