The saturation current from a thoriated-tungsten cathode at 2000K… - Vidyadip

Question

The saturation current from a thoriated$-$tungsten cathode at $2000K$ is $100mA.$ What will be the saturation current for a pure$-$tungsten cathode of the same surface area operating at the same temperature? The constant $A$ in the Richardson$-$Dushman equation is $60 \times 10^4Am^{-2}K^{-2}$ for pure tungsten and $3.0 \times 10^4Am^{-2}k^{-2}$ for thoriated tungsten. The work function of pure tungsten is $4.5eV$ and that of thoriated tungsten is $2.6eV.$

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Answer

$\text{i}=\text{AST}^2\text{ e}^{\frac{-\phi}{\text{KT}}}$
$\text{i}_1=\text{i},\ \text{i}_2=100\text{mA}$
$\text{A}_1=60\times10^4,\ \text{A}_2=3\times10^4$
$\text{S}_1=\text{S},\ \text{S}_2=\text{S}$
$\text{T}_1=2000,\ \text{T}_2=2000$
$\phi_1=4.5\text{eV},\ \phi_2 =2.6\text{eV}$
$\text{i}=(60\times10^{4})(\text{S})\times(2000)^2\frac{-45\times1.6\times10^{-19}}{\text{e}^{1.38\times10^{-23}\times2000}}$
$100=(3\times10^{4})(\text{S})\times(2000)^2\frac{-2.6\times1.6\times10^{-19}}{\text{e}^{1.38\times10^{-23}\times2000}}$
Dividing the equation,
$\Rightarrow\frac{\text{i}}{100}=\text{e}^{\big[\frac{-4.5\times1.6\times10}{1.38\times2}\big(\frac{-2.6\times1.6\times10}{1.38\times20}\big)\big]}$
$\Rightarrow\frac{\text{i}}{100}=20\times\text{e}^{-11.014}$
$\Rightarrow\frac{\text{i}}{100}=20\times0.000016$
$\Rightarrow\text{i}=20\times0.0016=0.0329\text{mA}=33\mu\text{A}$

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