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

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Solution

Pure tungsten
$ϕ = 2.6 e V A = 60 \times 10^{4} A / m^{2} - k^{2} i = A S T^{2} e^{- ϕ \sqrt{K T}}$
Thoriated tungsten
$ϕ = 2.6 e V A = 3 \times 10^{4} A / m^{2} - k^{2} i_{Thoriated tungsten} = 5000 i_{Tungesten} S o, 5000 \times S \times 60 \times 10^{4} \times T^{2} + e^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38 \times T \times 10^{- 23}}} = S^{3} \times 10^{4} \times T^{2} \times e^{\frac{- 2.6 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 28} \times T}} \Rightarrow 1 + 10^{4} \times e^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38} \times 10^{- 23} \times T} = e^{\frac{- 2.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times T}}$
Taking 'In' both sides
$\Rightarrow 9.21 T = 22029$
$T = \frac{22029}{9.21} = 2391.856$

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