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`        A tungsten cathode and a thoriated-tungsten cathode have the save 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.`
3 years ago

Kevin Nash
332 Points
```										Sol. Pure tungsten Thoriated tungsten
∅ = 4.5 eV ∅ = 2.6 eV
A = 60 * 10^4 A/m^2 – k^2 A = 3 * 10^4 A/m^2 – k^2
i = AST^2 e-∅ /KT
I base Thoriated Tungsten = 5000 I base Tungsten
So, 5000 * S * 60 * 10^4 * T^2 *
⇒ S * 3 * 10^4 * T^2 * e^-2.65*1.6*10^-19/1.38*T*10^23
⇒ 3 * 10^8 * e^-4.5*1.6*10^-19/1.38*T*10^23 e^-2.65*1.6*10^-19/1.38*T*10^23 * 3 * 10^4
Taking ‘ln’
⇒ 9.21 T = 220.29
⇒ T = 22029 / 9.21 = 2391.856 K

```
3 years ago
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