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.

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