Aditi Chauhan
Last Activity: 11 Years ago
Sol. A = 60 * 10^4 A/m^2 – k^2
∅ = 4.5 eV σ = 6 * 10^–8 ω/m^2 – k^4
S = 2 * 10^–5 m^2 K = 1.38 * 10^–23 J/K
H = 24 ω’
The Cathode acts as a black body, i.e. emissivity = 1
∴ E = σ A T^4 (A is area)
⇒ T^4 = E/σA = 24/6 * 10^-8 *2 * 10^-5 = 2 * 10^13K = 20 *10^12K
⇒ T = 2.1147 * 10^3 = 2114.7 K
Now, i = AST^2 e^-∅/KT
= 6 * 10^5 * 2 * 10^-5 * (2114.7)^2 * e^-4.5*1.6*10^-19/1.38*T*10^23
= 1.03456 * 10^-3 A = 1 mA