Question

The constant A in the Richardson-Dushman equation for tungsten is $60\times {10}^{4}A{m}^{-2}{K}^{-2}$ The work function of tungsten is 4.5 eV. A tungsten cathode having a surface area $2.0\times {10}^{-5}{m}^2$ is heated by the heater and the temperature becomes constant. Assuming that the cathode radiates like a blackbody, calculate the saturation current due to thermions. Take stefan constant$=6\times {10}^{-8}W{m}^{-2}{K}^{-4}.$ Assume that the termions take only a small fraction of th heat suplied.

Answer 1

Given, $A=60\times {10}^{4\frac{A}{m2-k^2}}\varphi =4.5eV,\delta =6\times {10}^{-8}\omega /m^2-k^{4}S=2\times {10}^{-5}{m}^2,k=1.38\times {10}^{-23}J/KH=24\omega$
The cathode acts as a black body,
i.e. emissivity = 1
$\therefore E=\delta A{T}^4\Rightarrow T^4=\frac{E}{\sigma A}=\frac{24}{6\times {10}^{-8}\times 2\times {10}^{-5}}=2\times {10}^{13}K=20\times {10}^{12}K\Rightarrow T=2.1147\times {10}^3=2114.7K.Now,i=AS{T}^2{e}^{-\varphi \sqrt{KT}}=6\times {10}^5\times 2\times {10}^{-5}\times (2114.7{)}^2\times e^{\frac{-4.5\times 1.6\times {10}^{-19}}{1.38\times {10}^{-29}\times 2114.7}}=1.03456\times {10}^{-3}A=1mA.$