Saturation current of thoriated tungsten cathode

The saturation current of a thermionic cathode is given by the Richardson-Dushman equation:

$$J = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad I = J \times A$$

where $A$ is the emitting (curved) surface area of the cylinder, $A = \pi d l$.

Given: $T = 1500 + 273.15 = 1773.15\ \text{K}$, $l = 5\ \text{cm} = 0.05\ \text{m}$, $d = 1.5\ \text{mm} = 1.5 \times 10^{-3}\ \text{m}$, $B_0 = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\Phi = 2.6\ \text{eV}$.

Emitting area

$$A = \pi d l = \pi \times 1.5 \times 10^{-3} \times 0.05 = 2.356 \times 10^{-4}\ \text{m}^2$$

Exponent

$$\frac{\Phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 1773.15} = 17.01, \qquad e^{-17.01} = 4.10 \times 10^{-8}$$

Current density

$$J = 3 \times 10^{4} \times (1773.15)^2 \times 4.10 \times 10^{-8} = 3.87 \times 10^{3}\ \text{A/m}^2$$

Saturation current

$$I = J A = 3.87 \times 10^{3} \times 2.356 \times 10^{-4} = 0.911\ \text{A}$$

Answer: Saturation current $I \approx 0.91\ \text{A}$ (current density $\approx 3.87 \times 10^{3}\ \text{A/m}^2$).