Saturation current of a 4 m cathode

The saturation current is given by the Richardson-Dushman equation:

$$J = A_0 T^2 \exp\left(-\frac{\phi}{kT}\right), \qquad I = J \times (\pi d l)$$

Given (as stated): $l = 4\ \text{m}$, $d = 2\ \text{mm}$, $T = 160 + 273.15 = 433.15\ \text{K}$, $A_0 = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\phi = 2.6\ \text{eV}$.

Emitting area

$$A = \pi d l = \pi \times 2 \times 10^{-3} \times 4 = 2.51 \times 10^{-2}\ \text{m}^2$$

Exponential term

$$\frac{\phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 433.15} = 69.63, \qquad e^{-69.63} = 5.75 \times 10^{-31}$$

Current density and current

$$\begin{aligned} J &= 3 \times 10^{4} \times (433.15)^2 \times 5.75 \times 10^{-31} = 3.24 \times 10^{-21}\ \text{A/m}^2 \\ I &= J A = 3.24 \times 10^{-21} \times 2.51 \times 10^{-2} = 8.1 \times 10^{-23}\ \text{A} \end{aligned}$$

So at 160°C there is practically no thermionic emission: $kT$ (0.037 eV) is far too small compared with the 2.6 eV work function.

Note: a transmitter tube cathode normally runs near 1600°C and is a few cm long. With $T = 1873.15$ K and $l = 4$ cm: $\phi/kT = 16.10$, $J = 1.07 \times 10^{4}\ \text{A/m}^2$, $A = 2.51 \times 10^{-4}\ \text{m}^2$, giving $I \approx 2.69\ \text{A}$.

Answer (data as given): $I \approx 8.1 \times 10^{-23}\ \text{A}$, i.e. negligible. (If 1600°C and 4 cm are meant, $I \approx 2.69$ A.)