Why the anode current dips every 4.9 V in mercury vapour.
-Voltage U (V)
-Anode current (relative)
-Electrons passing the retarding field (%)
-Collisions possible (floor U / E_ex)
-Mean free path (um)
-Vapour pressure (kPa)
-Free paths across the tube
-Excitation energy (eV)
The science
Electrons leave a hot cathode and are accelerated toward a grid by the voltage U, gaining kinetic energy e U if nothing gets in the way. The tube holds mercury vapour with mean free path lambda = k T / (sigma p). Collisions are elastic (almost no energy lost) until an electron carries at least the excitation energy E_ex = 4.9 eV; then it can hand exactly that amount to an atom, which later emits ultraviolet light.
Between the grid and the anode a small retarding voltage turns back electrons with less energy than e V_r. As U rises the current climbs, then falls sharply when electrons start to lose 4.9 eV just before the grid, and climbs again. The maxima are spaced by E_ex / e, evidence that atomic energy levels are discrete (Franck and Hertz, 1914).
The current plotted is the passing fraction multiplied by a space-charge envelope (U/10)^1.5 (Child-Langmuir) so the maxima rise with voltage as in real data.
Honest simplifications: one dimension (collisions do not change direction), a single excitation level, a simple linear ramp of the excitation probability above threshold, no contact potential, no ionisation, and a fitted vapour-pressure formula valid between about 100 and 300 degrees C.