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A useful Franck-Hertz energy-loss model separates three operations: accelerating an Electron, losing a discrete energy $\Delta$ in an inelastic collision, and selecting the surviving Electrons with a retarding collector. The collector measures an accepted fraction of an outgoing energy distribution. This makes the experiment a small inverse problem: the electric current depends on both the collision physics and the readout.

For a deliberately simple one-channel model, let $u\ge 0$ be the incoming kinetic energy, and let $p(u)$ be the probability that one encounter excites the chemical atom. Set $p(u)=0$ for $u<\Delta$ and $0\le p(u)\le 1$ otherwise. The conditional probability distribution of outgoing kinetic energy $E$ is the discrete loss kernelHere $\delta$ is the Dirac delta function. The kernel has unit integral, no negative outgoing energy, and mean outgoing kinetic energy $u-p(u)\Delta$. It neglects the small recoil loss in an elastic collision, other excitation channels and ionization. The threshold permits excitation; it does not require $p(u)=1$ above threshold. In particular, choosing a single $\Delta =4.9\mathrm{eV}$ here isolates the characteristic channel discussed above without claiming it is mercury's lowest possible excitation.

For Electrons directed normally toward an ideal planar collector, a retarding electric potential difference $V_r>0$ imposes the barrier $e{V}_r$. Write its transfer as $T(E)=H(E-e{V}_r)$, where $H$ is the Heaviside step function and $e$ the positive elementary charge. Thus a perfectly sharp loss need not produce a sharp drop in electric current: the drop also depends on where the outgoing energy lies relative to the barrier. For an oblique incident Electron at angle $\theta$ to the normal, the same planar model instead tests $E{\cos }^2\theta >e{V}_r$; angular scattering can change collection even at unchanged total kinetic energy.

For example, choose a monoenergetic incoming beam with $u=5.6\mathrm{eV}$, $p=0.75$ and $\Delta =4.9\mathrm{eV}$. After one encounter, $25\%$ have $5.6\mathrm{eV}$ and $75\%$ have $0.7\mathrm{eV}$. A $1.0\mathrm{V}$ retarder collects only $0.25I_0$, whereas a $0.5\mathrm{V}$ retarder collects $I_0$; here $I_0$ is the incident electric current with fixed geometrical acceptance and no background. The mean outgoing kinetic energy is $1.925\mathrm{eV}$, independently checked by $5.6-0.75(4.9)$. These are chosen inputs and calculated fractions, not historical measurements.

More generally, let $f(E;V)$ be the normalized outgoing probability density function at a fixed accelerating electric potential difference $V$, for the normally incident idealization. A retarding scan givesThe second equation follows by differentiating the lower integration limit, with Dirac delta functions understood for sharp beams. It explains why an integrated collector readout can reveal an energy distribution. With angles present it probes the distribution of normal kinetic energy; finite instrumental resolution smooths it, and numerical differentiation amplifies noise. Keep the incoming beam fixed during this scan. A scan of $V$ changes $f$ itself and cannot be differentiated by the same inversion formula.