= Franck-Hertz energy-loss model
{c}
{title2=$\Delta,\ d=\Delta/e$}
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 $u\geq0$ 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 $0\leq p(u)\leq1$ otherwise. The conditional <probability distribution> of outgoing <kinetic energy> $E$ is the discrete loss kernel
$$
K(E\mid u)=[1-p(u)]\delta(E-u)+p(u)\delta(E-u+\Delta).
$$
Here $\delta$ 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.
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