= Cathode-ray attenuation with finite detector acceptance
{title2=$U(x)=e^{-(a+b)x},\quad W(x)=e^{-ax}$}
An original toy model separates two ways to lose a central <cathode ray> signal. Let $x$ be forward depth into a uniform <gas>. Normalize the entering <Electron> population to one. Write $U(x)$ for the fraction still in a narrow forward channel, and $D(x)$ for the fraction diverted into a wider collection channel. Assume constant removal rate $a$ per unit depth in both channels, and constant transfer rate $b$ from $U$ to $D$, with no return. Removal means that an <Electron> stops or is trapped before reaching the collecting plane; it does not mean that its <electric charge> disappears. The wide ideal collector accepts all surviving members of both channels. This is a stipulated two-population <differential equation> model, not a general law of <Electron> transport:
$$
\frac{dU}{dx}=-(a+b)U,\qquad
\frac{dD}{dx}=bU-aD,\qquad
U(0)=1,\quad D(0)=0.
$$
The first equation gives <exponential decay>. Adding the equations gives $W'= -aW$ for $W=U+D$, hence
$$
U(x)=e^{-(a+b)x},\qquad
W(x)=e^{-ax},\qquad
D(x)=e^{-ax}\left(1-e^{-bx}\right).
$$
These formulas also give a conservation check: $U+D+(1-W)=1$. The fraction $1-W$ has been removed before the plane, while $D$ remains present but misses the narrow collector. An exponential fit to the narrow signal therefore measures $a+b$, not the removal rate $a$ by itself. This is an <identifiability> issue: one declining signal cannot distinguish the two unknown rates.
For a worked example, choose $a=0.15\,\mathrm{cm^{-1}}$ and $b=0.35\,\mathrm{cm^{-1}}$. These are invented teaching inputs, not measured coefficients for air or Lenard's apparatus. At $x=2.00\,\mathrm{cm}$,
$$
U=e^{-1}\simeq0.3679,\qquad
W=e^{-0.3}\simeq0.7408,\qquad
D\simeq0.3729,\qquad
1-W\simeq0.2592.
$$
Thus about $63.2\%$ misses the narrow collector, although only about $25.9\%$ has been removed in this model. Treating the whole narrow-signal deficit as stopping would confuse redistribution with removal.
\Image[/cathode-ray-attenuation-and-acceptance-model.png]
{title=Two collecting geometries distinguish redistribution from removal in a toy <cathode ray> model}
{id=image-cathode-ray-attenuation-and-acceptance-model}
{description=Original calculation with chosen constant rates $a=0.15\,\mathrm{cm^{-1}}$ and $b=0.35\,\mathrm{cm^{-1}}$. Left: fractions in the narrow channel, diverted channel, and removed population sum to one. Right: logarithmic plots of the narrow and wide surviving fractions have different slopes. The marked depth is $2\,\mathrm{cm}$. This model omits <Electron> slowing and return between channels; neither panel contains historical measurements.}
Suppose a calibrated readout is proportional to arriving <Electron> number with the same response for either channel. Choose an empty-path signal $R_0=1020$ arbitrary units, of which $B=20$ is background; the two collection geometries have the same calibrated reference. The calculated readings at $2\,\mathrm{cm}$ are $R_{\rm n}=20+1000e^{-1}\simeq387.88$ and $R_{\rm w}=20+1000e^{-0.3}\simeq760.82$. After <background subtraction>, define $T_j=(R_j-B)/(R_0-B)$ for each geometry. Then the original <inverse problem> has the solution
$$
a=-\frac{\ln T_{\rm w}}{x}=0.15\,\mathrm{cm^{-1}},\qquad
b=-\frac{\ln T_{\rm n}}{x}-a=0.35\,\mathrm{cm^{-1}}.
$$
A single narrow reading would instead allow every nonnegative pair with $a+b=0.50\,\mathrm{cm^{-1}}$. Using the raw ratios without subtracting $B$ gives an erroneous narrow rate of about $0.4834\,\mathrm{cm^{-1}}$. In a real <fluorescent screen>, the assumed response must be tested: deposited <energy> and conversion to <light> need not remain constant as <Electrons> slow. Repeating the comparison at several depths tests the constant slopes, although straight slopes alone do not establish the microscopic mechanism.
Actual <Electrons> change <kinetic energy> and direction. <Electron stopping power> $S(K)=-dK/ds>0$ describes mean <kinetic energy> loss per travelled <path length> $s$; collisional losses include <ionization>, while radiative losses include <bremsstrahlung>. The <continuous slowing-down approximation range> integrates reciprocal total <electron stopping power>, neglecting fluctuations. It approximates travelled <path length>, not forward penetration depth.<Stopping-Power and Range Tables for Electrons, Protons, and Helium Ions by M. J. Berger and others/Appendix: Significance of Calculated Quantities>[Stopping-Power and Range Tables for Electrons, Protons, and Helium Ions by M. J. Berger and others; Appendix: Significance of Calculated Quantities, collision/radiative/total stopping power, CSDA range, projected range and detour factor]{ref}
$$
R_{\mathrm{CSDA}}(K_0)=\int_0^{K_0}\frac{dK}{S(K)}.
$$
As a separate original limiting example, stipulate a straight path, constant $S=2\,\mathrm{keV\,cm^{-1}}$, and entering <kinetic energy> $K_0=6\,\mathrm{keV}$. Then $K(s)=6\,\mathrm{keV}-Ss$ until stopping, and $R_{\mathrm{CSDA}}=3\,\mathrm{cm}$. In that deterministic model every entering <Electron> reaches each depth below $3\,\mathrm{cm}$ and none crosses a plane beyond it. There is no exponential population tail. Real fluctuations and <electron scattering> make the situation more complicated, but the example already proves that continuous <energy> loss does not imply constant-rate removal.
The window also changes the entering population: source <kinetic energy> before the foil need not equal $K_0$ just outside it. For an <Electron> initially at rest, an ideal static accelerating <electric potential difference> $V>0$ supplies electrical <work> $eV$, where $e$ is the <elementary charge>. If losses before and within the foil total $\Delta K\geq0$, a transmitted <Electron> enters the observation chamber with $K_0=eV-\Delta K\leq eV$. A screen's brightness determines neither $V$ nor $\Delta K$. These formulas specify a modern ideal <work-energy theorem> argument; no numerical historical <voltage> or foil thickness is inferred from the diagram.
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