Lenard cathode-ray window experiment

The Lenard cathode-ray window experiment lets cathode rays leave a gas-discharge tube through a thin aluminium foil while preserving the tube's vacuum. This separates production from propagation.

Philipp Lenard's lecture, delivered on 28 May 1906, describes sealing a small hole in a metal plate with aluminium foil. Luminous phosphor grains first rested on the window, then glowed above it in air. His improved tube separated production and observation chambers; coated paper served as a fluorescent screen. An aperture revealed spreading beyond a narrow straight beam. Evacuating the observation chamber reduced spreading and attenuation without interrupting production. Interposed layers and stepped foil stacks gave shadow comparisons; approximate scaling with mass had exceptions.^[ref]

The experimental advantage is unusually clean: reducing the gas content in the observation chamber tests whether gas is needed to carry the cathode rays, while retaining the conditions needed to generate them. The window is a boundary that transmits some Electrons while confining the source environment. A spot outside establishes a transmitted effect; the aperture and changed observation environment provide additional ways to distinguish propagation from the source's behavior. This logic is useful whenever a complicated source would otherwise be changed by the very intervention intended to test the emitted beam.

Several inferences require care. A darker fluorescent screen can mean fewer arriving Electrons, less deposited energy per Electron, or redistribution by electron scattering beyond the observed area. A visible endpoint depends on the screen's sensitivity and background. It is not an intrinsic Electron range independent of initial kinetic energy, gas density, and observation geometry. For a modern comparison, measure a reference at fixed source conditions, use background subtraction, and compare the whole luminous patch as well as a small central region. A wider collection area tests redistribution; changing source-to-screen separation also changes collection geometry. These are proposed controls for interpreting a readout, not additional historical observations attributed to Lenard.

Nor is fluorescence a unique signature of cathode rays: X-rays can also excite it. Lenard notes that a later tube could transmit both, and that slow cathode rays failed to pass its window.^[ref] In modern terms, cathode rays are Electrons carrying negative electric charge, whereas X-rays are photons with zero electric charge. Magnetic bending of the incident Electrons, interpreted through the Lorentz force, supplies a discrimination that brightness alone cannot. Passage through foil alone determines neither the sign of electric charge nor its magnitude, and it does not measure charge-to-mass ratio. The separate Thomson charge-to-mass experiment explains that inference. Even a measured charge-to-mass ratio needs an independent electric charge measurement to determine mass.

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:The first equation gives exponential decay. Adding the equations gives $W^{\prime }=-aW$ for $W=U+D$, henceThese 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 $a=0.15\mathrm{c}{\mathrm{m}}^{-1}$ and $b=0.35\mathrm{c}{\mathrm{m}}^{-1}$. These are invented teaching inputs, not measured coefficients for air or Lenard's apparatus. At $x=2.00\mathrm{cm}$,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.

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 $R_{\mathrm{n}}=20+1000e^{-1}\simeq 387.88$ and $R_{\mathrm{w}}=20+1000e^{-0.3}\simeq 760.82$. After background subtraction, define $T_j=(R_j-B)/(R_0-B)$ for each geometry. Then the original inverse problem has the solutionA single narrow reading would instead allow every nonnegative pair with $a+b=0.50\mathrm{c}{\mathrm{m}}^{-1}$. Using the raw ratios without subtracting $B$ gives an erroneous narrow rate of about $0.4834\mathrm{c}{\mathrm{m}}^{-1}$. 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.^[ref]As a separate original limiting example, stipulate a straight path, constant $S=2\mathrm{keV}\mathrm{c}{\mathrm{m}}^{-1}$, 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 $\Delta K\ge 0$, a transmitted Electron enters the observation chamber with $K_0=eV-\Delta K\le eV$. 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.