X-ray attenuation ($I(x)=I_0{e}^{-\mu x}$)

For a narrow beam of monoenergetic photons crossing homogeneous material of thickness $x$, the uncollided wave intensity obeys $I(x)=I_0{e}^{-\mu x}$. Here $\mu$ is the linear attenuation coefficient, counting removal from the incident beam by photon absorption and scattering of light. It depends on photon energy and material; it is not universally independent of energy.^[ref]

An original derivation makes the model explicit. If each thin layer removes fraction $\mu dx$ of the surviving beam, with the same $\mu$ throughout, thenThus equal added thicknesses multiply transmission by the same factor. The ordinary differential equation describes uncollided beam survival. To test it with a detector, keep source output, distance, selected photon energy band and detector response fixed. Use a collimator and suitable detector acceptance to exclude scattering of light into the detector. Subtract background, measure an empty-path reference, and keep the detector in its linear operating range. A changing source distance adds geometric dilution and is not an attenuation measurement at fixed geometry.

For a stipulated example, take $\mu =0.40\mathrm{m}{\mathrm{m}}^{-1}$ and $x=2.5\mathrm{mm}$. Transmission is $e^{-1}\simeq 0.368$. If the empty-path detector rate is $1010{\mathrm{s}}^{-1}$, including background $b=10{\mathrm{s}}^{-1}$, the model predicts sample rate $10+1000e^{-1}\simeq 378{\mathrm{s}}^{-1}$. Recover transmission as $(R-b)/(R_0-b)$; the raw ratio $R/R_0\simeq 0.374$ is biased. These chosen values are a calculation, not measurements of a particular material or a historical apparatus.

The linear attenuation coefficient is the infinitesimal probability of removal from the uncollided beam per unit path length. Its units are inverse length. In the homogeneous monoenergetic model, $\mu =-d\ln I/dx$; $1/\mu$ is the thickness leaving fraction $e^{-1}$ uncollided. The mass attenuation coefficient is $\mu /\rho$ for material mass density $\rho$. Removing a photon by scattering of light does not imply deposition of all its energy at that point; attenuation and energy absorption are different quantities.^[ref]

The linear attenuation coefficient divided by the material mass density.

The half-value layer is the thickness reducing the specified transmitted signal by one half. For constant monoenergetic linear attenuation coefficient $\mu$, solving $e^{-\mu x_{1/2}}=1/2$ gives $x_{1/2}=\ln 2/\mu$. The stipulated $\mu =0.40\mathrm{m}{\mathrm{m}}^{-1}$ gives $x_{1/2}\simeq 1.73\mathrm{mm}$; two such layers transmit one quarter. A broad frequency spectrum need not retain the same half-value layer after filtering.

An original two-energy model demonstrates the failure of a single constant linear attenuation coefficient for a mixed beam. Suppose the two components contribute equally to the empty-path detector signal and have stipulated ${\mu }_1=0.40\mathrm{m}{\mathrm{m}}^{-1}$ and ${\mu }_2=0.10\mathrm{m}{\mathrm{m}}^{-1}$. With fixed linear detector response and no accepted scattering of light,The initial slope corresponds to $\mu =0.25\mathrm{m}{\mathrm{m}}^{-1}$, but at $x=5\mathrm{mm}$, $T\simeq 0.371$, rather than $e^{-0.25(5)}\simeq 0.287$. The more strongly attenuated component disappears faster. If that component has lower photon energy, the surviving beam has a higher mean energy: beam hardening. The formula is explicitly a two-component example, not a universal rule that attenuation decreases monotonically with photon energy. Detector weighting also matters when interpreting a broadband transmission signal.

An absorption edge is an abrupt change in photon absorption as incident photon energy crosses an electronic excitation threshold. For inner-shell photoionization, a new vacancy-production channel opens near the shell's binding energy. This threshold concerns incident energy; a Characteristic X-ray concerns a subsequent difference of atomic energies. They are different observations. Barkla's lecture explicitly treats characteristic emission as requiring a critical excitation condition, while discussing characteristic radiation separately from scattering.^[ref]

In the stipulated X-ray fluorescence model, the K threshold is $10.0\mathrm{keV}$, whereas the K lines are $8.5$ and $9.7\mathrm{keV}$. An absorber made of a different material could have its own absorption edge between those line energies, preferentially removing the upper line. Actual near-edge structure and attenuation values require material data; the following deliberately chosen coefficients are not such data.