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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.

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.

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.

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.

For a fixed homogeneous absorber in a narrow-beam geometry, define the X-ray spectral attenuation operator on L1 space over a bounded positive energy interval. A physical input $f(E)\ge 0$ is an integrable function giving incident photon rate density. Let its linear attenuation coefficient $\mu (E)\ge 0$ be a measurable function, with $x\ge 0$. The transmitted uncollided density isHere $w(E)\ge 0$ is a fixed bounded detector response described by a measurable function; background has been subtracted. This is a positive linear operator on L1 space. Since $0\le e^{-\mu (E)x}\le 1$, it cannot increase the integrated absolute incident rate. Multiplication of exponentials proves $A_{x+y}=A_x{A}_y$ for successive layers of the same material. Ignore secondary radiation entering the detector, saturation and changes to the source. Discrete spectral lines use the finite-sum version below instead of an ordinary rate density.

For a chosen filter, stipulate ${\mu }_{\alpha }=0.20\mathrm{m}{\mathrm{m}}^{-1}$, ${\mu }_{\beta }=0.80\mathrm{m}{\mathrm{m}}^{-1}$ and input line rates $(N_{\alpha },N_{\beta })=(4,1)$ in a common arbitrary rate unit. Assume equal calibrated detector weight for both lines. At $x=2.0\mathrm{mm}$,The selected $8.5\mathrm{keV}$ line's fraction rises from $4/5=0.800$ to $0.929974$, but its absolute rate falls by about $33.0\%$. Filtering improves purity at the cost of signal. It multiplies existing components and cannot create a new emitted spectral line.

With just one integrated reading, $S=N_{\alpha }+N_{\beta }$, the two unknown line rates are underdetermined. If the source is unchanged between an empty-path reading and a filtered reading, the ideal inverse problem becomesThe determinant $e^{-1.6}-e^{-0.4}\simeq -0.468424$ is nonzero, so these two ideal readings determine the two stipulated line rates uniquely. Identical coefficients would give zero determinant and destroy that separation. This small linear algebra example explains why absorber contrast supplies information, while also showing its limit: unknown scattered components or unknown detector weights add unknowns that two readings cannot determine.