A normally incident X-ray of energy 140 keV passes through the tissue phantom shown. Tissue P has attenuation coefficient 1 cm⁻¹ and thickness 1 cm; tissue Q has coefficient 10 cm⁻¹ and thickness 2 cm. Calculate detected intensity, expressed as energy in μeV under the source’s attenuation model, to one decimal place.
Apply exponential attenuation successively: Eout=140exp[−(1×1+10×2)] keV=140e⁻²¹ keV. One keV equals 10⁹ μeV, so Eout=140×10⁹e⁻²¹≈106.156 μeV, or 106.2 μeV. This source convention attenuates total beam energy/intensity; it does not mean each surviving photon loses energy by this exponential factor.
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