The linear attenuation coefficient for a material used to construct "compensators" for radiation therapy is 0.154 cm^-1 for a 6 MV beam. What percentage of radiation from this beam is transmitted through 15 mm of this material? What is the HVL? Please show all work!!
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Step 1
Given: - Linear attenuation coefficient (c) = 0.154 cm^-1 - Thickness of material (X) = 1.5 cm Use the formula N/N0 = e^(-cX) to calculate the percentage of radiation transmitted. N/N0 = e^(-0.154 * 1.5) N/N0 = e^(-0.231) N/N0 ≈ 0.793 or 79.3% Show more…
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