The half-value layer (HVL) for annihilation radiation in lead is 4.1 cm.Approximately what percentage of a of 511 keV photon beam will penetrate 15 cm of lead?

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Quizplus · Accepted Answer

The HVL is the thickness of the absorber necessary to diminish the intensity of the radiation to half its initial strength.Because 4.1 cm is equal to one HVL, 8 cm is equal to approximately 2 HVLs, thereby diminishing the intensity of radiation to one-fourth of the original intensity.The equation is: I = I₀e^-^0.693x/HVL

The Half-Value Layer (HVL) for Annihilation Radiation in Lead Is