At 100 kV, a 0.5-mm Pb equivalent apron attenuates what percentage of the scattered beam?

At 100 kV, a 0.5-mm Pb equivalent apron attenuates what percentage of the scattered beam?

• A. 50%
• B. 25%
• C. 75%
• D. 100%

Answer: (C)

Attenuation is governed by the half-value layer concept: the thickness that reduces the beam’s intensity by half. For lead at diagnostic energies around 100 kVp, the HVL is about 0.25 mm Pb. A 0.5-mm Pb equivalent apron is roughly two HVLs thick (0.5 ÷ 0.25 ≈ 2). Each HVL halves the beam, so after two HVLs the transmitted intensity is about (1/2)² = 1/4 of the original. That means roughly 25% gets through, and about 75% is attenuated. Thus the apron would attenuate about 75% of the scattered beam at 100 kV.

Attenuation is governed by the half-value layer concept: the thickness that reduces the beam’s intensity by half. For lead at diagnostic energies around 100 kVp, the HVL is about 0.25 mm Pb. A 0.5-mm Pb equivalent apron is roughly two HVLs thick (0.5 ÷ 0.25 ≈ 2). Each HVL halves the beam, so after two HVLs the transmitted intensity is about (1/2)² = 1/4 of the original. That means roughly 25% gets through, and about 75% is attenuated. Thus the apron would attenuate about 75% of the scattered beam at 100 kV.