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of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop half of the heidbrinkions emitted from a source. calculate the absorber thickness for lead.

User Sharpper
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Final answer:

To absorb all but one in 1000 of the gamma rays, 10 layers or 1.70 mm of lead is required, with each layer being 0.170 mm and representing a half-value layer.

Step-by-step explanation:

Absorber Thickness Calculation for Lead

To determine the thickness of lead needed to absorb all but one in 1000 of gamma (γ) rays, we can use the concept of half-value layers.

A half-value layer is the thickness of a material that reduces the intensity of radiation to half its original value.

If 0.170 mm of lead absorbs half of the γ rays, then for each additional 0.170 mm of lead, half of the remaining γ rays will be absorbed.

To reduce to one in 1000 of the original intensity, we need the amount of radiation to be 0.1% of the original.

Since each 0.170 mm reduces the intensity by half, we can express 0.1% as a power of ½ (which is 2-10 = 0.0977).

This means 10 half-value layers would be required to achieve this level of absorption.

Therefore, the total thickness of lead required is:
10 layers × 0.170 mm/layer = 1.70 mm.

User Fschmitt
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