The linear absorption coefficient for 0.05 nm x-rays in lead and in iron are respectively, 5.8x10(4) /m and 1.1x10 (4) /m. How thick should an iron shield be in order to provide the same protection from these x-rays as 10 nm of lead?

The product of thickness L and absorption coefficient k should be the same for both lead and iron, to obtain equivalent protection.

In other words, the e^(-kL) values must be the same.

Iron must therefore be 5.27 times thicker than lead.

To determine the thickness of the iron shield needed to provide the same protection as 10 nm of lead, we need to use the concept of linear absorption coefficient.

The linear absorption coefficient (μ) represents the probability of an x-ray photon being absorbed per unit thickness of the material. It is measured in units of 1/m (inverse meters).

Given:
Linear absorption coefficient for 0.05 nm x-rays in lead (μ_lead) = 5.8 x 10^4 /m
Linear absorption coefficient for 0.05 nm x-rays in iron (μ_iron) = 1.1 x 10^4 /m
Thickness of lead shield (d_lead) = 10 nm

To find the thickness of the iron shield (d_iron), we can use the formula:

d_iron = (μ_lead / μ_iron) * d_lead

Substituting the given values into the formula:

d_iron = (5.8 x 10^4 / 1.1 x 10^4) * 10 nm

Simplifying:

d_iron = (5.27) * 10 nm
d_iron ≈ 52.7 nm

Therefore, the iron shield should have a thickness of approximately 52.7 nm to provide the same protection from the 0.05 nm x-rays as 10 nm of lead.