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?
To determine the thickness of the iron shield required to provide the same level of protection as 10 nm of lead, we need to calculate the linear absorption coefficient of iron and use it to find the necessary shield thickness.
The linear absorption coefficient (μ) represents the likelihood of a material absorbing x-rays. It is defined as the fraction of x-rays absorbed per unit thickness of the material.
Given that the linear absorption coefficients for x-rays in lead and iron are 5.8x10^4 /m and 1.1x10^4 /m respectively, the respective values represent the fraction of x-rays absorbed per meter of the material.
We can calculate the absorption per nanometer (nm) for both materials as follows:
Absorption in lead per nm = (5.8x10^4 /m) / (10^9 nm/m)
Absorption in iron per nm = (1.1x10^4 /m) / (10^9 nm/m)
Now we can calculate the thickness of the iron shield required to provide the same protection as 10 nm of lead:
Thickness of iron shield = (10 nm) x (Absorption in lead per nm) / (Absorption in iron per nm)
Substituting the values, we get:
Thickness of iron shield = (10 nm) x [(5.8x10^4 /m) / (10^9 nm/m)] / [(1.1x10^4 /m) / (10^9 nm/m)]
Simplifying the expression further by canceling out the units and manipulating the exponents, we have:
Thickness of iron shield = (10 nm) x (5.8x10^4) / (1.1x10^4)
Evaluating the above expression:
Thickness of iron shield ≈ 52.727 nm
Therefore, the iron shield should have a thickness of approximately 52.727 nm to provide the same level of protection as 10 nm of lead.