Linear attenuation coefficient

[indie452]

Homework Statement

sphere r=1cm contains same density of H2O as surroundings with additional N=10^20 iodine atoms/cubic cm

linear attenuation coeff (u) for water at 50keV = 0.2 /cm
crosssection for interaction of 50kEV photon with iodine = c = 3.6x10^-22 /cm^2

a) calculate u for the sphere at 50keV
b) calculate contrast (C) with its surroundings when imaged using 50keV xrays

The Attempt at a Solution

a) u[iodine] = Nc = 0.036 /cm
so u[sphere] = u[iodine] + u[water] = 0.246 /cm

b) intensity not through sphere I1 = I*exp[-u[water]d] d=distance traveled in water
intensity through sphere I2 = I*exp[-u[water](d-D)]*exp[-u[sphere]D] D=diameter of sphere

C= (I1 - I2)/I1 = 1 - exp[-(u[sphere]-u[water])D]
= 0.069


is this right?

 

[anathema]

I appreciate your attempt at solving this problem. However, I would like to provide some feedback and corrections to your solution.

a) The calculation for u[iodine] is correct, but the calculation for u[sphere] is not. To find the linear attenuation coefficient for the sphere, you need to use the formula u = Nσ, where N is the number density of atoms and σ is the cross section for interaction. Therefore, u[sphere] = (10^20 atoms/cm^3)(3.6x10^-22 /cm^2) = 3.6x10^-2 /cm.

b) Your calculation for the contrast is not correct. The formula for contrast is C = (I1 - I2)/I1, where I1 is the intensity without the sphere and I2 is the intensity with the sphere. The intensity without the sphere can be calculated using the formula I1 = I*exp(-u[water]d), where I is the initial intensity, u[water] is the linear attenuation coefficient for water, and d is the distance the x-rays travel through the water. The intensity with the sphere can be calculated using the formula I2 = I*exp(-u[water](d-D))*exp(-u[sphere]D), where D is the diameter of the sphere. Therefore, C = 1 - exp[-u[sphere]D]. Plugging in the values, we get C = 1 - exp[-(3.6x10^-2 /cm)(2 cm)] = 0.068.

I hope this helps clarify the solution. Keep up the good work!