Question

Compute the contrast between soft tissue (approximated as having the same
attenuation coefficient as water) and a 2 mm diameter artery filled with an iodine
contrast medium at a density 400 milligram/cm 3 .

Use photon energies of 40 Kev just above the iodine K absorption edge. Use values of: µ mwater = 0.36 cm 2 /g and µ mIodine = 22.38 cm 2 /g for 40 keV x-rays with

(a) no scattering and

(b) 50% scattering fraction. Scattering fraction = (scattered events)/(transmitted events+scattered events)

Answer 1

Given: diameter of the artery, d = 2 mm, the density of iodine contrast medium, = 400 mg/cc, E = 40 keV, of water = 0.36 cm²/g, of iodine = 22.38 cm²/g

a) no scattering

where I₁ is the x-ray intensity transmitted through tissue and I₂ is the x-ray intensity transmitted through the artery.

x₂ = d

b) 50% scattering fraction

F = 50 % = 0.5

I₁ is the transmitted intensity, RI₁ is the scattered intensity

Contrast due to scattering is,

C_s = 0.82/(1+1) = 0.82/2 = 0.41