Question

Compute the contrast between normal fatty breast tissue and a 0.1-mm microcalcification for (a) 20 keV and (b) 60 keV x-rays. We can approximate the breast tissue as having roughly the same linear attenuation coefficient as fat, while the microcalcification is modeled as having the same linear attenuation coefficient as mineralized bone. Do the same calculation for a 0.1-cm lump having the same linear attenuation coefficient as water in fatty breast tissue for (c) 20 keV and (d) 60 keV x-rays. Note which cases would be detectable for a film-screen image receptor sensi- tive only to contrasts greater than 2%.

Answer 1

For this question I am making use of the following table:

X-ray energy (keV)	μ air (cm-1)	μ water (cm-1)	μ fat (cm -1)	μ bone (cm-1)
20	6.4 x 10 -4	0.76	0.5	4.8
60	3.7 x 10-5	0.2	0.17	.47 to 0.55

x = 0.1 mm = 10^-2 cm

(a)

)

)

)

(b)

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)

In both the cases, C is positive. This means, the microcalcification is more absorbing than fat, i.e., I₁ > I₂.

At 20 keV, the difference is higher. This tells us that an image obtained at 20 keV will be able to distinguish the microcalcification, as the film screen image receptor is sensitive only to contrasts greater than 2%.