Soft tissue has a linear attenuation coefficient in the range of = 0.35 cm1 at 3

Question

Soft tissue has a linear attenuation coefficient in the range of = 0.35 cm1 at 30 keV and = 0.16 cm1 at 100 keV. For this problem we use = 0.21 cm1, which applies at around 50 keV incident X-ray energy. (a) What fraction of X-ray photons at 50 keV are passing through a person's body? Hint: the person's body thickness is about 22 cm. Use a soft tissue approximation, i.e., neglecting bones. N N0 = (b) We compare bone and soft tissue of 5 cm thickness each. Using bone = 0.57 cm1 at 50 keV, what fraction of a 50 keV incident X-ray beam is stopped in bone and soft tissue respectively? bone N N0 = soft tissue N N0 =

Explanation / Answer

Part a ) For this exercise we use the expression

I = Io e-x

I/Io = e-x

Data

= 0.21 cm-1

x = 22 cm

I/Io = e - 0.21 22

I/Io = 9.85 10-3

Part b)

x2 = 5 cm

soft tissu

I/Io t = e-0.21 5= 0.3499

bone

= 0.57 cm-1

I/Io b = e-0.57 5= 0,0578

These are the fractions of radiation passing in each type of tissue and bone

The fraction is stoping

soft tissu 1- 0.3499 = 0.6501

bone 1- 0.0578 = 0.9422

the rate between tissue and bone

It/Ib = 0.3499/0.0578

It / Ib = 6.05

This means that radiation times 6.05 passing for soft tissue that in the bone

stoping

f stop = 0.6501/0.9422

f Stop = 0.69

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