Soft tissue has a linear attenuation coefficient in the range of = 0.35 cm1 at 3
ID: 1446584 • Letter: S
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 19 cm. Use a soft tissue approximation, i.e., neglecting bones.
N N0 =
(b) We compare bone and soft tissue of 3 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
a) I = Io * e(-µx)
=> fraction of X-ray photons passing through a person's body = e(-µx)
= e(- 0.21 * 19)
= 0.0185
b) fraction of X-ray beam stopped in bone = 1 - e(- 0.57 * 3)
= 0.8191
fraction of X-ray beam stopped in tissue = 1 - e(- 0.21 * 3)
= 0.4674
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