Soft tissue has a linear attenuation coefficient in the range of u = 0.35 cm ^-1

Question

Soft tissue has a linear attenuation coefficient in the range of u = 0.35 cm^-1 at 30 keV and u= 0.16 cm^-1 at 100 keV. For this problem we use u= 0.16 cm^-1, which applies at around 80 keV incident X-ray energy.

(a) What fraction of X-ray photons at 80 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.

(b) We compare bone and soft tissue of 3 cm thickness each. Using ubone = 0.34 cm^-1 at 80 keV, what fraction of a 80 keV incident X-ray beam is stopped in bone and soft tissue respectively?

N over No = ? Soft tissue has a linear attenuation coefficient in the range of u = 0.35 cm^-1 at 30 keV and u= 0.16 cm^-1 at 100 keV. For this problem we use u= 0.16 cm^-1, which applies at around 80 keV incident X-ray energy. (a) What fraction of X-ray photons at 80 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. (b) We compare bone and soft tissue of 3 cm thickness each. Using u bone = 0.34 cm^-1 at 80 keV, what fraction of a 80 keV incident X-ray beam is stopped in bone and soft tissue respectively?

Explanation / Answer

Fraction of photons passing through( for large thickness) = e^(-u*t)

Fraction of photons stoppeed ( for large thickness) = 1- e^(-u*t)

where u = linear attenuation coefficient and t = thickness

a)  fraction of X-ray photons at 80 keV are passing through a person's body = e^(-0.16*19)= e^-3.04=0.0478

b) For bone the fraction stopped = 1 -  e^(-u*t) = 1 - e^(-0.34*3) =0.6394

For soft tissue the fraction stopped = 1 -  e^(-u*t) = 1 - e^(-0.16*3)=0.3812

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