Question

ssume kVp= 150kV. At this kVp, mass attenuation coefficients of bone and soft-tissue are0.14 cm2g−1and 0.15 cm2g−1, respectively. Assuming the densities of bone and soft-tissue are3.8 g cm−3and 1.5 g cm−3, respectively, calculate the numbers (relative to the incident number)of X-ray photons that will be transmitted through the left (bone and tissue) and right (tissueonly) halves of the body. (Points: 4)

Answer 1

We know that,

`I=I_oe^(-\mu R)`

With the dates that we have, we can calculate \mu for the bone and for the tissue, that is

`\mu_(tissue)=0.15*1.5cm^(-1)`

`\mu_(tissue)=0.225cm^(-1)`

And for the bone,

`\mu_(bone)=0.14*3.8cm^(-1)`

`\mu_(bone)=0.532cm^(-1)`

So, calculate I for left side,

For the tissue part,

`I_1 = I_0e^(-0.225*1)\\I_1=I_0e^(-0.225)`

FOr the bone part,

`I_2=I_0e^(-0.225)e^(-0.532*1)\\I_2=I_0e^(0.225+0.532)\\I_2=I_0e^(-0.757)`

And,

`I_3=I_0e^(-0.757)e^(-0.225)\\I_3=I_0e^(-0.982)`

Final Intensity for the left side is given by,

`I'=I_0e^(-0.982)\\(I')/(I_0) = e^(-0.982)\\(I')/(I_0)=0.374`

Moreover for the right side

`I'=I_0e^(-0.225*3)\\I'=I_0e^(-0.675)`

`(I')/(I_0)=0.51`