Question

A normally incident X-ray of energy 140 keV passes through a tissue phantom and is detected by the detector as shown. The phantom consists of tissue \(P\) with an absorption coefficient of \(1~\mathrm{cm^{-1}}\) and a thickness of \(1~\mathrm{cm}\), and tissue \(Q\) with an absorption coefficient of \(10~\mathrm{cm^{-1}}\) and a thickness of \(2~\mathrm{cm}\). Calculate the intensity (in \(\mu\mathrm{eV}\)) detected by the detector. (Round off the answer to one decimal place.)

Hint:

- For layered media, multiply the linear attenuation coefficient \(\mu\) by each layer thickness and add the exponents: \(I=I_0e^{-\sum \mu_i d_i}\).
- Convert energy units at the end to avoid carrying large/small numbers through intermediate steps.

Answer 1

Correct Answer: 106

Step 1: X-ray intensity decays exponentially through matter: \[ I=I_0\,e^{-\mu_P d_P-\mu_Q d_Q}. \] Here, \(\mu_P=1~\mathrm{cm^{-1}},~d_P=1~\mathrm{cm}\) and \(\mu_Q=10~\mathrm{cm^{-1}},~d_Q=2~\mathrm{cm}\). Thus the total attenuation exponent is \[ \mu_P d_P+\mu_Q d_Q = 1\times 1 + 10\times 2 = 21. \] Hence, \[ \frac{I}{I_0}=e^{-21}\approx 7.5826\times 10^{-10}. \] Step 2: The incident energy per photon is \(I_0=140~\text{keV}=1.4\times 10^5~\text{eV}\). Therefore, \[ I = 1.4\times 10^5~\text{eV}\times 7.5826\times 10^{-10} \approx 1.061\times 10^{-4}~\text{eV} = 106.1~\mu\text{eV}. \]