Define the attenuation coefficient as applied to a beam of X-rays travelling through a medium.

Derive the relationship between the attenuation coefficient μ and the half-value thickness \({x_{\frac{1}{2}}}\).

Aluminium is often used to filter out the low energy X-rays in a beam of X-rays. The following data are available for a particular X-ray beam.

Assuming equal initial intensities, determine, after the X-ray beam has passed through an aluminium sheet 6.0 mm thick, the following ratio.

\[\frac{{{\rm{intensity of 15keV X - rays}}}}{{{\rm{intensity of 30keV X - rays}}}}\]

Outline why X-rays are not suitable to image an organ such as the liver.

the intensity of the beam I falls by a constant amount dI through each distance dx travelled such that;
\(\frac{{dI}}{I} =  - \mu {\rm{d}}x\) where μ is the attenuation coefficient;

or

I=I₀e^-μx where I₀ is the intensity of the beam as it enters the medium;
I is the intensity after the beam has travelled a distance x in the medium and μ is the attenuation coefficient;

or

the probability per unit length;
that a photon will be absorbed;

\(I = \frac{{{I_0}}}{2} = {I_0}{e^{ - \mu {x_{\frac{1}{2}}}}}\) so \({e^{ - \mu {x_{\frac{1}{2}}}}} = \frac{1}{2}\);
\(\mu  = \frac{{\ln 2}}{{{x_{\frac{1}{2}}}}}\);

μ for 15keV = \(\left( {\frac{{\ln 2}}{{0.7}} = } \right)1.0{\rm{m}}{{\rm{m}}^{ - 1}}\) and μ for 30keV = \(\left( {\frac{{\ln 2}}{{3.5}} = } \right)0.20{\rm{m}}{{\rm{m}}^{ - 1}}\);
\(\left( {\frac{{{\rm{intensity of 15}}\left( {{\rm{keV}}} \right){\rm{ X - rays}}}}{{{\rm{intensity of 30}}\left( {{\rm{keV}}} \right){\rm{ X - rays}}}}} \right) = \frac{{{I_0}{e^{ - 1.0 \times 6.0}}}}{{{I_0}{e^{ - 0.2 \times 6.0}}}}\);
=8.2×10^-3;
Allow similar approach using half value thickness directly in the exponent, without intermediate calculation of μ, to find ratio.
Without the same rounding error this gives 8.6×10^-3.
Award [3] for a bald correct answer between 8.0 and 8.6×10^-3.

the attenuation coefficient of the liver and of the surrounding tissue for the X-rays have approximately the same value;
so no contrast can be obtained between liver and surrounding tissue / liver cannot be distinguished from surrounding tissue;