The diagram below shows X-rays being used to scan a sample of bone and muscle.

(i) Outline how the arrangement differentiates between bone and muscle.

(ii) Use the data below to determine the ratio \(\frac{{{I_b}}}{{{{\mathop{\rm I}\nolimits} _m}}}\) where I_b and I_m are the intensity of X-rays reaching the photographic plate through the bone and the muscle, respectively.
                                                                       Thickness x of sample                                 = 10.0 cm
                                                                       Linear attenuation coefficient of bone μ_b     = 0.53 cm^–1
                                                                       Linear attenuation coefficient of muscle μ_m = 0.30 cm^–1

(iii) The half-value thickness of a material increases as the energy of the radiation increases.

Discuss, with reference to penetration and effect on tissue, why using low energy X-rays in medical imaging is highly desirable but is rare in practice.

The same sample is now investigated with an ultrasound A-scan from the side as shown.

(i) State one advantage of ultrasound over X-ray imaging.

(ii) State why gel is needed at the transducer-muscle boundary.

(iii) A short pulse is directed from the transducer into the sample at time t = 0.
The graph shows how the intensity of the reflected signal from the muscle-bone boundary varies as a function of time. The speed of sound in muscle is 1.6×10³ m s^–1.

Calculate the thickness y of the sample of muscle.

(i) X-rays are absorbed more by bone as it is denser / X-rays are transmitted more by muscle as it is less dense;
denser material/bone leads to less exposure / less dense material/muscle leads to more exposure (on the photographic plate); 

(ii) I_b=I₀e^-0.53×10=0.0050×I₀;
I_m=I₀e^-0.30×10=0.0498×I₀;
\(\frac{{{I_b}}}{{{{\mathop{\rm I}\nolimits} _m}}} = \frac{{0.0050 \times {I_0}}}{{0.0498 \times {I_0}}} = 0.10\);

(iii) low energy X-rays cause less damage to tissue;
but do not penetrate as deeply;

(i) no exposure to radiation / OWTTE;

(ii) for impedance matching / OWTTE; 

(iii) \(y = \frac{{vt}}{2}\);
\(y = \frac{{1600 \times 10 \times {{10}^{ - 5}}}}{2} = 0.096\left( {\rm{m}} \right)\left( {{\rm{ = 9.6cm}}} \right)\);

(a)(i) Many candidates had the idea of the difference between muscle and bone, but not all mentioned density or exposure on the plate. There were many mistakes in algebra in (a)(ii), especially with the exponent. Many candidates got confused with units, changing the thickness to m but not changing the attenuation coefficient to m^-1. (a)(iii) was well answered by most candidates.

(b)(i) and (b)(ii) were well answered though answers to (b)(ii) were not always precise. (b)(iii) was very poorly answered. This was surprising as it was basically reading a value from a graph and applying the equation speed = distance / time. Many candidates misread the graph value and many forgot to divide by two to take into account travel and return time.