The intensity of a parallel X-ray beam is reduced to 50% of its initial intensity when it passes through bone of thickness 1.2 cm. Determine the thickness of bone needed to reduce the intensity of the same X-ray beam to 15% of its initial value.

$\mu  = \frac{{\ln 2}}{{1.2}}{\text{ }}\left( { = 0.58{\text{ c}}{{\text{m}}^{ - 1}}} \right)$;

$0.15 = {{\text{e}}^{ - 0.58x}}$;

$x = 3.3{\text{ (cm)}}$;

Award [3] for a bald correct answer.

Watch for valid working in metres.

or

${(0.5)^n} = 0.15$ (where n is number of half value thicknesses travelled);

so $n = 2.74$;

$x = (2.74 \times 1.2{\text{ cm}} = ){\text{ 3.3 (cm)}}$;

Award [3] for a bald correct answer.

In (c) there were two popular methods used and many correct answers. Mistakes included using 0.65 instead of 0.35 and logarithmic errors.