| Date | May 2015 | Marks available | 5 | Reference code | 15M.3.HL.TZ2.20 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Estimate, Show that, and State | Question number | 20 | Adapted from | N/A |
Question
This question is about X-rays.
(i) X-rays travelling in a medium experience attenuation. State what is meant by attenuation.
(ii) Show that the half-value thickness \({x_{\frac{1}{2}}}\) is related to the attenuation coefficient \(\mu \) by
\[\mu {x_{\frac{1}{2}}} = 1{\rm{n}}2\]
(iii) Estimate the fraction of the incident intensity of an X-ray beam that has travelled through 2.0 cm of muscle. The half-value thickness of muscle is 0.73 cm.
Markscheme
(i) the absorption of energy/loss of power from the beam;
(ii) correct substitution \(\frac{{{I_0}}}{2} = {I_0}{e^{ - \mu {x_{\frac{1}{2}}}}}\);
taking natural logs \(1{\rm{n}}\frac{1}{2} = - \mu {x_{\frac{1}{2}}}\);
\(\left( {1{\rm{n}}2 = \mu {x_{\frac{1}{2}}}} \right)\)
Answer given, marks are for correct working.
(iii) \(\mu = \left( {\frac{{1{\rm{n}}2}}{{0.73}} = } \right)0.95{\rm{c}}{{\rm{m}}^{ - 1}}\);
\(I = \left( {{I_0}{e^{ - 0.95 \times 2.0}} = } \right)0.15{I_0}\);
or
number of half-value thicknesses \( = \frac{2}{{0.73}} = 2.74\);
\(I = {0.5^{2.74}} = 0.15{I_0}\);
Award [2] for a bald correct answer.
Examiners report
Part (a) contains two standard questions and was well answered. In comparing the processes of computed tomography (CT) and conventional X-ray imaging many candidates did well. Common problems included not mentioning the fact that CT images are taken at all angles during rotation and that CT involves a far greater absorbed dose.
