Production and use of X-rays — Paper 4 · A Level Physics 9702

9 (a) Explain the principles behind the use of X-rays for imaging internal body structures. … … … … … … … [4] (b) Describe how the image produced during CT scanning differs from that produced by X-ray imaging. … … … … … … … [5]

9 marks

Mark scheme: 9 (a) X-ray beam directed through body onto detector (plate) B1 different tissues absorb/attenuate beam by different amounts B1 giving ‘shadow’ image of structures B1 any other detail e.g. comment re sharpness or contrast B1 [4] (b) X-ray image is flat OR 2-dimensional (1) CT scan takes many images of a slice at different angles (1) these build up an image of a slice through the body (1) series of images of slices is made (1) so that 3D image can be built up (1) image can then be rotated (1) 1 mark for each point, max 5 B5 [5]

This question in 9702/41 May/June 2007

11 (a) Distinguish between the images produced by CT scanning and X-ray imaging. For Examiner’s … Use … … … [3] (b) By reference to the principles of CT scanning, suggest why CT scanning could not be developed before powerful computers were available. … … … … … … … [5]

8 marks

Mark scheme: 11 (a) CT image: (thin) slice (through structure) B1 any further detail e.g. built up from many ‘slices’ / 3-D image B1 X-ray image: ‘shadow’ image (of whole structure) / 2-D image B1 [3] (b) X-ray image of slice taken from many different angles (1) these images are combined (and processed) (1) repeated for many different slices (1) to build up a 3-D image (1) 3-D image can be rotated (1) computer required to store and process huge quantity of data (1) (any five, 1 each to max 5) B5 [5]

This question in 9702/41 Oct/Nov 2008

10 (a) A typical spectrum of the X-ray radiation produced by electron bombardment of a metal For target is illustrated in Fig. 10.1. Examiner’s Use intensity 0 wavelength Fig. 10.1 Explain why (i) a continuous spectrum of wavelengths is produced, … … … … [3] (ii) the spectrum has a sharp cut-off at short wavelengths. … … [1] (b) The variation with photon energy E of the linear absorption coefficient of X-rays in soft For tissue is illustrated in Fig. 10.2. Examiner’s Use 300 µ / cm–1 200 100 0 10 30 50 70 E / keV Fig. 10.2 (i) Explain what is meant by linear absorption coefficient … … … … [3] (ii) For one particular application of X-ray imaging, electrons in the X-ray tube are accelerated through a potential difference of 50 kV. Use Fig. 10.2 to explain why it is advantageous to filter out low-energy photons from the X-ray beam. … … … … [3]

10 marks

Mark scheme: 10 (a) (i) e.m. radiation / photons is produced whenever a charged particle is accelerated … M1 wavelength depens on magnitude of acceleration … A1 electrons have a distribution of accelerations … A1 so continuous spectrum … A0 [3] (ii) either when electron loses all its energy in one collision or when energy of electron produces a single photon … B1 [1] (b) (i) parallel beam (in matter) … B1 I = I0 exp(-µx) … M1 I, I0, (µ) and x explained … A1 [3] (ii) either low-energy photons absorbed (much) more readily or low-energy photons (far) less penetrating … B1 low-energy photons do not contribute to X-ray image … B1 low energy photons could cause tissue damage … B1 [3] [Total: 10] GCE A/AS LEVEL – October/November 2009 9702 42

This question in 9702/42 Oct/Nov 2009

10 (a) Briefly explain the principles of CT scanning. For Examiner’s … Use … … … … … … … … [6] (b) A simple section through a body consists of four voxels, as illustrated in Fig. 10.1. For Examiner’s section Use directions of viewing Fig. 10.1 An X-ray image of the section is obtained by viewing along each of the directions shown in Fig. 10.1. The detector readings for each direction of viewing are summed to give the pattern of readings shown in Fig. 10.2. 25 22 34 31 Fig. 10.2 For any one direction, the total of the detector readings is 16. (i) For the pattern of readings of Fig. 10.2, state the magnitude of the background reading. background reading = … [1] (ii) On Fig. 10.1, mark the pattern of pixels for the four-voxel section. [2]

9 marks

Mark scheme: 10 (a) X-ray taken of slice / plane / section B1 repeated at different angles B1 images / data is processed B1 combined / added to give (2-D) image of slice B1 repeated for successive slices B1 to build up a 3-D image B1 image can be viewed from different angles / rotated B1 max 6 [6] (b) (i) 16 A1 [1] (ii) evidence of deducting 16 then dividing by 3 C1 to give A1 [2] 3 2 6 5

This question in 9702/41 May/June 2010

9 (a) (i) State, with reference to X-ray images, what is meant by sharpness. … … [1] (ii) Describe briefly two factors that affect the sharpness of an X-ray image. 1. … … 2. … … [3] (b) An X-ray image is taken of the skull of a patient. Another patient has a CT scan of his head. By reference to the formation of the image in each case, suggest why the exposure to radiation differs between the two imaging techniques. … … … … … [4]

8 marks

Mark scheme: 9 (a) (i) edges can be (clearly) distinguished B1 [1] (ii) e.g. size of X-ray source / anode / target / aperture scattering of X-ray beam pixel size (any two, 1 each) B2 further detail e.g. use of lead grid B1 [3] (b) X-ray image involves a single exposure B1 CT scan: exposure of a slice from many different angles M1 repeated for different slices A1 CT scan involves a (much) greater exposure B1 [4]

This question in 9702/43 Oct/Nov 2010

11 The linear attenuation (absorption) coefficient µ for X-ray radiation in bone, fat and muscle is For given in Fig. 11.1. Examiner’s Use µ / cm–1 bone 2.9 fat 0.90 muscle 0.95 Fig. 11.1 (a) A parallel X-ray beam of intensity I0 is incident either on some bone or on some muscle. The emergent beam has intensity I. I Calculate the ratio for a thickness of I0 (i) 1.5 cm of bone, ratio = … [2] (ii) 4.6 cm of muscle. ratio = … [1] (b) Suggest why, on an X-ray plate, the contrast between bone and muscle is much greater than that between fat and muscle. … … … … [3]

6 marks

Mark scheme: 11 (a) (i) I / I0 = exp(–1.5 × 2.9) C1 = 0.013 A1 [2] (ii) I / I0 = exp(–4.6 × 0.95) = 0.013 A1 [1] (b) attenuation (coefficients) in muscle and in fat are similar B1 attenuation (coefficients) in bone and muscle / fat are different B1 contrast depends on difference in attenuation B1 [3] GCE A LEVEL – October/November 2010 9702 43

This question in 9702/43 Oct/Nov 2010

11 (a) Distinguish between sharpness and contrast in X-ray imaging. For Examiner’s sharpness: … Use … contrast: … … [2] (b) A student investigates the absorption of X-ray radiation in a model arm. A cross-section of the model arm is shown in Fig. 11.1. 4.0 cm muscle bone B B M M 8.0 cm Fig. 11.1 Parallel X-ray beams are directed along the line MM and along the line BB. The linear absorption coefficients of the muscle and of the bone are 0.20 cm–1 and 12 cm–1 respectively. Calculate the ratio intensity of emergent X-ray beam from model intensity of incident X-ray beam on model for a parallel X-ray beam directed along the line (i) MM, ratio = … [2] (ii) BB. For Examiner’s Use ratio = … [3] (c) State whether your answers in (b) would indicate that the X-ray image (i) is sharp, … [1] (ii) has good contrast. … [1]

9 marks

Mark scheme: 11 (a) sharpness: ease with which edges of structures can be seen B1 contrast: difference in degree of blackening between structures B1 [2] (b) (i) I = I0 e–µx C1 I/I0 = exp(–0.20 × 8) = 0.20 A1 [2] (ii) I/I0 = exp(–µ1 × x1) × exp(–µ 2 × x2) (could be three terms) C1 I/I0 = exp(–0.20 × 4) × exp(–12 × 4) C1 I/I0 = 6.4 × 10–22 or I/I0 ≈ 0 A1 [3] (c) (i) sharpness unknown/no B1 [1] (ii) contrast good/yes (ecf from (b)) B1 [1] GCE AS/A LEVEL – October/November 2011 9702 43

This question in 9702/43 Oct/Nov 2011

11 High-speed electrons are incident on a metal target. The spectrum of the emitted X-ray For radiation is shown in Fig. 11.1. Examiner’s Use intensity 0 wavelength Fig. 11.1 (a) Explain why (i) there is a continuous distribution of wavelengths, … … … [2] (ii) there is a sharp cut-off at short wavelength. … … … [2] (b) State (i) what is meant by the hardness of an X-ray beam, … … … [2] (ii) how hardness is controlled. … … … [2] (c) (i) Suggest why, when producing an X-ray image, long-wavelength X-ray radiation For poses a greater hazard to health than short-wavelength radiation. Examiner’s Use … … [1] (ii) Suggest how this hazard is minimised. … … [1]

10 marks

Mark scheme: 11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1 electrons hitting target have distribution of accelerations A1 [2] (ii) either wavelength shorter/shortest for greater/greatest acceleration or λmin = hc/ Emax or minimum wavelength for maximum energy B1 all electron energy given up in one collision/converted to single photon B1 [2] (b) (i) hardness measures the penetration of the beam C1 greater hardness, greater penetration A1 [2] (ii) controlled by changing the anode voltage C1 higher anode voltage, greater penetration/hardness A1 [2] (c) (i) long-wavelength radiation more likely to be absorbed in the body/less likely to penetrate through body B1 [1] (ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1]

This question in 9702/41 May/June 2012

10 (a) An aluminium block is placed near to a small source of X-ray radiation, as shown in For Fig. 10.1. Examiner’s Use aluminium block X-ray A B source Fig. 10.1 X-rays from the source are detected at point A and at point B. State two reasons why the intensity of the X-ray beam at point B is not as great as the intensity at point A. 1. … … 2. … … [2] (b) A cross-section through a model of a finger is shown in Fig. 10.2. 2.4 cm 1.1 cm bone C D A B soft tissue Fig. 10.2 The thickness of the model is 2.4 cm and that of the bone in the model is 1.1 cm. Parallel beams of X-rays are incident on the model in the directions AB and CD, as shown in Fig. 10.2. Data for the linear attenuation (absorption) coefficient μ for the bone and the soft tissue For in the model are given in Fig. 10.3. Examiner’s Use μ / cm–1 bone 3.00 soft tissue 0.27 Fig. 10.3 Calculate the ratio intensity of X-ray beam incident on the model intensity of X-ray beam emergent from the model for (i) the beam AB, ratio = … [2] (ii) the beam CD. ratio = … [2] (c) Use your answers in (b) to suggest why, for this model, an X-ray image with good contrast may be obtained. … … [1]

7 marks

Mark scheme: 10 (a) e.g. beam is divergent/obeys inverse square law absorption (in block) scattering (of beam in block) reflection (at boundaries) (any two sensible suggestions, 1 each) B2 [2] (b) (i) I = I0 exp(–µx) C1 I0/I = exp(0.27 × 2.4) = 1.9 A1 [2] (ii) I0/I = exp(0.27 × 1.3) × exp(3.0 × 1.1) C1 = 1.42 × 27.1 = 38.5 A1 [2] (c) either much greater absorption in bone than in soft tissue or Io / I much greater for bone than soft tissue B1 [1]

This question in 9702/42 May/June 2012

11 High-speed electrons are incident on a metal target. The spectrum of the emitted X-ray For radiation is shown in Fig. 11.1. Examiner’s Use intensity 0 wavelength Fig. 11.1 (a) Explain why (i) there is a continuous distribution of wavelengths, … … … [2] (ii) there is a sharp cut-off at short wavelength. … … … [2] (b) State (i) what is meant by the hardness of an X-ray beam, … … … [2] (ii) how hardness is controlled. … … … [2] (c) (i) Suggest why, when producing an X-ray image, long-wavelength X-ray radiation For poses a greater hazard to health than short-wavelength radiation. Examiner’s Use … … [1] (ii) Suggest how this hazard is minimised. … … [1]

10 marks

Mark scheme: 11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1 electrons hitting target have distribution of accelerations A1 [2] (ii) either wavelength shorter/shortest for greater/greatest acceleration or λmin = hc/ Emax or minimum wavelength for maximum energy B1 all electron energy given up in one collision/converted to single photon B1 [2] (b) (i) hardness measures the penetration of the beam C1 greater hardness, greater penetration A1 [2] (ii) controlled by changing the anode voltage C1 higher anode voltage, greater penetration/hardness A1 [2] (c) (i) long-wavelength radiation more likely to be absorbed in the body/less likely to penetrate through body B1 [1] (ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1]

This question in 9702/43 May/June 2012

10 (a) Distinguish between sharpness and contrast in X-ray imaging. For Examiner’s sharpness: … Use … contrast: … … [2] (b) State two causes of loss of sharpness of an X-ray image. 1. … … 2. … … [2] (c) Data for the linear attenuation (absorption) coefficient μ of X-ray photons are given in Fig. 10.1. μ/ cm–1 bone 2.85 muscle 0.95 Fig. 10.1 A parallel beam of X-rays is incident, separately, on a thickness of 3.5 cm of bone and on a muscle of thickness 8.0 cm. (i) Calculate the ratio intensity of X-ray beam transmitted through bone . intensity of X-ray beam transmitted through muscle ratio = … [3] (ii) Use your answer in (i) to suggest whether an X-ray image of the bone and muscle For would show good or poor contrast. Examiner’s Use … … … [2]

9 marks

Mark scheme: 10 (a) sharpness: how well the edges (of structures) are defined B1 contrast: difference in (degree of) blackening between structures B1 [2] (b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid large penumbra on shadow/large area anode/wide beam large pixel size (any two sensible suggestions, 1 each) B2 [2] (c) (i) I = I0e–µx C1 ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1 = (4.65 × 10–5) / (5.00 × 10–4) = 0.093 A1 [3] (ii) either large difference (in intensities) or ratio much less than 1.0 M1 so good contrast A1 [2] (answer given in (c)(ii) must be consistent with ratio given in (c)(i)) GCE AS/A LEVEL – May/June 2013 9702 41

This question in 9702/41 May/June 2013

11 (a) Explain how the hardness of an X-ray beam is controlled by the accelerating voltage in For the X-ray tube. Examiner’s Use … … … [2] (b) The attenuation of a parallel beam of X-ray radiation is given by the expression I = e–μx I0 where μ is the linear attenuation (absorption) coefficient and x is the thickness of the material through which the beam passes. (i) State 1. what is meant by attenuation, … … [1] 2. why the expression applies only to a parallel beam. … … … [2] (ii) The linear attenuation coefficients for X-rays in bone and in soft tissue are 2.9 cm–1 and 0.95 cm–1 respectively. Calculate, for a parallel X-ray beam, the ratio I fraction of intensity transmitted through bone of thickness 2.5 cm I0 . I fraction of intensity transmitted through soft tissue of thickness 6.0 cm I0 ratio = … [2]

7 marks

Mark scheme: 11 (a) changing voltage changes energy / speed of electrons M1 changing electron energy changes maximum X-ray photon energy A1 [2] (b) (i) 1. loss of power / energy / intensity B1 [1] 2. intensity changes when beam not parallel C1 decreases when beam is divergent A1 [2] (ii) ratio = (exp {–2.9 × 2.5}) / (exp {–0.95 × 6.0}) C1 = 0.21 (min. 2 sig. fig.) A1 [2] (values of both lengths incorrect by factor of 10–2 to give ratio of 0.985 scores 1 mark) GCE AS/A LEVEL – May/June 2013 9702 42

This question in 9702/42 May/June 2013

10 (a) Distinguish between sharpness and contrast in X-ray imaging. For Examiner’s sharpness: … Use … contrast: … … [2] (b) State two causes of loss of sharpness of an X-ray image. 1. … … 2. … … [2] (c) Data for the linear attenuation (absorption) coefficient μ of X-ray photons are given in Fig. 10.1. μ/ cm–1 bone 2.85 muscle 0.95 Fig. 10.1 A parallel beam of X-rays is incident, separately, on a thickness of 3.5 cm of bone and on a muscle of thickness 8.0 cm. (i) Calculate the ratio intensity of X-ray beam transmitted through bone . intensity of X-ray beam transmitted through muscle ratio = … [3] (ii) Use your answer in (i) to suggest whether an X-ray image of the bone and muscle For would show good or poor contrast. Examiner’s Use … … … [2]

9 marks

Mark scheme: 10 (a) sharpness: how well the edges (of structures) are defined B1 contrast: difference in (degree of) blackening between structures B1 [2] (b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid large penumbra on shadow/large area anode/wide beam large pixel size (any two sensible suggestions, 1 each) B2 [2] (c) (i) I = I0e–µx C1 ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1 = (4.65 × 10–5) / (5.00 × 10–4) = 0.093 A1 [3] (ii) either large difference (in intensities) or ratio much less than 1.0 M1 so good contrast A1 [2] (answer given in (c)(ii) must be consistent with ratio given in (c)(i)) GCE AS/A LEVEL – May/June 2013 9702 43

This question in 9702/43 May/June 2013

11 (a) Distinguish between an X-ray image of a body structure and a CT scan. X-ray image: … … … … … CT scan: … … … … … [5] (b) Data for the linear absorption (attenuation) coefficient μ of X-ray radiation of energy 80 keV are given in Fig. 11.1. metal μ / mm−1 aluminium 0.46 copper 0.69 Fig. 11.1 A parallel X-ray beam is incident on a copper filter, as shown in Fig. 11.2. copper filter incident beam emergent intensity I0 beam Fig. 11.2 The intensity of the incident beam is I0. (i) Calculate the thickness of copper required to reduce the intensity of the emergent beam to 0.25 I0. thickness = … mm [2] (ii) An aluminium filter of thickness 2.4 mm is now placed in the X-ray beam, together with the copper filter in (i). Calculate the fraction of the incident intensity that emerges after passing through the two filters. fraction = … [2] (iii) Express your answer in (ii) as a gain in decibels (dB). gain = … dB [3]

12 marks

Mark scheme: 11 (a) X-ray: flat / shadow / 2D image B1 regardless of depth of object / depth not indicated B1 CT scan: built up from (many) images at different angles B1 image is three-dimensional B1 image can be rotated / viewed at different angles B1 [5] (b) (i) I R I0 e–µx C1 0.25 R e–0.69x x R2.0 mm (allow 1 s.f.) A1 [2] (ii) for aluminium, I / I0 R e–0.46 × 2.4 R 0.33 C1 fraction R 0.33 × 0.25 = = = R 0.083 A1 [2] (iii) gain / dB R 10 lg(I / I0) C1 R 10 lg(0.083) = = = R (–) 10.8 dB (allow 2 s.f.) A1 with negative sign B1 [3]

This question in 9702/42 May/June 2014

11 The use of X-rays in medical diagnosis gives rise to an increased exposure of the patient to radiation. Explain why (a) an aluminium filter may be placed in the X-ray beam when producing an X-ray image of a patient, … … … … … [3] (b) the radiation dose received by a patient is different for a CT scan from that for a simple X-ray image. … … … … … … [4]

7 marks

Mark scheme: 11 (a) X-ray beam contains many wavelengths B1 aluminium filter absorbs long wavelength X-ray radiation M1 that would be absorbed by the body (and not contribute to the image) A1 [3] (b) CT scan consists of (many) X-ray images of a slice M1 and there are many slices A1 X-ray image is a single exposure B1 (so much) greater exposure with CT scan B1 [4]

This question in 9702/41 Oct/Nov 2014

11 The use of X-rays in medical diagnosis gives rise to an increased exposure of the patient to radiation. Explain why (a) an aluminium filter may be placed in the X-ray beam when producing an X-ray image of a patient, … … … … … [3] (b) the radiation dose received by a patient is different for a CT scan from that for a simple X-ray image. … … … … … … [4]

7 marks

Mark scheme: 11 (a) X-ray beam contains many wavelengths B1 aluminium filter absorbs long wavelength X-ray radiation M1 that would be absorbed by the body (and not contribute to the image) A1 [3] (b) CT scan consists of (many) X-ray images of a slice M1 and there are many slices A1 X-ray image is a single exposure B1 (so much) greater exposure with CT scan B1 [4]

This question in 9702/42 Oct/Nov 2014

11 (a) State and explain how, in an X-ray tube, the hardness of the X-ray beam is controlled. … … … … [3] (b) A parallel beam of X-rays has intensity I0 and is incident on a medium having a linear absorption (attenuation) coefficient μ. (i) State an equation for the variation of the intensity I with the thickness x of the medium. … [1] (ii) Data for the linear absorption (attenuation) coefficient μ for an X-ray beam in blood and in muscle is shown in Fig. 11.1. μ/ cm−1 blood 0.23 muscle 0.22 Fig. 11.1 Suggest why, if this X-ray beam is used to image blood vessels in muscle, contrast on the image would be poor. … … … [2]

6 marks

Mark scheme: 11 (a) change/increase/decrease anode/tube voltage B1 electrons striking anode have changed (kinetic) energy/speed B1 X-ray/photons/beam have different wavelength/frequency B1 [3] (b) (i) I = I0 e–µx B1 [1] (ii) contrast is difference in degree of blackening (of regions of the image) B1 µ (very) similar so similar absorption of radiation (for same thickness) so little contrast A1 [2]

This question in 9702/41 May/June 2015

11 (a) State and explain how, in an X-ray tube, the hardness of the X-ray beam is controlled. … … … … [3] (b) A parallel beam of X-rays has intensity I0 and is incident on a medium having a linear absorption (attenuation) coefficient μ. (i) State an equation for the variation of the intensity I with the thickness x of the medium. … [1] (ii) Data for the linear absorption (attenuation) coefficient μ for an X-ray beam in blood and in muscle is shown in Fig. 11.1. μ/ cm−1 blood 0.23 muscle 0.22 Fig. 11.1 Suggest why, if this X-ray beam is used to image blood vessels in muscle, contrast on the image would be poor. … … … [2]

6 marks

Mark scheme: 11 (a) change/increase/decrease anode/tube voltage B1 electrons striking anode have changed (kinetic) energy/speed B1 X-ray/photons/beam have different wavelength/frequency B1 [3] (b) (i) I = I0 e–µx B1 [1] (ii) contrast is difference in degree of blackening (of regions of the image) B1 µ (very) similar so similar absorption of radiation (for same thickness) so little contrast A1 [2]

This question in 9702/43 May/June 2015

10 (a) Explain what is meant by the sharpness and by the contrast of an X-ray image. sharpness: … … contrast: … … [2] (b) A parallel X-ray beam of intensity I is incident on a medium of thickness x, as illustrated in Fig. 10.1. x incident transmitted intensity I intensity IT medium Fig. 10.1 The transmitted intensity is IT. Data for the linear absorption (attenuation) coefficient μ for 80 keV X-rays in bone and in muscle are given in Fig. 10.2. μ/ cm–1 bone 3.0 muscle 0.27 Fig. 10.2 (i) State, with reference to the production of X-rays, what is meant by 80 keV X-rays. … … … [2] (ii) Calculate the ratio IT / I for 80 keV X-rays passing through a thickness of 1.4 cm of bone. ratio = … [2] (c) An X-ray image of the upper leg of a student is produced. Part of the X-ray beam passes through a comparatively large thickness of muscle and part through some muscle and the leg bone. Use data from Fig. 10.2 to suggest whether the image has good contrast. … … … … … [3]

9 marks

Mark scheme: 10 (a) sharpness: clarity of edges/resolution (of image) B1 contrast: difference in degree of blackening (of structures) B1 [2] (b) (i) X-rays produced when (high speed) electrons hit target/anode B1 either electrons have been accelerated through 80 kV or electrons have (kinetic) energy of 80 keV B1 [2] (ii) IT / I = e–3.0 × 1.4 C1 = 0.015 A1 [2] (c) for good contrast, µx or eµx or e–µx must be very different B1 µx or eµx or e–µx for bone and muscle will be different than that for muscle M1 so good contrast A1 [3]

This question in 9702/41 Oct/Nov 2015

10 (a) Explain what is meant by the sharpness and by the contrast of an X-ray image. sharpness: … … contrast: … … [2] (b) A parallel X-ray beam of intensity I is incident on a medium of thickness x, as illustrated in Fig. 10.1. x incident transmitted intensity I intensity IT medium Fig. 10.1 The transmitted intensity is IT. Data for the linear absorption (attenuation) coefficient μ for 80 keV X-rays in bone and in muscle are given in Fig. 10.2. μ/ cm–1 bone 3.0 muscle 0.27 Fig. 10.2 (i) State, with reference to the production of X-rays, what is meant by 80 keV X-rays. … … … [2] (ii) Calculate the ratio IT / I for 80 keV X-rays passing through a thickness of 1.4 cm of bone. ratio = … [2] (c) An X-ray image of the upper leg of a student is produced. Part of the X-ray beam passes through a comparatively large thickness of muscle and part through some muscle and the leg bone. Use data from Fig. 10.2 to suggest whether the image has good contrast. … … … … … [3]

9 marks

Mark scheme: 10 (a) sharpness: clarity of edges/resolution (of image) B1 contrast: difference in degree of blackening (of structures) B1 [2] (b) (i) X-rays produced when (high speed) electrons hit target/anode B1 either electrons have been accelerated through 80 kV or electrons have (kinetic) energy of 80 keV B1 [2] (ii) IT / I = e–3.0 × 1.4 C1 = 0.015 A1 [2] (c) for good contrast, µx or eµx or e–µx must be very different B1 µx or eµx or e–µx for bone and muscle will be different than that for muscle M1 so good contrast A1 [3]

This question in 9702/42 Oct/Nov 2015

11 (a) An X-ray source is placed on one side of a metal plate, as shown in Fig. 11.1. metal plate X-ray source A B Fig. 11.1 The intensity of the X-ray beam is measured at points A and B. State two reasons, other than absorption of X-ray photons in the metal, for the intensity at point A to be different to that at point B. 1. … … 2. … … [2] (b) A specimen of muscle and bone undergoes X-ray examination. Parallel beams of X-rays are incident on the specimen, as shown in Fig. 11.2. X-ray beams 1.5 cm bone muscle 4.0 cm Fig. 11.2 The specimen has a total thickness of 4.0 cm. One section contains a bone of thickness 1.5 cm. Data for the linear absorption (attenuation) coefficient μ for the bone and for the muscle in the specimen are given in Fig. 11.3. μ/ cm−1 bone 3.0 muscle 0.27 Fig. 11.3 (i) Calculate the ratio intensity of X-ray beam incident on the specimen intensity of X-ray beam emerging from the specimen for the beam passing through 1. the 4.0 cm thickness of muscle alone, ratio = … [2] 2. the bone and the muscle. ratio = … [2] (ii) Using your answers in (i), suggest and explain whether an X-ray image of this specimen is likely to have good contrast. … … … [2]

8 marks

Mark scheme: 11 (a) e.g. scattering (in metal) non-parallel beam (not just “A closer than B”) reflection (from metal) diffraction in the metal/lattice any two B2 [2] (b) (i) 1. ratio = eµx = exp(0.27 × 4.0) C1 = 2.94 (2.9) A1 [2] 2. ratio = exp(0.27 × 2.5) × exp(3.0 × 1.5) C1 = 1.96 × 90 = 177 (180) A1 [2] (do not penalise unit error more than once) (ii) each ratio gives measure of transmission B1 ratios (in (i)) very different so good contrast B1 [2]

This question in 9702/43 Oct/Nov 2015

12 (a) In an X-ray tube, the hardness of an X-ray beam may be controlled. (i) State what is meant by the hardness of the beam. … … [1] (ii) State how the hardness of the beam may be decreased. … … [1] (b) State one advantage and one disadvantage of producing a CT scan image of a person rather than a standard X-ray image. advantage: … … disadvantage: … … [2] [Total: 4]

4 marks

Mark scheme: 12 (a) (i) the penetration of the beam B1 [1] (ii) either decrease the accelerating voltage or decrease voltage between cathode and anode B1 [1] (b) advantage: image gives depth / image is 3D / final image can be viewed from any angle B1 disadvantage: greater exposure / more risk to health / more expensive / person must remain stationary B1 [2]

This question in 9702/42 Feb/March 2016

10 (a) State (i) what is meant by the hardness of an X-ray beam, … … … [2] (ii) how the hardness of an X-ray beam from an X-ray tube is increased. … … [1] (b) The same parallel beam of X-ray radiation is incident, separately, on samples of bone and of muscle. Data for the thickness x of the samples of bone and of muscle, together with the linear attenuation (absorption) coefficients μ of the radiation in bone and in muscle, are given in Fig. 10.1. x / cm μ/ cm–1 bone 1.5 2.9 muscle 4.0 0.95 Fig. 10.1 Determine the ratio intensity transmitted through bone . intensity transmitted through muscle ratio = … [2] [Total: 5]

5 marks

Mark scheme: 10(a)(i) penetration of beam M1 greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1 10(a)(ii) greater accelerating potential difference or greater p.d. between anode and cathode B1 10(b) I = I0 exp(–µx) ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55}) C1 = 0.58 A1

This question in 9702/41 May/June 2017

10 (a) State (i) what is meant by the hardness of an X-ray beam, … … … [2] (ii) how the hardness of an X-ray beam from an X-ray tube is increased. … … [1] (b) The same parallel beam of X-ray radiation is incident, separately, on samples of bone and of muscle. Data for the thickness x of the samples of bone and of muscle, together with the linear attenuation (absorption) coefficients μ of the radiation in bone and in muscle, are given in Fig. 10.1. x / cm μ/ cm–1 bone 1.5 2.9 muscle 4.0 0.95 Fig. 10.1 Determine the ratio intensity transmitted through bone . intensity transmitted through muscle ratio = … [2] [Total: 5]

5 marks

Mark scheme: 10(a)(i) penetration of beam M1 greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1 10(a)(ii) greater accelerating potential difference or greater p.d. between anode and cathode B1 10(b) I = I0 exp(–µx) ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55}) C1 = 0.58 A1

This question in 9702/43 May/June 2017

9 (a) In computed tomography (CT scanning), it is necessary to take a series of many X-ray images. Outline briefly the principles of CT scanning. … … … … … … … … … … [4] (b) A student creates a model for CT scanning. A section is divided into four voxels, with pixel numbers A, B, C and D, as shown in Fig. 9.1. A B V1 D C V2 V4 V3 Fig. 9.1 The section is viewed from four different directions V1, V2, V3 and V4, as shown in Fig. 9.1. The detector readings for each direction are noted and then summed. The result is shown in Fig. 9.2. 47 59 44 32 Fig. 9.2 The background count is 26. Determine the pixel numbers A, B, C and D as shown in Fig. 9.1. A … B … D … C … [3] [Total: 7]

7 marks

Mark scheme: 9(a) image of one slice/section (B1) images (of one slice) taken from different angles (M1) to give 2D image (of one slice) (A1) (repeated for) many slices (M1) to build up 3D image (of whole body/structure) (A1) Max. 4 marks total 4 9(b) evidence of subtraction of background (–26) C1 evidence of division by three C1 7 11 6 2 A1

This question in 9702/41 Oct/Nov 2017

9 (a) In computed tomography (CT scanning), it is necessary to take a series of many X-ray images. Outline briefly the principles of CT scanning. … … … … … … … … … … [4] (b) A student creates a model for CT scanning. A section is divided into four voxels, with pixel numbers A, B, C and D, as shown in Fig. 9.1. A B V1 D C V2 V4 V3 Fig. 9.1 The section is viewed from four different directions V1, V2, V3 and V4, as shown in Fig. 9.1. The detector readings for each direction are noted and then summed. The result is shown in Fig. 9.2. 47 59 44 32 Fig. 9.2 The background count is 26. Determine the pixel numbers A, B, C and D as shown in Fig. 9.1. A … B … D … C … [3] [Total: 7]

7 marks

Mark scheme: 9(a) image of one slice/section (B1) images (of one slice) taken from different angles (M1) to give 2D image (of one slice) (A1) (repeated for) many slices (M1) to build up 3D image (of whole body/structure) (A1) Max. 4 marks total 4 9(b) evidence of subtraction of background (–26) C1 evidence of division by three C1 7 11 6 2 A1

This question in 9702/43 Oct/Nov 2017

12 (a) Suggest two causes of lack of sharpness of an X-ray image. 1. … … 2. … … [2] (b) The thickness of a sheet of metal is examined using a parallel X-ray beam, as illustrated in Fig. 12.1. 3.2 mm parallel X-ray beam metal x sheet Fig. 12.1 (not to scale) Part of the beam passes normally through the metal of thickness 3.2 mm. Another part of the beam passes normally through the metal of thickness x mm. The linear attenuation (absorption) coefficient for the X-ray beam in the metal is 1.5 cm–1. The ratio intensity of X-ray beam transmitted through 3.2 mm of metal intensity of X-ray beam transmitted through x mm of metal is found to be 0.81. (i) Calculate the thickness x. x = … mm [2] (ii) The ratio of the intensities is also the ratio of the powers of the X-ray beams. Calculate this ratio in decibels. ratio = … dB [2] [Total: 6]

6 marks

Mark scheme: 12(a) Any 2 from: scattering of X-ray beam / no lead grid lack of collimation of beam / aperture large anode area large beam p.d. low / photon energy low / X-ray soft B2 12(b)(i) 0.81 = (e–1.5 × 0.32) / (e–1.5 × x) C1 x = 1.8 mm A1 Question Answer Marks 12(b)(ii) ratio/dB = 10 lg(0.81) C1 = (–) 0.92 dB A1

This question in 9702/42 Feb/March 2018

12 An X-ray beam is used to produce an image of a model of a thumb. A parallel beam of X-ray radiation of intensity I0 is incident on the model, as illustrated in Fig. 12.1. soft tissue bone incident beam 1.3 cm emergent beam intensity I0 2.8 cm Fig. 12.1 Data for the attenuation (absorption) coefficient μ in bone and in soft tissue are shown in Fig. 12.2. μ/ cm–1 bone 3.0 soft tissue 0.90 Fig. 12.2 (a) Calculate, in terms of the incident intensity I0 of the X-ray beam, the intensity of the beam after passing through (i) a thickness of 2.8 cm of soft tissue, intensity = … I0 [2] (ii) the bone and soft tissue, as shown in Fig. 12.1. intensity = … I0 [2] (b) (i) State what is meant by the contrast of an X-ray image. … … … [2] (ii) By reference to your answers in (a), suggest whether the X-ray image of the model has good contrast. … … … [1] [Total: 7]

7 marks

Mark scheme: 12(a)(i) C1 = I0 exp (–0.90 × 2.8) = 0.080 I0 A1 12(a)(ii) I = I0 exp [(–0.90 × 1.5) × (–3.0 × 1.3)] C1 = I0 (0.259 × 0.20) = 0.0052 I0 A1 12(b)(i) difference in degrees of blackening M1 between structures A1 12(b)(ii) large difference in intensities so good contrast B1

This question in 9702/41 May/June 2018

11 (a) Describe the basic principles of CT scanning (computed tomography). … … … … … … … … … … … … [5] (b) By reference to your answer in (a), suggest why (i) CT scanning was not possible before fast computers with large memories were available, … … … [1] (ii) the radiation dose for a CT scan is much larger than for an X-ray image of a leg bone. … … … [1] [Total: 7]

7 marks

Mark scheme: 11(a) X-ray image(s) taken of one slice M1 (many images) taken from different angles A1 (computer) produces 2D image of slice B1 (this is) repeated for (many) slices M1 to build up a 3D image (of structure) A1 11(b)(i) combining of images involves (very) large number of calculations B1 11(b)(ii) CT scan consists of (very) many (single X-ray) images B1

This question in 9702/42 May/June 2018

12 An X-ray beam is used to produce an image of a model of a thumb. A parallel beam of X-ray radiation of intensity I0 is incident on the model, as illustrated in Fig. 12.1. soft tissue bone incident beam 1.3 cm emergent beam intensity I0 2.8 cm Fig. 12.1 Data for the attenuation (absorption) coefficient μ in bone and in soft tissue are shown in Fig. 12.2. μ/ cm–1 bone 3.0 soft tissue 0.90 Fig. 12.2 (a) Calculate, in terms of the incident intensity I0 of the X-ray beam, the intensity of the beam after passing through (i) a thickness of 2.8 cm of soft tissue, intensity = … I0 [2] (ii) the bone and soft tissue, as shown in Fig. 12.1. intensity = … I0 [2] (b) (i) State what is meant by the contrast of an X-ray image. … … … [2] (ii) By reference to your answers in (a), suggest whether the X-ray image of the model has good contrast. … … … [1] [Total: 7]

7 marks

Mark scheme: 12(a)(i) C1 = I0 exp (–0.90 × 2.8) = 0.080 I0 A1 12(a)(ii) I = I0 exp [(–0.90 × 1.5) × (–3.0 × 1.3)] C1 = I0 (0.259 × 0.20) = 0.0052 I0 A1 12(b)(i) difference in degrees of blackening M1 between structures A1 12(b)(ii) large difference in intensities so good contrast B1

This question in 9702/43 May/June 2018

10 (a) The root-mean-square (r.m.s.) value of the voltage of a sinusoidal alternating supply is 9.9 V. The frequency of the supply is 50 Hz. Derive an expression for the variation with time t (in second) of the potential difference V (in volt) of the supply. V = … [2] (b) Explain the function of the non-uniform magnetic field superposed on the large constant magnetic field in diagnosis using magnetic resonance imaging (NMRI). … … … … … … … [3] (c) A parallel beam of X-rays of intensity I0 is incident normally on some soft tissue and bone, as illustrated in Fig. 10.1. 0.40 cm incident transmitted intensity I0 bone intensity I soft tissue 1.8 cm Fig. 10.1 The bone is 0.40 cm thick and the total thickness of the bone and the soft tissue is 1.8 cm. The intensity of the transmitted beam is I. Data for the linear attenuation (absorption) coefficient μ of bone and of soft tissue are given in Fig. 10.2. μ/ cm–1 bone 2.9 soft tissue 0.92 Fig. 10.2 Calculate, in dB, the ratio transmitted intensity I . incident intensity I0 ratio = … dB [4]

9 marks

Mark scheme: 10(a) and ω = 2πf = 2π × 50 (= 314 rad s–1) C1 V = 14 sin 314t A1 10(b) enables (resonating) nuclei to be located B1 resonant frequency depends on magnetic field strength B1 Any one from: • non-uniform field is (accurately) calibrated • (non-uniform) field may be varied to enable detection in different positions • unique (magnetic) field strength/frequency at each point B1 10(c) I = I0 exp(–µx) C1 I = I0 [exp(–µx)bone × exp(–µx)soft tissue] I = I0 [exp(–2.9 × 0.40) × exp(–0.92 × 1.4)] C1 I / I0 = 0.0865 C1 ratio / dB = 10 lg 0.0865 = –11 dB A1

This question in 9702/42 Oct/Nov 2018

9 Outline the principles of computed tomography (CT) scanning. … … … … … … … … … … … [5] [Total: 5]

5 marks

Mark scheme: 9 X-rays (are used) B1 (object is) scanned in sections / slices B1 either: scans taken at many angles / directions or images of each section / slice are 2-dimensional B1 scans of many sections / slices are combined B1 (to give) 3-dimensional image (of whole structure) B1

This question in 9702/42 Feb/March 2019

7 Describe the principles of computed tomography (CT) scanning. … … … … … … … … [5]

5 marks

Mark scheme: 7 X-rays are used B1 section (of object) is scanned B1 scans/images taken at many angles/directions or images of each section are 2-dimensional B1 images of (many) sections are combined B1 (to give) 3-dimensional image of (whole) structure B1

This question in 9702/42 Oct/Nov 2019

11 Electrons are accelerated through a potential difference of 100 kV. They are then incident on a metal target, they decelerate, and X-ray photons are emitted. (a) Calculate the maximum possible frequency of the emitted X-ray photons. frequency = … Hz [2] (b) Explain why an aluminium filter may be placed in the X-ray beam when producing an X-ray image of a patient. … … … … … … [3] (c) The linear attenuation (absorption) coefficients μ for X-rays in bone, blood and muscle are given in Table 11.1. Table 11.1 μ/ cm−1 bone 3.0 blood 0.23 muscle 0.22 (i) A beam of these X-rays is incident on a person. Calculate the percentage of the intensity of the X-ray beam that has been absorbed after passing through 0.80 cm of blood. percentage of intensity absorbed = … % [2] (ii) In an X-ray image, white regions show greater absorption of X-rays than dark regions. State and explain the difference between the X-ray image of bone compared to that of muscle. … … … [2] [Total: 9]

9 marks

Mark scheme: 11(a) eV = hf f = 1.60 × 10–19 × 100 000 / 6.63 × 10–34 = 2.41 × 1019 Hz A1 11(b) (aluminium filter) absorbs (most) low energy X-rays B1 Any 2 from • X-ray beam contains many wavelengths • so low energy X-rays are not absorbed in the body • low energy X-rays can can cause harm but do not contribute to the image B2 11(c)(i) I / Io = e– μx e–0.23 × 0.80 = 0.83 C1 17% absorbed A1 11(c)(ii) bone is seen as lighter / muscle is seen as darker B1 either bone has a higher µ value so absorbs more or muscle has a lower µ value so transmits more B1

This question in 9702/42 Feb/March 2020

10 (a) Outline briefly the principles of computed tomography (CT scanning). … … … … … … … … … … [5] (b) One section of a model designed to illustrate CT scanning is divided into four voxels. The pixel numbers K, L, M and N of the voxels are shown in Fig. 10.1. D3 D2 D4 K L D1 N M Fig. 10.1 The section is viewed, in turn, from four different directions D1, D2, D3 and D4, as shown in Fig. 10.1. The detector readings for each direction are noted and these are summed to give the values shown in Fig. 10.2. 42 45 51 30 Fig. 10.2 The background reading is 24. Determine the pixel numbers K, L, M and N shown in Fig. 10.1. K = … L = … M = … N = … [3] [Total: 8]

8 marks

Mark scheme: 10(a) X-rays are used B1 section (of object) is scanned B1 scans/images taken at many angles/directions or images of each section are 2-dimensional B1 (images of (many)) sections are combined B1 (to give) 3-dimensional image of (whole) structure B1 10(b) K = 6 L = 7 M = 2 N = 9 3 marks: all four correct 2 marks: three correct and one incorrect or all correct with two numbers transposed 1 mark: two correct and two incorrect B3

This question in 9702/41 Oct/Nov 2020

7 Electrons in a beam are travelling at high speed in a vacuum. The electrons are incident on a metal target, causing X-ray radiation to be emitted. The variation with wavelength λ of the intensity I of the emitted X-ray radiation is shown in Fig. 7.1. I 0 0 λ Fig. 7.1 Explain why: (a) there is a continuous distribution of wavelengths … … … … … [3] (b) at certain wavelengths, there are narrow peaks of increased intensity. … … … … … [3] [Total: 6]

6 marks

Mark scheme: 7(a) X-ray photon produced when electron is decelerated B1 larger acceleration results in larger photon energy B1 continuous range of accelerations so continuous spectrum of wavelengths/frequencies B1 7(b) electron in (inner shell of) target atom is excited (on collision) B1 electron de-excites causing emission of a photon B1 discrete energy levels so discrete photon wavelengths B1

This question in 9702/42 Oct/Nov 2020

10 (a) Outline briefly the principles of computed tomography (CT scanning). … … … … … … … … … … [5] (b) One section of a model designed to illustrate CT scanning is divided into four voxels. The pixel numbers K, L, M and N of the voxels are shown in Fig. 10.1. D3 D2 D4 K L D1 N M Fig. 10.1 The section is viewed, in turn, from four different directions D1, D2, D3 and D4, as shown in Fig. 10.1. The detector readings for each direction are noted and these are summed to give the values shown in Fig. 10.2. 42 45 51 30 Fig. 10.2 The background reading is 24. Determine the pixel numbers K, L, M and N shown in Fig. 10.1. K = … L = … M = … N = … [3] [Total: 8]

8 marks

Mark scheme: 10(a) X-rays are used B1 section (of object) is scanned B1 scans/images taken at many angles/directions or images of each section are 2-dimensional B1 (images of (many)) sections are combined B1 (to give) 3-dimensional image of (whole) structure B1 10(b) K = 6 L = 7 M = 2 N = 9 3 marks: all four correct 2 marks: three correct and one incorrect or all correct with two numbers transposed 1 mark: two correct and two incorrect B3

This question in 9702/43 Oct/Nov 2020

11 (a) Electrons are accelerated through a potential difference of 15 kV. The electrons collide with a metal target and a spectrum of X-rays is produced. (i) Explain why a continuous spectrum of energies of X-ray photons is produced. … … … … … [3] (ii) Calculate the wavelength of the highest energy X-ray photon produced. wavelength = … m [3] (b) A beam of X-rays has an initial intensity Io. The beam is directed into some body tissue. After passing through a thickness x of tissue the intensity is I. The graph in Fig. 11.1 shows the variation with x of ln (I/Io). x / cm 0 1 2 3 4 5 6 7 8 9 10 11 0 –0.2 –0.4 –0.6 In (I/Io) –0.8 –1.0 –1.2 –1.4 –1.6 –1.8 –2.0 –2.2 –2.4 Fig. 11.1 (i) Determine the linear attenuation (absorption) coefficient μ for this beam of X-rays in the tissue. μ = … cm–1 [2] (ii) Determine the thickness of tissue that the X-ray beam must pass through so that the intensity of the beam is reduced to 5.0% of its initial value. thickness = … cm [2] [Total: 10]

10 marks

Mark scheme: 11(a)(i) electrons decelerate (on hitting target) so X-ray photons produced B1 range of decelerations B1 photon energy depends on (magnitude of) deceleration B1 11(a)(ii) hc eV λ = C1 34 8 19 6.63 10 3.0 10 1.6 10 15000 λ − − × × × = × × C1 11 8.3 10 m − = × A1 or E = hf and c = fλ and electron energy = eV or E = hc / λ and electron energy = eV electron energy = 1.6 × 10–19 × 15000 = 2.4 × 10–15 (C1) 34 8 15 6.63 10 3.0 10 2.4 10 λ − − × × × = × (C1) 11 8.3 10 m λ − = × (A1) 11(b)(i) μ = – gradient or ln (I / Io) = x μ − C1 (e.g. 2.08 / 10.0) = 0.21 cm–1 A1 Question Answer Marks 11(b)(ii) ( ) ln 0.05 x μ = − C1 ln0.05 x μ = − . . 14 e g x cm = A1

This question in 9702/42 Feb/March 2021

11 (a) State how, in a modern X-ray tube, the intensity of the X-ray beam and its hardness are controlled. intensity: … … hardness: … … [2] (b) A model of a limb consists of soft tissue and bone, as illustrated in Fig. 11.1. 3.0 cm I0 IC incident transmitted intensity intensity I0 IS bone soft tissue 9.0 cm Fig. 11.1 The soft tissue has a thickness of 9.0 cm. The bone within the soft tissue has a thickness of 3.0 cm. Data for the linear attenuation (absorption) coefficient μ of X-rays in soft tissue and in bone are shown in Table 11.1. Table 11.1 μ / cm−1 soft tissue 0.92 bone 2.90 A parallel beam of X-rays of intensity I0 is incident normally on the model. Calculate, in terms of I0: (i) the transmitted intensity IS through soft tissue alone IS = … I0 [2] (ii) the transmitted intensity IC through soft tissue and bone. IC = … I0 [2] (c) By reference to your answers in (b), suggest, with a reason, whether good contrast on an X-ray image would be obtained. … … [1] [Total: 7]

7 marks

Mark scheme: 11(a) intensity: vary filament current/p.d. across filament B1 hardness: vary accelerating potential difference B1 11(b)(i) I = I0e –μx C1 IS = I0 exp(–0.92 × 9.0) = 2.5 × 10–4 I0 A1 11(b)(ii) IC = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I0 C1 = 6.7 × 10–7 I0 A1 11(c) conclusion consistent with values in (b)(i) and (b)(ii) e.g. IS ≫ IC so good contrast B1

This question in 9702/41 May/June 2021

11 (a) State the purpose of computed tomography (CT scanning). … … [1] (b) Outline the principles of CT scanning. … … … … … … … … … … [5] [Total: 6]

6 marks

Mark scheme: 11(a) to produce a 3-dimensional image of structure/body B1 11(b) X-rays (are used) B1 scanning in sections B1 scanning from many angles B1 image of each section is 2-dimensional B1 scanning repeated for many sections or images of many sections combined together B1

This question in 9702/42 May/June 2021

11 (a) State how, in a modern X-ray tube, the intensity of the X-ray beam and its hardness are controlled. intensity: … … hardness: … … [2] (b) A model of a limb consists of soft tissue and bone, as illustrated in Fig. 11.1. 3.0 cm I0 IC incident transmitted intensity intensity I0 IS bone soft tissue 9.0 cm Fig. 11.1 The soft tissue has a thickness of 9.0 cm. The bone within the soft tissue has a thickness of 3.0 cm. Data for the linear attenuation (absorption) coefficient μ of X-rays in soft tissue and in bone are shown in Table 11.1. Table 11.1 μ / cm−1 soft tissue 0.92 bone 2.90 A parallel beam of X-rays of intensity I0 is incident normally on the model. Calculate, in terms of I0: (i) the transmitted intensity IS through soft tissue alone IS = … I0 [2] (ii) the transmitted intensity IC through soft tissue and bone. IC = … I0 [2] (c) By reference to your answers in (b), suggest, with a reason, whether good contrast on an X-ray image would be obtained. … … [1] [Total: 7]

7 marks

Mark scheme: 11(a) intensity: vary filament current/p.d. across filament B1 hardness: vary accelerating potential difference B1 11(b)(i) I = I0e –μx C1 IS = I0 exp(–0.92 × 9.0) = 2.5 × 10–4 I0 A1 11(b)(ii) IC = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I0 C1 = 6.7 × 10–7 I0 A1 11(c) conclusion consistent with values in (b)(i) and (b)(ii) e.g. IS ≫ IC so good contrast B1

This question in 9702/43 May/June 2021

11 (a) State, for an X-ray image, what is meant by: (i) sharpness … … [1] (ii) contrast. … … [1] (b) A parallel X-ray beam passes through a thickness of 2.3 cm of soft body tissue. The intensity of the emerging beam is 12% of the intensity of the incident beam. Calculate the linear attenuation (absorption) coefficient μ of the soft body tissue. Give a unit with your answer. μ = … unit … [3] (c) In medical diagnosis, X-rays may be used to produce a single X-ray image or may be used in computed tomography (CT scanning). Suggest an advantage and a disadvantage of CT scanning compared with single X-ray imaging for diagnosis. advantage: … … disadvantage: … … [2] [Total: 7]

7 marks

Mark scheme: 11(a)(i) ease with which edges can be distinguished B1 11(a)(ii) difference in degrees of blackening B1 11(b) I = I0 exp (–μx) C1 0.12 = exp (–μ × 2.3) ln 0.12 = –2.3 × μ C1 μ = 0.92 cm–1 A1 11(c) advantage: produces 3-dimensional image B1 disadvantage: (much) greater exposure to radiation B1

This question in 9702/41 Oct/Nov 2021

11 (a) State, for an X-ray image, what is meant by: (i) sharpness … … [1] (ii) contrast. … … [1] (b) A parallel X-ray beam passes through a thickness of 2.3 cm of soft body tissue. The intensity of the emerging beam is 12% of the intensity of the incident beam. Calculate the linear attenuation (absorption) coefficient μ of the soft body tissue. Give a unit with your answer. μ = … unit … [3] (c) In medical diagnosis, X-rays may be used to produce a single X-ray image or may be used in computed tomography (CT scanning). Suggest an advantage and a disadvantage of CT scanning compared with single X-ray imaging for diagnosis. advantage: … … disadvantage: … … [2] [Total: 7]

7 marks

Mark scheme: 11(a)(i) ease with which edges can be distinguished B1 11(a)(ii) difference in degrees of blackening B1 11(b) I = I0 exp (–μx) C1 0.12 = exp (–μ × 2.3) ln 0.12 = –2.3 × μ C1 μ = 0.92 cm–1 A1 11(c) advantage: produces 3-dimensional image B1 disadvantage: (much) greater exposure to radiation B1

This question in 9702/43 Oct/Nov 2021

10 In an X-ray tube, electrons are accelerated through a potential difference of 75 kV. The electrons then strike a tungsten target of effective mass 15 g. The electron energy is converted into the energy of X-ray photons with an efficiency of 5.0%. The rest of the energy is converted into thermal energy. (a) The X-ray tube produces an image using a current of 0.40 A for a time of 20 ms. The specific heat capacity of tungsten is 130 J kg–1 K–1. Determine the temperature rise ΔT of the tungsten target. ΔT = … K [3] (b) The linear attenuation coefficient of the X-ray photons in muscle is 0.22 cm–1. Calculate the thickness t of muscle that will absorb 80% of the incident X-ray intensity. t = … cm [2] (c) Table 10.1 shows the linear attenuation coefficient μ for the X-ray photons in different tissues. Table 10.1 μ/ cm–1 bone 3.0 blood 0.23 muscle 0.22 Two X-ray images are taken, one of equal thicknesses of bone and muscle and another of equal thicknesses of blood and muscle. Explain why one of these images has good contrast, but the other does not. … … … … [2] [Total: 7]

7 marks

Mark scheme: 10(a) C1 energy = ItV 0.40 0.020 75 000 0.95 ( T =) 0.015 130 × × × Δ × C1 =290 K A1 10(b) t oe μ − = I I 0.22t 0.20 = e− C1 t = 7.3 cm A1 Question Answer Marks 10(c) either (linear) attenuation coefficients / μ very different for bone and muscle M1 (very) different amounts (of X-rays) absorbed so good contrast or (very) different intensities transmitted so good contrast A1 or (linear) attenuation coefficients / μ similar for blood and muscle (M1) similar amounts (of X-rays) absorbed so poor contrast or similar intensities transmitted so poor contrast (A1)

This question in 9702/42 Feb/March 2022

9 (a) (i) Explain how X-rays are produced for use in medical diagnosis. … … … … [3] (ii) State why X-ray images are taken of multiple sections of the body during computed tomography (CT) scanning. … … [1] (b) An X-ray image is taken of the structure shown in Fig. 9.1. 2.4 cm soft tissue bone Q incident detected X-rays X-rays P 5.6 cm Fig. 9.1 The linear attenuation coefficient of bone is 3.4 cm–1. The linear attenuation coefficient of soft tissue is 0.89 cm–1. The incident X-rays are parallel and have a uniform intensity I0 across the structure. Determine, in terms of I0, the intensity of the detected X-rays from: (i) point P detected intensity = … I0 [2] (ii) point Q. detected intensity = … I0 [2] (c) Explain, with reference to your answers in (b), whether the X-ray image of the structure in Fig. 9.1 has good contrast. … … … [1] [Total: 9]

9 marks

Mark scheme: 9(a)(i) electrons are accelerated (by an applied p.d.) B1 electrons hit target B1 X-rays produced when electrons decelerate B1 9(a)(ii) images of the multiple sections are combined to create a 3-D image B1 9(b)(i) I = I0 exp (– μx) C1 = I0 exp (– 0.89  5.6) = 0.0068 I0 A1 9(b)(ii) I = I0 exp (– 2.4  3.4)  exp (– 0.89  3.2) C1 = 1.7  10–5 I0 A1 9(c) comparison of intensities or values in (b) leading to conclusion consistent with these values B1

This question in 9702/41 May/June 2022

9 (a) (i) Explain how X-rays are produced for use in medical diagnosis. … … … … [3] (ii) State why X-ray images are taken of multiple sections of the body during computed tomography (CT) scanning. … … [1] (b) An X-ray image is taken of the structure shown in Fig. 9.1. 2.4 cm soft tissue bone Q incident detected X-rays X-rays P 5.6 cm Fig. 9.1 The linear attenuation coefficient of bone is 3.4 cm–1. The linear attenuation coefficient of soft tissue is 0.89 cm–1. The incident X-rays are parallel and have a uniform intensity I0 across the structure. Determine, in terms of I0, the intensity of the detected X-rays from: (i) point P detected intensity = … I0 [2] (ii) point Q. detected intensity = … I0 [2] (c) Explain, with reference to your answers in (b), whether the X-ray image of the structure in Fig. 9.1 has good contrast. … … … [1] [Total: 9]

9 marks

Mark scheme: 9(a)(i) electrons are accelerated (by an applied p.d.) B1 electrons hit target B1 X-rays produced when electrons decelerate B1 9(a)(ii) images of the multiple sections are combined to create a 3-D image B1 9(b)(i) I = I0 exp (– μx) C1 = I0 exp (– 0.89  5.6) = 0.0068 I0 A1 9(b)(ii) I = I0 exp (– 2.4  3.4)  exp (– 0.89  3.2) C1 = 1.7  10–5 I0 A1 9(c) comparison of intensities or values in (b) leading to conclusion consistent with these values B1

This question in 9702/43 May/June 2022

10 (a) X-rays for use in medical diagnosis are produced in an X-ray tube. In the X-ray tube, charged particles are accelerated towards a metal target by an applied potential difference (p.d.). (i) State the name of the charged particles that are accelerated by the applied p.d. … [1] (ii) Explain how X-rays are produced at the metal target. … … … … [2] (iii) Calculate the minimum wavelength of X-rays produced when the applied p.d. is 5.80 kV. wavelength = … m [3] (b) X-rays pass through a medium that has an attenuation coefficient of 1.4 cm–1. Calculate the percentage of the X-ray energy that is absorbed by a 2.8 cm thickness of this medium. percentage absorbed = … % [3] [Total: 9]

9 marks

Mark scheme: 10(a)(i) electrons B1 10(a)(ii) electrons are decelerated / stopped on impact with the target B1 (kinetic) energy lost by electrons emitted as (X-ray) photons B1 10(a)(iii) eV = hc /  C1  = (6.63  10–34  3.00  108) / (1.60  10–19  5800) C1 = 2.14  10–10 m A1 10(b) I = I0 exp (–x) C1 IT / I0 = exp (–(1.4  2.8)) = 0.020 C1 % absorbed = (1.000 – 0.0198)  100 = 98% A1

This question in 9702/42 May/June 2023

10 Ultrasound and X-rays are both types of wave that are used in medical diagnosis to form images of internal body structures. (a) Complete Table 10.1 to state, for each type of wave: ● the method of production of the wave ● whether the wave that is detected and used to form the image is the wave that has been absorbed, reflected or transmitted by the internal body structure. Table 10.1 ultrasound X-rays method of … … production … … detected wave (absorbed, reflected … … or transmitted) [4] (b) (i) For one type of wave passing through tissue, the wave has 72% of its initial intensity after it has passed through 6.2 cm of the tissue. Calculate the linear attenuation coefficient μ of the tissue for this wave. μ = … cm–1 [2] (ii) Another wave of the same type as in (b)(i) passes through 9.3 cm of the same tissue. Calculate the percentage of the initial intensity of the wave that is attenuated by the tissue. percentage attenuated = … % [2]

8 marks

Mark scheme: 10(a) ultrasound production: vibrating quartz crystal B1 X-ray production: electrons hitting metal target B1 ultrasound detected wave: reflected B1 X-ray detected wave: transmitted B1 10(b)(i) I = I0 exp (–x) C1 ln (0.72) = –6.2 A1 = 0.053 cm–1 10(b)(ii) I / I0 = exp (–9.3  0.053) C1 ( = 0.61) percentage attenuated = 100  (1.00 – 0.61) A1 = 39%

This question in 9702/41 Oct/Nov 2023

10 Ultrasound and X-rays are both types of wave that are used in medical diagnosis to form images of internal body structures. (a) Complete Table 10.1 to state, for each type of wave: ● the method of production of the wave ● whether the wave that is detected and used to form the image is the wave that has been absorbed, reflected or transmitted by the internal body structure. Table 10.1 ultrasound X-rays method of … … production … … detected wave (absorbed, reflected … … or transmitted) [4] (b) (i) For one type of wave passing through tissue, the wave has 72% of its initial intensity after it has passed through 6.2 cm of the tissue. Calculate the linear attenuation coefficient μ of the tissue for this wave. μ = … cm–1 [2] (ii) Another wave of the same type as in (b)(i) passes through 9.3 cm of the same tissue. Calculate the percentage of the initial intensity of the wave that is attenuated by the tissue. percentage attenuated = … % [2]

8 marks

Mark scheme: 10(a) ultrasound production: vibrating quartz crystal B1 X-ray production: electrons hitting metal target B1 ultrasound detected wave: reflected B1 X-ray detected wave: transmitted B1 10(b)(i) I = I0 exp (–x) C1 ln (0.72) = –6.2 A1 = 0.053 cm–1 10(b)(ii) I / I0 = exp (–9.3  0.053) C1 ( = 0.61) percentage attenuated = 100  (1.00 – 0.61) A1 = 39%

This question in 9702/43 Oct/Nov 2023

8 (a) State what is meant by a photon. … … … [2] (b) Fig. 8.1 shows a tube in which X-rays are produced at a metal target. X Y particles filament vacuum glass tube metal target Fig. 8.1 Particles are accelerated from the filament to the target by a constant high voltage applied across the terminals X and Y. (i) State the name of the particles. … [1] (ii) On Fig. 8.1, use + and – signs to label terminals X and Y to indicate the polarity of the high voltage. [1] (c) For an accelerating voltage of 32 kV in Fig. 8.1, determine: (i) the maximum energy, in MeV, of an X-ray photon produced at the target maximum photon energy = … MeV [1] (ii) the maximum momentum of an X-ray photon produced at the target maximum photon momentum = … N s [2] (iii) the minimum wavelength of X-rays produced at the target. minimum wavelength = … m [3] (d) Explain why X-rays can be used to produce images of internal body structures that have good contrast. … … … … … [3] [Total: 13]

13 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) electron(s) B1 8(b)(ii) X labelled – and Y labelled + B1 8(c)(i) 0.032 MeV A1 8(c)(ii) momentum = E / c C1 momentum = (0.032 × 1.60  10–13) / (3.00  108) = 1.7  10–23 N s A1 8(c)(iii) E = hf and  = c / f C1  = hc / E = (6.63  10–34 × 3.00  108) / (0.032  1.60 × 10–13) C1  = 3.9  10–11 m A1 8(d) discussion of bone and soft tissue B1 discussion of different attenuation (coefficients) or discussion differences in penetration / transmission / absorption B1 transmitted intensities (by bone and tissue) are very different (leading to good contrast images) B1

This question in 9702/41 May/June 2024

8 (a) State what is meant by a photon. … … … [2] (b) Fig. 8.1 shows a tube in which X-rays are produced at a metal target. X Y particles filament vacuum glass tube metal target Fig. 8.1 Particles are accelerated from the filament to the target by a constant high voltage applied across the terminals X and Y. (i) State the name of the particles. … [1] (ii) On Fig. 8.1, use + and – signs to label terminals X and Y to indicate the polarity of the high voltage. [1] (c) For an accelerating voltage of 32 kV in Fig. 8.1, determine: (i) the maximum energy, in MeV, of an X-ray photon produced at the target maximum photon energy = … MeV [1] (ii) the maximum momentum of an X-ray photon produced at the target maximum photon momentum = … N s [2] (iii) the minimum wavelength of X-rays produced at the target. minimum wavelength = … m [3] (d) Explain why X-rays can be used to produce images of internal body structures that have good contrast. … … … … … [3] [Total: 13]

13 marks

Mark scheme: 8(a) packet / quantum of energy M1 of electromagnetic radiation A1 8(b)(i) electron(s) B1 8(b)(ii) X labelled – and Y labelled + B1 8(c)(i) 0.032 MeV A1 8(c)(ii) momentum = E / c C1 momentum = (0.032 × 1.60  10–13) / (3.00  108) = 1.7  10–23 N s A1 8(c)(iii) E = hf and  = c / f C1  = hc / E = (6.63  10–34 × 3.00  108) / (0.032  1.60 × 10–13) C1  = 3.9  10–11 m A1 8(d) discussion of bone and soft tissue B1 discussion of different attenuation (coefficients) or discussion differences in penetration / transmission / absorption B1 transmitted intensities (by bone and tissue) are very different (leading to good contrast images) B1

This question in 9702/43 May/June 2024

6 Two parallel metal plates X and Y are separated by a distance of 0.041 m, as shown in Fig. 6.1. X Y electron vacuum 0.041 m Fig. 6.1 There is a vacuum between the plates. An electron is at rest at the centre of plate X. A potential difference (p.d.) of 58 kV is applied across the plates. This causes the electron to accelerate towards plate Y. (a) On Fig. 6.1, use the symbols + and – to indicate which of plates X and Y is the positive plate and which is the negative plate. [1] (b) (i) Calculate the electric field strength E between the plates. Give a unit with your answer. E = … unit … [2] (ii) Determine the acceleration of the electron. acceleration = … m s–2 [2] (c) Many electrons are now accelerated from rest from plate X to plate Y in Fig. 6.1. When the electrons hit plate Y, the absorption of their kinetic energies results in the emission of electromagnetic waves. (i) Show that the minimum wavelength of these electromagnetic waves is 21 pm. [3] (ii) State the region of the electromagnetic spectrum that contains these waves. … [1] (iii) Explain how these electromagnetic waves may be used to form images of internal body structures. … … … … [2] [Total: 11]

11 marks

Mark scheme: 6(a) plate X marked as negative and plate Y marked as positive B1 6(b)(i) E = V / x C1 = (58 × 103) / 0.041 A1 = 1.4 × 106 N C–1 6(b)(ii) ma = eE C1 a = (1.60 × 10–19 × 1.41 × 106) / (9.11 × 10–31) A1 = 2.5 × 1017 m s–2 6(c)(i) eV = hc /  C1 or eV = hf and f = c /  (1.60 × 10–19 × 58 × 103) = (6.63 × 10–34 × 3.00 × 108) /  M1 clear conversion from m to pm leading to  = 21 pm A1 6(c)(ii) X-rays B1 6(c)(iii) Any two points from: B2 • waves are passed into structure and transmitted waves detected • different parts of the structure absorb different fractions of energy • difference in detected / transmitted intensities used (to form image)

This question in 9702/42 May/June 2025

10 (a) State what is meant by contrast in an X-ray image. … … [1] (b) X-rays of intensity I0 are incident normally on a structure, as shown in Fig. 10.1. 2.1 cm material P material Q A incident X-rays, detected intensity I0 X-rays B 5.8 cm Fig. 10.1 Material P has a linear attenuation coefficient of 0.35 cm–1. The X-rays emerging from the structure in region A have an intensity of 0.053I0. (i) Show that the intensity of the X-rays emerging in region B is 0.13I0. [1] (ii) Determine the linear attenuation coefficient μ of material Q. μ = … cm–1 [3] (iii) Use the information in (b)(i) to suggest why the X-rays emerging from the structure form an image that has poor contrast. … … … [1] (c) Explain how X-rays are used in computed tomography (CT) scanning to produce a three-dimensional image of an internal structure. … … … … … [3] [Total: 9]

9 marks

Mark scheme: 10(a) difference in degrees of blackening B1 10(b)(i) I = I0 exp (–x) A1 = I0 exp (– 5.8  0.35) = 0.13 I0 10(b)(ii) use of exp {–(0.35  3.7)} factor C1 0.053I0 = I0 exp {–[(0.35  3.7) + 2.1]} C1 = 0.78 cm–1 A1 10(b)(iii) factor of only 2.5 between the (detected) intensities (so not good contrast) B1 10(c) (structure) scanned in (thin) sections B1 (many) scans (of each section) taken from different angles B1 scanning repeated for all sections and (data) compiled (to form 3D image) B1

This question in 9702/42 Oct/Nov 2025