Show that the attenuation coefficient for bone of density 1800 kg m^–3, for X-rays of 20 keV, is about 7 cm^–1.

The density of muscle is 1200 kg m^–3. Calculate the ratio of intensities to compare, for a beam of 20 keV, the attenuation produced by 1 cm of bone and 1 cm of muscle.

Suggest why more energetic beams of about 150 keV would be unsuitable for imaging a bone–muscle section of a body.

reads value on graph at 20 keV as 4 «cm² g^–1»

«4 cm² g^–1 × 1800 kg m^–3 × $\frac{1000}{1000000}$ = » 7.2 «cm^–1»

Ensure that the calculation has right POT conversion.

Answer must be to at least two significant figures.

ALTERNATIVE 1
(finds intensity ratios for muscle and bone separately)
Watch for ECF

for muscle: obtains μ = 0.96 cm⁻¹
Allow answers in the range of 0.90 to 1.02 cm^–1.

$\frac{I}{I_0}$ = e^−μx so for muscle 0.38

Allow answers in the range of 0.36 to 0.41.

Allow answers in dB. Muscle -4dB, Bone -30 or -31dB

for bone: $\frac{I}{I_0}$ = 7.5 × 10⁻⁴ «if μ = 7.2 is used»

OR

9.1×10⁻⁴ «if μ=7 is used»

ALTERNATIVE 2
for muscle: obtains μ = 0.96 cm⁻¹
Allow answers in the range of 0.90 to 1.02 cm^–1.

$\frac{I_{\mathrm{M}\mathrm{U}\mathrm{S}\mathrm{C}\mathrm{L}\mathrm{E}}}{I_{\mathrm{B}\mathrm{O}\mathrm{N}\mathrm{E}}}=\frac{e^{-0.96}}{e^{-7.2}}$

Frequently the POT will be incorrect for MP1. Allow ECF from MP1 to MP2.
Allow +/- 26 or 27dB
Award [2 max] if μ=960 as they will get $\frac{I_{\mathrm{M}\mathrm{U}\mathrm{S}\mathrm{C}\mathrm{L}\mathrm{E}}}{I_{\mathrm{B}\mathrm{O}\mathrm{N}\mathrm{E}}}$ = 0.

ratio is about 500 «513»
Allow range 395 to 546
If 7 used, ratio is about 420, if 7.2 is used, ratio is about 510
Allow answer I_BONE/I_MUSCLE from a range 0.0017 to 0.0026.

similar absorption so poor contrast

State a typical frequency used in medical ultrasound imaging.

Describe how an ultrasound transducer produces ultrasound.

Calculate the acoustic impedance Z of muscle.

Ultrasound of intensity 0.012 W$$cm^–2 is incident on a water–muscle boundary. The acoustic impedance of water is 1.50 x 10⁶ kg$$m^–2$$s^–1.

The fraction of the incident intensity that is reflected is given by

$\frac{{\left(Z_2-Z_1\right)}^2}{{\left(Z_2+Z_1\right)}^2}$

where Z₁ and Z₂ are the acoustic impedances of medium 1 and medium 2.

Calculate the intensity of the reflected signal.

accept any value between 1 MHz to 20 MHz

[1 mark]

an alternating electrical signal is applied to a crystal

crystal vibrates emitting sound

frequency of vibration of crystal is the same as the frequency of the ac

mention of piezoelectric effect/crystal

[3 marks]

Z_muscle = 1.71 x 10⁶ «kgm^–2$$s^–1»

[1 mark]

«$\frac{I_2}{I_1}=\frac{{\left(Z_2-Z_1\right)}^2}{{\left(Z_2+Z_1\right)}^2}$» = 4.3 x 10^–3

 I₂ = «0.012 x (4.3 x 10^–3) =» 5.1 x 10^–5 «W$$cm^–2»

Allow ECF from (c)(i).

Allow ECF from MP1 to MP2.

[2 marks]

Outline why the fracture in a broken bone can be seen in a medical X-ray image.

The diagram shows X-rays incident on tissue and bone.

The thicknesses of bone and tissue are both 0.054 m.

The intensity of X-rays transmitted through bone is I_b and the intensity transmitted through tissue is I_t.

The following data are available.

Mass absorption coefficient for bone = mass absorption
coefficient for tissue = 1.2 × 10^–2$$m²$$kg^–1
Density of bone = 1.9 × 103 kg$$m^–3
Density of tissue = 1.1 × 103 kg$$m^–3

Calculate the ratio $\frac{I_{\text{b}}}{I_{\text{t}}}$.

the large uniform magnetic field applied to the patient.

the radio-frequency signal emitted towards the patient.

the non-uniform magnetic field applied to the patient.

bone and tissue absorb different amounts of X-rays
OR
bone and tissue have different attenuation coefficients

so boundaries and fractures are delineated in an image

[2 marks]

$\frac{I_{\text{bone}}}{I_{\text{tissue}}}=\frac{I_0{\text{e}}^{-{\mu }_{\text{b}}x}}{I_0{\text{e}}^{-{\mu }_{\text{t}}x}}={\text{e}}^{-\left({\mu }_{\text{b}}-{\mu }_{\text{t}}\right)x}$

$\frac{I_{\text{bone}}}{I_{\text{tissue}}}={\text{e}}^{-1.2\times {10}^{-2}\times \left(1.9-1.1\right)\times {10}^3\times 5.4\times {10}^{-2}}$

$\frac{I_{\text{bone}}}{I_{\text{tissue}}}=0.60$

[3 marks]

to split the energy level of protons in the body
OR
to cause protons in the body to align with the field / precess at Larmor frequency

[1 mark]

to force/excite protons that are in the spin up/parallel state

into a transition to the spin down/antiparallel state

[2 marks]

the emitted radio frequency signal has a frequency that depends on the magnetic field

with a gradient field different parts of the body have different frequencies and so can be identified

[2 marks]

Outline how ultrasound is generated for medical imaging.

Describe one advantage and one disadvantage of using high frequencies ultrasound over low frequencies ultra sound for medical imaging.

Suggest one reason why doctors use ultrasound rather than X-rays to monitor the development of a fetus. 

Calculate the density of skin.

Explain, with appropriate calculations, why a gel is used between the transducer and the skin.

crystal vibration /piezo-electric effect

caused by an alternating potential difference is applied across a crystal

[2 marks]

ADVANTAGES

the wavelength must be less than the size of the object being imaged to avoid diffraction effects

the frequency must be high to ensure several full wavelengths in the pulse

DISADVANTAGES

the depth of the organ being imaged must be considered (no more than 200 wavelengths)

attenuation increases at higher frequencies

[1] for advantages, [1] for disadvantages.

[2 marks]

X-rays are an ionizing radiation and so might cause harm to the developing fetus.

OR

there are no known harmful effects when using ultrasound

Ignore “moving images by ultrasound”.

[1 mark]

ρ = $\frac{1.99\times {10}^6}{1.73\times {10}^3}$ = 1.15 × 10³ «kgm^–3»

[1 mark]

F = $\frac{{(1.99\times {10}^6-4.3\times {10}^2)}^2}{{(1.99\times {10}^6+4.3\times {10}^2)}^2}$ = 1.0

F = $\frac{{(1.48\times {10}^6-1.99\times {10}^6)}^2}{{(1.48\times {10}^6+1.99\times {10}^6)}^2}$ = 0.02

almost 100% of the ultrasound will be reflected from the air-skin surface OR almost none is transmitted

whereas only 2% will be reflected from the gel-skin surface and so a much greater proportion is transmitted

Need to explain that more is transmitted through gel-skin surface for MP4.

[4 marks]

Outline the formation of a B scan in medical ultrasound imaging.

State what is meant by half-value thickness in X-ray imaging.

A monochromatic X-ray beam of energy 20 keV and intensity I₀ penetrates 5.00 cm of fat and then 4.00 cm of muscle.

Calculate, in terms of I₀, the final beam intensity that emerges from the muscle.

Compare the use of high and low energy X-rays for medical imaging.

many/array of transducers send ultrasound through body/object

B scan made from many A scans in different directions

the reflection from organ boundaries gives rise to position

the amplitude/size gives brightness to the B scan

2D/3D image formed «by computer»

[3 marks]

the thickness of tissue that reduces the intensity «of the X-rays» by a half

OR

$x_{\frac{1}{2}}=\frac{\ln 2}{\mu }$ where $x_{\frac{1}{2}}$ is the half value thickness and μ is attenuation coefficient

Symbols must be defined for mark to be awarded

[1 mark]

after fat layer, I_fat = I₀e^{–0.4499 × 5.00}

after muscle layer, I = I_fate^{–0.8490 × 4.00}

I = 0.003533 I₀ or 0.35%

[3 marks]

«high energies factors:»

less attenuation/more penetration

more damage to the body

«so» stronger signal leaves the body

OR

«so» used in «most» medical imaging techniques

«low energy factors:»

must be used with enhancement techniques

greater attenuation/less penetration

«so» more damage to the body «on surface layers»

OR

«so» unwanted in «most» medical imaging techniques

[3 marks]

Calculate the maximum angle β for light to travel through the fibre.

Refractive index of core       = 1.50
Refractive index of cladding = 1.48

Outline how the combination of core and cladding reduces the overall dispersion in the optic fibres.

realization that 𝜃 min is the critical angle

𝜃 = «sin^–1 $\frac{1.48}{1.5}$ =» 80.6 «°»

Accept 1.4 rad

β = «90 – 80.6 =» 9.4 «°»

Accept 0.16 rad

because the critical angle is nearly 90°

then only rays that are «almost» parallel to the fibre pass down it

so pulse broadening is reduced

OWTTE

Explain how attenuation causes the contrast between soft tissue and bone in the image.

X-ray images of other parts of the body require the contrast to be enhanced. State one technique used in X-ray medical imaging to enhance contrast.

bone «denser so» absorb rays «and appear white in the negative» ✓

larger attenuation for bone ✓

muscles have less attenuation, so rays pass through «and appear darker» ✓

Accept the reversed argument

collimation ✓

fluorescent screens «each side of photographic plate» ✓

barium/magnesium meal ✓

Candidates successfully answered in terms of absorption and managed to score two marks but not all of them.

Almost all candidates were familiar with different techniques to enhance contrast. A barium meal was the most popular one.

Describe the effect of the RF signal on the protons in the body.

Outline the measurement that needs to be made after the RF signal is turned off.

Describe how the measurement in (b) provides diagnostic information for the doctor.

protons spin direction changes

OR

proton energy state changes ✔

Relaxation time «of signal/proton spin» ✔

Location/time delay of the emitted RF signal ✔

Relaxation time gives information on tissue type/density/health/OWTTE✔

Location information provides 3D image/OWTTE✔

Outline how ultrasound, in a medical context, is produced.

Suggest the advantage in medical diagnosis of using ultrasound of frequency 1 MHz rather than 0.1 MHz.

Ultrasound can be used to measure the dimensions of a blood vessel. Suggest why a B scan is preferable to an A scan for this application.

mention of AC voltage OR to piezo-electric crystal ✔

crystal vibrates «at its resonant frequency» ✔

1 MHz waves have shorter wavelength than 0.1 MHz ✔

can probe smaller size areas of organs/have higher resolution ✔

a B scan is a computer generated combination of a large number of A scans ✔

allowing a measurement in different directions/two dimensional image ✔

Production of ultrasound. This is not well known by many candidates. Quite a high number wrote generally correct information or statements about ultrasound, but these were not related or well considered responses directed to the question, for example, “use of gel in ultrasound medical applications”.

In (b) the higher resolution was mentioned by only the best candidates. Some candidates mentioned the notion of “smaller penetration at higher frequencies” – correct, though again not well related to what the question was asking.

Part c) was also difficult to answer for most of the candidates.

Show that the attenuation coefficient of lead is 60 cm^–1.

A technician operates an X-ray machine that takes 100 images each day. Estimate the width of the lead screen that is required so that the total exposure of the technician in 250 working days is equal to the exposure that the technician would receive from one X-ray exposure without the lead screen.

evidence of finding the gradient

μ = «– gradient =» 59.9 «cm^–1»

I = $\frac{I_0}{25000}$

«ln 25000 = μx» x = 0.17 «cm» or 1.7 «mm»

State one advantage and one disadvantage of using ultrasound imaging in medicine compared to using x-ray imaging.

Advantage: 

Disadvantage: 

Suggest why ultrasound gel is necessary during an ultrasound examination.

Ultrasound of intensity 50 mW m^-2 is incident on a muscle. The reflected intensity is 10 mW m^-2. Calculate the relative intensity level between the reflected and transmitted signals.

The acoustic impedance of soft tissue is 1.65 × 106 kg m^-2 s^-1. Show that the speed of sound in the soft tissue is approximately 1500 m s^–1.

Estimate, using data from the graph, the depth of the organ represented by the dashed line.

In the ultrasound scan the frequency is chosen so that the distance between the transducer and the organ is at least 200 ultrasound wavelengths. Estimate, based on your response to (b)(ii), the minimum ultrasound frequency that is used.

A physician has a range of frequencies available for ultrasound. Comment on the use of higher frequency sound waves in an ultrasound imaging study.

Advantage of ultrasound compared to X-rays:

no exposure to radiation
OR
relatively harmless
OR
can be performed in a doctor’s office
OR
can be used to measure blood flow rate
OR
Video image possible <<eg heart, foetus>> ✔

Accept any reasonable advantage.

Disadvantage:

limited resolution
OR
difficulty imaging lungs or gastrointestinal system
OR
difficulty imaging any body part with a gas in it ✔

Accept any reasonable disadvantage.

Do not allow answers that contradict each other.

gel has similar Z to skin

OR

gel prevents acoustic mismatch ✔

without gel much ultrasound is reflected at skin

OR

gel increases ultrasound transmission ✔

OWTTE

$c=«\frac{Z}{\rho }=\frac{1.65\times {10}^6}{1090\text{kg}{\text{m}}^{-3}}=»1514$«ms^–1»  ✔

«≈1500ms^–1»

Answer 1500 is given, check working or look for at least 3 significant figures.

4.5 × 10⁻² «m»✔

$\lambda =\frac{4.5\times {10}^{-2}}{200}=2.25\times {10}^{-4}$ «m» ✔

$f=\frac{v}{\lambda }=\frac{1500}{2.25\times {10}^{-4}}=6.7\times {10}^6$ «Hz»  ✔

«compared to lower frequencies, higher frequencies»

have better resolution ✔

have greater attenuation ✔

used for superficial structures/organs ✔

have greater heating effect ✔

OWTTE

Award [0] for contradictory comments or for any incorrect comment

The question was well answered by almost all candidates.

Most candidates mentioned that the gel improves the transmission of ultrasound. On quite a few occasions candidates seemed to confuse acoustic impedance and refractive index.

The question was generally well answered with a few candidates simply taking the ratio of intensities instead of 10x log ratio (Intensity level)

Almost all candidates managed to obtain the result given.

Many candidates did not seem to know how to start answering the question. The factor of two was often omitted when finding the depth of the organ in the A scan.

Few candidates managed to understand how to approach the problem and to obtain the correct answer. ECF from bii was frequently need.

Most candidates mentioned that the resolution would be better at higher frequencies.

Determine, in terms of I₀, the intensity of ultrasound that is incident on the muscle–bone boundary.

Determine, in terms of I₀, the intensity of ultrasound that is reflected at the muscle–bone boundary.

Determine, in terms of I₀, the intensity of ultrasound that returns to the muscle–gel boundary.

I₀e^{−23 × 0.041} ✔

= 0.39 I₀ ✔

R = «${\left(\frac{6.3\times {10}^6-1.7\times {10}^6}{6.3\times {10}^6+1.7\times {10}^6}\right)}^2$ =» 0.33 ✔

so reflected intensity is 0.33 × 0.39I₀ = 0.13I₀ ✔

0.13I₀ × 0.39 = 0.05I₀ ✔

Explain the cause of the radio-frequency emissions from a patient’s body during nuclear magnetic resonance (NMR) imaging.

Outline how a gradient field allows NMR to be used in medical resonance imaging.

Identify one advantage of NMR over ultrasound in medical situations.

use of strong magnetic field ✓

protons are aligned ✓

radio wave at «nuclear» resonant frequency flips «some of» them into higher energy state ✓

proton de-excites emitting energy at known «radio» wavelength/frequency/Larmor frequency ✓

«which can be located and detected»

mention of gradient field «added to the NMR uniform magnetic field» ✓

reference to «the total field that determines» the output «Larmor» frequency from the de-excitation ✓

different positions «in the body» give rise to different frequencies ✓

«and this can be mapped»

NMR higher resolution ✓

NMR less attenuation ✓

Accept the reverse argument

Candidates proved to be familiar with the use of strong magnetic fields to produce proton alignment and its consequence on excitation de-excitation of protons by emission of radio frequencies (Larmor frequency).

Most candidates continued successfully here by referring to a gradient field eventually leading to a mapping of the position of the protons from (a).

Higher resolution was the most popular answer in a high scoring question overall.

A parallel beam of X-rays travels through 7.8 cm of tissue to reach the bowel surface. Calculate the fraction of the original intensity of the X-rays that reach the bowel surface. The linear attenuation coefficient for tissue is 0.24 cm^–1.

The fluid in the bowel has a similar linear attenuation coefficient as the bowel surface. Gases have much lower linear attenuation coefficients than fluids. Explain why doctors will fill the bowel with air before taking an X-ray image.

I₀e^{−0.24 × 7.8} ✔

0.15I₀ ✔

Award [2] for bald correct answer.

to produce an X-ray image there must be constrast/a difference in the intensity of the beam transmitted through tissue and the bowel ✔

introduction of air will produce contrast ✔

Air as a contrast medium. Part (a) was well calculated by most of the prepared candidates.

In (b) only the best candidates well identified the importance of the change in contrast of the image (resulting from different attenuation values) needed to locate the organ.

Show that the ratio $\frac{I_{\mathrm{b}}}{I_{\mathrm{t}}}$ is close to 1.

State and explain, with reference to you answer in (a)(i), what needs to be done to produce a clear image of the leg artery using X-rays.

In nuclear magnetic resonance (NMR) protons inside a patient are made to emit a radio frequency electromagnetic radiation. Outline the mechanism by which this radiation is emitted by the protons.

${\mu }_{\mathrm{t}}=0.379\times 1.1\times {10}^3\times \frac{{10}^3}{{10}^6}=0.417 «{\mathrm{cm}}^{-1}»$  AND  ${\mu }_{\mathrm{b}}=0.408 «{\mathrm{cm}}^{-1}»$ ✔

$\frac{I_{\mathrm{b}}}{I_{\mathrm{t}}}=\frac{I_0{e}^{-{\mu }_{\mathrm{b}}x}}{I_0{e}^{-{\mu }_{\mathrm{t}}x}}=e^{-\left(0.408-0.417\right)\times 0.5}$ ✔

$\frac{I_{\mathrm{b}}}{I_{\mathrm{t}}}=1.004$ ✔

the difference between intensities is negligible so no contrast ✔

modifying the blood is easier than modifying the soft tissue ✔

increase absorption of X-rays in the blood ✔

by injecting/introducing a liquid/chemical/contrast medium ✔

with large mass absorption coefficient/nontoxic/higher density ✔

«a uniform» magnetic field is applied to align proton spins ✔

proton spins are excited by an «external» radio frequency signal/field
OR
protons change from spin-up to spin-down state due to «external» RF signal/field ✔

«radio frequency» radiation is emitted as the protons relax ✔

NOTE: For MP3 do not allow simplistic “protons emit RF radiation” as this is given in the question

State the property of protons used in nuclear magnetic resonance (NMR) imaging.

Explain how a gradient field and resonance are produced in NMR to allow for the formation of images at a specific plane.

«proton» spin

strong B field applied to align proton spins
OWTTE

cross-field applied to give gradient field
OR
each point in a plane has a unique B

RF field excites spins

protons emit RF at resonant/Larmor frequency dependent on Total B field  

RF detected shows position in the plane / is used to form 2D images

Allow features to be mentioned in any order

It is proposed to build an array of radio telescopes such that the maximum distance between them is 3800 km. The array will operate at a wavelength of 2.1 cm.

Comment on whether it is possible to build an optical telescope operating at 580 nm that is to have the same resolution as the array.

«use of $\frac{1.22\lambda }{d}$ to get» resolution of 6.7 × 10^–9 «rad»

$\frac{5.8\times {10}^{-7}}{6.7\times {10}^{-9}}$ = 87 «m»

some reference to difficulty in making optical mirrors/lenses of this size

Allow $\frac{5.8\times {10}^{-7}}{5.5\times {10}^{-9}}$ = 105 «m»

[3 marks]

In nuclear magnetic resonance (NMR) imaging radio frequency electromagnetic radiation is detected by the imaging sensors. Discuss the origin of this radiation.

«strong» magnetic field aligns proton «spins» 

an RF signal is applied to excite protons
OR
change spin up to spin down state

protons de-excite/return to lower energy state
OR
proton relaxation occurs

with emission of RF radiation «that is detected»

OWTTE

Treat any mention of the following as neutral as they are not strictly relevant to the question:
gradient field, Larmor frequency, precession, resonance, 3-D image

[3 marks]