CAIE A-Level Physics A2 24 Medical Physics Questions

Practise analysing diagnostic ultrasound, X-ray imaging, CT and PET using their physical principles, calculations, image formation and clinical trade-offs.

Exam points

explain and calculate ultrasound generation, reflection, impedance and attenuation in medical diagnosis
explain X-ray production, image contrast, attenuation and minimum-wavelength calculations
interpret CT image construction and compare its diagnostic trade-offs with single X-ray imaging
explain tracer-based PET from ?? decay through annihilation and opposing gamma detection
calculate annihilation photon energy and interpret detector data as tissue images

Covered syllabus points

24.1 Production and use of ultrasound 24.2 Production and use of X-rays 24.3 PET scanning

[Maximum number: 7]

Piezo-electric transducers are used for the generation of ultrasonic waves.

(a)

State one other use, apart from in ultrasound, of piezo-electric transducers.

[ 1 ]

(b)

Explain the main principles behind the use of ultrasound to obtain diagnostic information about internal body structures.

[ 6 ]

[Maximum number: 10]

(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.

[ 5 ]

(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.

[ 2 ]

$e V=\frac{h c}{\lambda}$ $\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{1.6 \times 10^{-19} \times 15000}$ $=8.3 \times 10^{-11} \mathrm{~m}$
or

E=h f and $c=f \lambda$ and electron energy =eV

or

$E=h c / \lambda$ and electron energy =eV
$\begin{aligned} \text { electron energy } & =1.6 \times 10^{-19} \times 15000 & =2.4 \times 10^{-15} \end{aligned}$ $\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{2.4 \times 10^{-15}}$ $\lambda=8.3 \times 10^{-11} \mathrm{~m}$

(b)

A beam of X-rays has an initial intensity $I_{0}$ . 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 $\ln \left(I / I_{\mathrm{o}}\right)$ .

Fig. 11.1

[ 5 ]

(i)

Determine the linear attenuation (absorption) coefficient $\mu$ for this beam of X-rays in the tissue.

\mu=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \ldots . . . . . . . . . . . . . . . . . . . . . ~ \mathrm{~cm}^{-1}

[ 3 ]

$\mu=-$ gradient or $\ln \left(I / I_{\mathrm{o}}\right)=-\mu x$
(e.g. 2.08 / 10.0) $=\mathbf{0 . 2 1} \mathrm{cm}^{-1}$

(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. cm

[ 2 ]

$\ln (0.05)=-\mu x$ $x=\frac{\ln 0.05}{-\mu}$

e.g. $x=14 \mathrm{~cm}$

[Maximum number: 5]

Fluorine-18 $\left({ }_{9}^{18} \mathrm{~F}\right)$ is a radioactive nuclide that is used as a tracer in positron emission tomography (PET scanning). Fluorine-18 decays to a nuclide of oxygen ( O ) according to

{ }_{9}^{18} \mathrm{~F} \longrightarrow{ }_{P}^{Q} \mathrm{X}+{ }_{8}^{R} \mathrm{O} \text {. }

(a)

State what is meant by a tracer.

[ 1 ]

(b)

Explain how the radioactive decay of fluorine-18 results in the emission from the body of the gamma-ray photons that are detected during a PET scan.

[ 2 ]

(c)

Explain how the detection of the gamma-ray photons is used to produce an image of the tissue being examined.

[ 2 ]

[Maximum number: 7]

Ultrasound is used to produce diagnostic information about internal body structures.

(a)

Explain how ultrasound waves are detected.

[ 3 ]

(b)

Table 9.1 contains information about air and soft tissue.

Table 9.1

[ 4 ]

(i)

Determine the unit for the specific acoustic impedance values shown in Table 9.1.

[ 1 ]

$\mathrm{kgm}^{-2} \mathrm{~s}^{-1}$

(ii)

Calculate the density of soft tissue.
density = $\mathrm{kg} \mathrm{m}^{-3}$

[ 1 ]

$\begin{aligned} \rho =Z / c=1.7 \times 10^{6} / 1600 =1100 \mathrm{~kg} \mathrm{~m}^{-3} \end{aligned}$

(iii)

Use data from Table 9.1 to explain why ultrasound cannot be used to produce an image inside an air-filled cavity such as the lungs.

[ 2 ]

intensity reflection coefficient $\approx 1$ or $Z_{1}$ and $Z_{2}$ are very different
almost no / no ultrasound transmitted (into air filled cavity)

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)