Oxford Revise AQA A Level Physics | Chapter P24 answers
P24: Medical imaging
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Question |
Answers |
Extra information |
Mark |
AO |
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01.1 |
Time = 0.7 s Pulse rate =\(\frac{60}{0.7}\)= 86 beats per minute |
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3.10.3.1 |
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01.2 |
Suitable scale added 0 to 1 mV Each small square is 0.2 mV |
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3.10.3.1 |
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01.3 |
Max 4 marks from: (should be in this order)
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A suitably labelled diagram can gain all marks |
max 4 |
3.10.3.1 |
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01.4 |
Max 2 marks from:
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1 |
3.10.3.1 |
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01.5 |
Hair and dead skin removed Use of conducting gel/no air gap |
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3.10.3.1 |
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02.1 |
Max 4 marks from:
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Must contain at least one point about received signal for full marks |
max 4 |
3.10.4.1 |
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02.2 |
Signal reflected/transmitted at boundary between two media Proportion reflected depends on the acoustic impedance of each medium |
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3.10.4.1 |
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02.3 |
Correctly identifying B and C T = 3 × 2.5 × 10–6 s D = s × t = 4080 × 3 × 2.5 × 10–6 = 0.031 m |
Allow 2 marks for correct working but wrong section identified |
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3.10.4.1 |
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02.4 |
\( \begin{array}{l} \textit{v} = f \lambda\\ \lambda = \frac{4080}{8 \times {10}^6} = 5.1 \times 10^{-4}\textrm{ m}\\ \end{array} \) |
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3.3.1.1 |
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02.5 |
The resolution is the ability to distinguish between objects close together (or wtte) The smaller the wavelength, the greater detail it is possible to see |
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3.10.4.1 |
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03.1 |
Max 2 marks from:
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max 2 |
3.10.5.1 |
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03.2 |
Filters out low-energy photons These cannot pass through tissue, so not needed for diagnosis/are absorbed by tissue increasing harm/risk |
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3.10.5.1 |
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03.3 |
Decrease in intensity, decrease increasing at 3 cm then more steadily at 4.5 cm Clear exponential drop shown for 1.5 cm of bone suitably labelled |
Values of intensity can be ignored – looking for contrast between bone and muscle |
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3.10.5.3 |
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03.4 |
Max 3 marks from:
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max 3 |
3.10.5.3 |
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04.1 |
\( \begin{array}{l} \sin C = \frac{{{n_2}}}{{{n_1}}}\\ \sin C = \frac{{1.43}}{{1.55}}\\ \textit{C} = 67^{\circ}\\ \end{array} \) |
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3.3.2.3 |
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04.2 |
This decreases the angle of incidence, making it more likely to be less than the critical angle Increases probability |
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3.10.4.2 |
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04.3 |
Max 3 marks from:
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max 3 |
3.10.4.2 |
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04.4 |
Cutting tissue/cauterising/stopping bleed/removing tumour Comparatively non-invasive/less scaring/less risk to patient |
Allow any sensible suggestion here |
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3.10.4.2 |
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05.1 |
Max 4 marks from:
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max 4 |
3.10.4.3 |
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05.2 |
Points to consider: Advantages CT scanner:
Disadvantages CT scanner:
Advantages MR scanner:
Disadvantages MR scanner:
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6 marks will clearly explain the advantages and disadvantages of both scanners in a range of situations 4 marks will include at least one advantage and disadvantage for each or consider one scanner in detail 2 marks will address at least one of the bullet points 1 mark will have any sensible comment 0 marks has no relevant physics |
max 6 |
AO2 |
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06.1 |
Collimator: Only gamma rays/photons which travel along the axes of lead tubes are detected/ensures only photons originating from directly below that point reach the scintillator Scintillator: A gamma ray photon produces thousands/many photons of (visible) light Photomultiplier: An electrical pulse is produced from each photon of visible light/one electron released, accelerated and then releases 4 more, etc., until electric pulse achieved |
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AO1 |
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06.2 |
Annihilation when electron meets positron Conservation of momentum means they have to travel in opposite directions |
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3.1.2.3 |
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06.3 |
\( \begin{array}{l} E = mc^{2}\\ E = 2 \times 9.11 \times 10^{-31} \times (3 \times 10^{8})^{2}\\ \textrm{Total }E = 1.6398 \times 10^{-13}\textrm{ J}\\ \end{array} \) \( \begin{array}{lll} E \textrm{ of 1 photon } & = & \frac{8.199 \times 10^{-14} \textrm{ J}}{1.6 \times 10^{-19}}\\ E \textrm{ of 1 photon } & = & 0.51\textrm{ MeV}\\ \end{array} \) |
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3.1.2.3 |
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06.4 |
Gamma photons are detected at different times at opposite points Computer calculates point of origin using distance = speed of light × time difference |
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3.10.6.1 |
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07.1 |
The time taken for half the initial sample of material to be excreted from the body by biological means |
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3.10.6.2 |
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07.2 |
\( \begin{array}{l} \frac{1}{T_{\rm{e}}} = \frac{1}{T_{\rm{p}}} + \frac{1}{T_{\rm{b}}}\\ \frac{1}{T_{\rm{e}}} = \frac{1}{21700}+\frac{1}{86400}\\ {T_{\rm{e}}} = 17\ 000\textrm{ s}\\ {T_{\rm{e}}} = 4.8\textrm{ hours}\\ \end{array} \) |
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3.10.6.2 |
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07.3 |
Allow time between 1 and 5 hours By 5 hours, activity halved so needs to be within that time But needs time to be taken up by tissues so greater than 1 hour |
Allow any sensible suggestion |
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3.10.6.2 |
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07.4 |
Longer half-life so that activity is higher for longer period (or wtte) Beta emitter so ionises local tissue/tumour only |
Must have description and explanation for mark |
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3.10.6.5 |
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08.1 |
Thermionic emission |
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3.10.5.1 |
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08.2 |
To prevent collisions of electrons with molecules To stop the target overheating/melting |
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3.10.5.1 |
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08.3 |
\( \begin{array}{l} \textrm{Use of } E = eV,\ E = \frac{hc}{\lambda }\textrm{ or } eV = \frac{hc}{\lambda }\\ \lambda = \frac{hc}{ev} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 180\ 000}\\ \lambda = 6.9 \times 10^{-12}\textrm{ m}\\ \end{array} \) |
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3.2.1.3 |
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08.4 |
\( \begin{array}{l} \textrm{Use of } Q = It \textrm{ or number of photons needed } = \frac{1.2}{eV} = 4.166 \times 10^{13}\\ \textrm{Number of electrons per second } = \frac{25 \times 10^{-3}}{1.6 \times 10^{-19}}\\ \textrm{1% photons produced } = 0.01 \times \frac{25 \times 10^{-3}}{1.6 \times 10^{-19}}\\ \textrm{Only 4% produced used}\\ \textrm{photons per second } = 0.04 \times 0.01\ \times \frac{25 \times 10^{-8}}{1.6 \times 10^{-19}} = 6.25 \times 10^{13}\textrm{ per second}\\ \rm{Time } =\frac{\textrm{number of photons}}{\textrm{photons per second}} = \frac{4.1666 \times 10^{18}}{6.25 \times 10^{15}} = 0.67\textrm{ s}\\ \end{array} \) |
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3.5.1.1 |
Skills box answers
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Question |
Answer |
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1 |
Oxygen atoms are found in many compounds used by the body, so radioactive 15O incorporated into water or glucose molecules will be carried by the blood and travel through the body to the brain. |
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2(a) |
\(^{15}{\rm{O}} \to \,^{15}\rm{N} + \rm{e}^+ \ + \rm{v}\left( \rm{position} + \rm{neutrino} \right)\) |
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2(b) |
\(\frac{\ln 2}{2.03 \textrm{ min}} = 0.34 \textrm{ min}^{-1} \textrm{ or} \frac{\ln 2}{(2.03 \times 60\textrm{ s})} = 5.7 \times 10^{-3} \textrm{s}^{-1}\) |
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3 |
Activity after 3 days:\(A = A_0 \textrm{ e}^{-\lambda t} = 14.8 \textrm{ MBq} \times \textrm{ e}^{( -0.053\textrm{h} – 1 \times (24 \times 3)\textrm{h})} = 0.33\textrm{ MBq}\). Activity after 11 days:\(A = 14.8 \textrm{ MBq} \times \rm{ e}^{(0.053\textrm{ h }-1 \times (24 \times 11 ) \rm{h} )} = 12 \textrm{Bq}\). Two days after the test, the patient will still be emitting gamma rays with an activity of 0.33MBq. If they were to sleep next to someone else for about 7 h this would increase the radiation dose of that person too, so they should keep at a distance for 11 days until the 123I has decayed to an activity of tens of becquerels. However, in public they would not be so close to another person for such long periods so it is not a risk to other people to go out while the activity is still slightly raised. |