Cambridge International A-Level Physics · topic 24
Medical physics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A sharper X-ray image with a dose trade-off
An X-ray beam passes through a uniform absorber. Measured transmitted intensity fractions are supplied. Assume exponential attenuation I/I0 = exp(−μx), ignore scatter, and use h = 6.63×10⁻³⁴ J s, c = 3.0×10⁸ m s⁻¹ and electron charge e = 1.60×10⁻¹⁹ C. The tube voltage is 60 kV. Supplied model: For the shortest X-ray wavelength use eV = hc/λmin. Supplied data (Thickness / cm; I/I₀): (0, 1), (1, 0.5), (2, 0.25), (3, 0.125). (a) Determine attenuation coefficient, transmission through 4 cm and the shortest possible tube-emission wavelength. [3 marks] (b) Explain the image contrast and evaluate raising beam intensity to improve a weak image. [3 marks]
Measured X-ray attenuation
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): I/I₀
Horizontal axis (x): Thickness / cm
Read the exact data as a table
| Thickness / cm | I/I₀ |
|---|---|
| 0 | 1 |
| 1 | 0.5 |
| 2 | 0.25 |
| 3 | 0.125 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Determine attenuation coefficient, transmission through 4 cm and the shortest possible tube-emission wavelength.
Reveal the worked solution
- The half-value thickness is 0.01 m, so μ = ln2/0.01 ≈69.3 m⁻¹. Four such thicknesses transmit (1/2)⁴ = 1/16.
- The greatest photon energy is the full electron energy eV = (1.60×10⁻¹⁹)(60000) = 9.60×10⁻¹⁵ J.
- Using E = hc/λ gives λmin = hc/(eV) ≈2.07×10⁻¹¹ m. Most emitted photons need not have this maximum energy.
- Different tissue thicknesses and attenuation coefficients give different transmitted intensities at the detector, creating contrast.
- If detected photons are independently counted, relative random counting uncertainty scales approximately as 1/√N; increasing N can improve statistical image quality.
- X-rays are ionising, so increased photon fluence can increase patient dose. The exposure should be justified by the diagnostic task, with collimation, shielding and detector efficiency considered; ultrasound may answer some questions without ionising radiation.
Answer (a) μ≈69.3 m⁻¹; transmission at 4 cm = 1/16 = 6.25%; λmin≈2.07×10⁻¹¹ m. (b) Different attenuation gives contrast; more incident photons can reduce relative counting noise but also increases exposure.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
An ultrasound image with two hidden boundaries
A pulse-echo scanner detects reflected pulses at 40 μs and 70 μs after transmission into tissue. Sound speed is 1500 m s⁻¹. A 5.0 MHz pulse lasts four cycles. At one boundary acoustic impedances are Z1 = 1.5×10⁶ and Z2 = 6.0×10⁶ kg m⁻² s⁻¹. Use reflected intensity fraction R = [(Z2−Z1)/(Z2+Z1)]². Supplied data (Time after transmission / μs; Relative echo amplitude): (0, 1), (5, 0), (35, 0), (40, 0.6), (45, 0), (65, 0), (70, 0.3), (75, 0). (a) Find both boundary depths and their separation, assuming the same wave speed throughout the path. [3 marks] (b) Estimate axial pulse resolution and the reflection fraction. Explain the role of coupling gel and one access or performance trade-off. [3 marks]
Echo timing evidence
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Relative echo amplitude
Horizontal axis (x): Time after transmission / μs
Read the exact data as a table
| Time after transmission / μs | Relative echo amplitude |
|---|---|
| 0 | 1 |
| 5 | 0 |
| 35 | 0 |
| 40 | 0.6 |
| 45 | 0 |
| 65 | 0 |
| 70 | 0.3 |
| 75 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find both boundary depths and their separation, assuming the same wave speed throughout the path.
Reveal the worked solution
- Each pulse travels to a boundary and back, so depth is d = vt/2.
- At 40 μs, d = 1500(40×10⁻⁶)/2 = 0.030 m. At 70 μs, d = 0.0525 m.
- Their separation is 0.0525−0.030 = 0.0225 m. Different tissue speeds or refraction would make this uniform-speed interpretation approximate.
- Four cycles at 5.0 MHz last 4/(5.0×10⁶) = 0.80 μs. The spatial pulse length is 1500(0.80×10⁻⁶) = 1.2 mm; two interfaces need about half this separation, 0.60 mm, to resolve distinct echoes.
- R = [(6.0−1.5)/(6.0+1.5)]² = (0.60)² = 0.36. Amplitude ratios and intensity ratios must not be confused.
- Gel removes an air gap that otherwise strongly reflects sound. Ultrasound avoids ionising radiation and can be portable, but penetration, operator skill and acoustic access limit some examinations; higher frequency can improve resolution while increasing attenuation.
Answer (a) Depths = 30 mm and 52.5 mm; separation = 22.5 mm. (b) Pulse duration = 0.80 μs; axial resolution≈0.60 mm; reflected intensity fraction = 0.36.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Choosing a tracer with physical and biological clearance
A diagnostic tracer has physical half-life 6 h and biological half-life 12 h in the relevant organ. Assume the two removal processes are independent exponentials and initial organ activity is 80 MBq. The supplied graph gives the expected organ activity for the combined removal. Use 1/Teffective = 1/Tphysical + 1/Tbiological. Supplied data (Time / h; Activity / MBq): (0, 80), (4, 40), (8, 20), (12, 10). (a) Find effective half-life and expected organ activity after 12 h. Explain why biological removal does not change a nucleus's physical decay law. [3 marks] (b) Evaluate replacing this tracer by one with the same biological behaviour but physical half-life 30 years. Name the evidence needed for a patient-benefit decision. [3 marks]
Expected organ activity
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Activity / MBq
Horizontal axis (x): Time / h
Read the exact data as a table
| Time / h | Activity / MBq |
|---|---|
| 0 | 80 |
| 4 | 40 |
| 8 | 20 |
| 12 | 10 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find effective half-life and expected organ activity after 12 h. Explain why biological removal does not change a nucleus's physical decay law.
Reveal the worked solution
- 1/Teff = 1/6+1/12 = 1/4 h⁻¹, so Teff = 4 h.
- Twelve hours is three effective half-lives, giving 80/2³ = 10 MBq remaining in the organ.
- Biological clearance moves radionuclides out of the measured organ or body; it does not alter their intrinsic decay constant. Cleared material can still be radioactive elsewhere.
- A useful diagnostic tracer should remain measurable for preparation and imaging, then have limited unnecessary retention and exposure; 'shortest possible' is not automatically best if it decays before use.
- The emitted radiation must escape sufficiently for detection, while energy, branching ratio, chemistry and organ uptake determine image quality and dose.
- Compare expected diagnostic benefit, administered activity, absorbed and effective dose estimates, alternatives, logistics and patient-specific circumstances. Activity in MBq alone is not a complete measure of risk or image quality.
Answer (a) Effective half-life = 4 h; organ activity after 12 h = 10 MBq. (b) A very long physical half-life may leave unnecessary prolonged activity; suitability also depends on emissions, uptake and diagnostic performance.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.