MYP Physics · topic MYP-12.05
X-ray imaging: benefit and exposure
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
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.