Cambridge AS & A Level9702

Medical physics

Physics 9702 Chapter Notes

What this chapter covers

Medical physics - Production and use of ultrasoundMedical physics - Production and use of X-raysMedical physics - PET scanning
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1. Nature and Production of X-rays

X-rays are high-frequency, short-wavelength electromagnetic radiation, typically with wavelengths from 0.01 to 10 nanometres. They are produced in an X-ray tube when high-energy electrons are rapidly decelerated. In a standard X-ray tube, a heated cathode filament releases electrons through thermionic emission. These electrons are accelerated through a very high potential difference (voltage) towards a metal target, usually made of tungsten, which acts as the anode. Upon striking the target, the electrons decelerate suddenly, converting their kinetic energy into other forms. About 99% is converted into heat, while about 1% is converted into X-ray photons. This process of producing radiation by decelerating charged particles is called Bremsstrahlung, or 'braking radiation'. The maximum energy (and thus minimum wavelength) of the X-ray photons produced is determined by the accelerating voltage, as the maximum energy a photon can have is equal to the entire kinetic energy of one electron: E_max = eV.

E_k = eV

E_max = hf_max

c = fλ

λ_min = hc / eV

Key term

Bremsstrahlung: Electromagnetic radiation produced by the deceleration of a charged particle, such as an electron, when deflected by another charged particle, such as an atomic nucleus.

Examiner insight

Examiners look for a clear link between the kinetic energy gained by the electron (eV) and the maximum energy of the resulting photon (hf_max).

Common pitfall

Confusing the 'hardness' of X-rays (their energy/penetrating power) with their intensity (the number of photons). Hardness is controlled by the accelerating voltage, while intensity is primarily controlled by the filament current.

Worked example 14 marks

In an X-ray tube, electrons are accelerated from rest through a potential difference of 80 kV. They strike a tungsten target. Calculate the maximum energy of the X-ray photons produced and their minimum wavelength. (e = 1.60 x 10⁻¹⁹ C, h = 6.63 x 10⁻³⁴ Js, c = 3.00 x 10⁸ m/s)

  1. 1

    Step 1: Calculate the maximum kinetic energy of an electron. This is equal to the work done on it by the electric field, E_k = eV.

  2. 2

    E_k = (1.60 x 10⁻¹⁹ C) x (80 x 10³ V) = 1.28 x 10⁻¹⁴ J.

  3. 3

    Step 2: The maximum energy of a photon (E_max) is produced when an electron converts all its kinetic energy into a single photon. So, E_max = 1.28 x 10⁻¹⁴ J.

  4. 4

    Step 3: Use the photon energy equation E = hc/λ to find the minimum wavelength (λ_min).

  5. 5

    λ_min = hc / E_max

  6. 6

    λ_min = (6.63 x 10⁻³⁴ Js x 3.00 x 10⁸ m/s) / (1.28 x 10⁻¹⁴ J)

  7. 7

    λ_min = 1.55 x 10⁻¹¹ m (or 0.0155 nm).

Recap

  • X-rays are high-frequency, short-wavelength electromagnetic waves.
  • They are produced when fast-moving electrons are rapidly decelerated by a metal target.
  • The accelerating potential difference determines the maximum energy of the X-ray photons.
  • The minimum X-ray wavelength is inversely proportional to the accelerating voltage.
  • Most of the electron's kinetic energy is converted to heat, not X-rays.

Quick check

  1. What is the name of the process where electrons are released from a heated filament?1 mark
  2. If the accelerating voltage in an X-ray tube is doubled, what happens to the minimum wavelength of the X-rays produced?1 mark

2. X-ray Attenuation

As a beam of X-rays passes through a material, its intensity decreases. This reduction in intensity is called attenuation. Attenuation occurs due to two main processes: absorption (e.g., photoelectric effect) and scattering. For a parallel, monochromatic X-ray beam, the intensity decreases exponentially with the thickness of the material it passes through. This is described by the equation I = I₀e^(-μx), where I is the transmitted intensity, I₀ is the initial intensity, x is the thickness of the material, and μ is the linear attenuation coefficient. The attenuation coefficient μ is a property of the material and depends on the energy of the X-ray photons and the atomic number of the material. Denser materials and materials with higher atomic numbers (like bone or lead) have a much higher attenuation coefficient than less dense materials (like soft tissue), which is why they show up differently on an X-ray image.

I = I₀e^(-μx)

Key term

Attenuation Coefficient (μ): A measure of the extent to which the intensity of an X-ray beam is reduced as it passes through a specific material, with units of m⁻¹ or cm⁻¹.

Examiner insight

Marks are often awarded for correctly rearranging the exponential attenuation equation to find μ or x. This involves using natural logarithms, so be sure to show the ln() step clearly in your working.

Common pitfall

Forgetting to ensure that the units of thickness (x) and the attenuation coefficient (μ) are consistent before performing the calculation. For example, if μ is in cm⁻¹, x must be in cm.

Worked example 13 marks

An X-ray beam with an initial intensity of 8.0 W m⁻² passes through 2.5 cm of tissue. The linear attenuation coefficient μ for this tissue is 0.95 cm⁻¹. Calculate the intensity of the beam after it has passed through the tissue.

  1. 1

    Step 1: State the formula for exponential attenuation.

  2. 2

    I = I₀e^(-μx)

  3. 3

    Step 2: Identify the given values: I₀ = 8.0 W m⁻², μ = 0.95 cm⁻¹, x = 2.5 cm.

  4. 4

    Step 3: Substitute the values into the formula. Ensure units of μ and x are consistent (they are both in cm, which is correct).

  5. 5

    I = 8.0 x e^(-0.95 x 2.5)

  6. 6

    I = 8.0 x e^(-2.375)

  7. 7

    I = 8.0 x 0.0930

  8. 8

    I = 0.744 W m⁻².

Worked example 23 marks

The half-value thickness (HVT) of a material is the thickness required to reduce the intensity of the X-ray beam to half its initial value. Show that HVT = ln(2) / μ.

  1. 1

    Step 1: Start with the attenuation equation: I = I₀e^(-μx).

  2. 2

    Step 2: By definition of HVT, when x = HVT, the intensity I = I₀/2.

  3. 3

    Step 3: Substitute these into the equation: I₀/2 = I₀e^(-μ * HVT).

  4. 4

    Step 4: Cancel I₀ from both sides: 1/2 = e^(-μ * HVT).

  5. 5

    Step 5: Take the natural logarithm (ln) of both sides: ln(1/2) = ln(e^(-μ * HVT)).

  6. 6

    Step 6: Using log rules, ln(1/2) = -ln(2) and ln(e^y) = y. So, -ln(2) = -μ * HVT.

  7. 7

    Step 7: Rearrange to make HVT the subject: HVT = ln(2) / μ.

Recap

  • Attenuation is the reduction of X-ray intensity as it passes through matter.
  • Intensity decreases exponentially with thickness, following I = I₀e^(-μx).
  • The attenuation coefficient μ depends on the material and the X-ray energy.
  • Different materials have different μ values, allowing for contrast in images (e.g., bone vs tissue).
  • Half-value thickness (HVT) is the thickness that reduces intensity by 50%.

Quick check

  1. What are the SI units of the attenuation coefficient μ?1 mark
  2. If material A has a higher attenuation coefficient than material B, which material is better at stopping X-rays?1 mark

3. Improving X-ray Images and CT Scans

Standard X-ray images are 2D projections and have limitations, especially in distinguishing between soft tissues which have similar attenuation coefficients. Image quality can be improved in two main ways. First, a contrast medium, such as barium or iodine, can be introduced into the body. These substances have high atomic numbers and significantly increase the attenuation in the specific tissues or organs they fill (like the digestive tract or blood vessels), making them stand out clearly. Second, image intensifiers can be used. These devices convert the few X-ray photons that pass through the patient into a much larger number of light photons, which are then converted into an electrical signal to create a brighter image, allowing for a lower initial X-ray dose.

Computerised Axial Tomography (CAT or CT) is a more advanced technique. Instead of one static image, a CT scanner takes a series of X-ray images from many different angles as the X-ray tube rotates around the patient. A computer then processes these multiple 2D 'slices' and reconstructs them into a detailed 3D image. This allows doctors to see the body's internal structures without the overlapping of organs found in a conventional 2D X-ray, providing far superior soft tissue detail and spatial information.

Key term

Contrast Medium: A substance with a high atomic number (like barium or iodine) used in medical imaging to increase the attenuation of X-rays in a specific region of the body, enhancing the visibility of internal structures.

Examiner insight

When comparing CT scans and conventional X-rays, be sure to mention both an advantage (e.g., 3D image, better soft tissue contrast for CT) and a disadvantage (e.g., higher radiation dose, higher cost for CT).

Fun fact

The first commercially viable CT scanner was invented by Sir Godfrey Hounsfield at EMI, the same company that signed The Beatles. Its development was partly funded by the band's success.

Worked example 14 marks

Explain why a contrast medium is necessary to obtain a clear X-ray image of the small intestine.

  1. 1

    Step 1: State the basic principle of X-ray imaging. An image is formed due to differential attenuation of X-rays by different tissues.

  2. 2

    Step 2: Explain the problem with soft tissues. The small intestine and surrounding soft tissues are composed of elements with similar, low atomic numbers.

  3. 3

    Step 3: This means their linear attenuation coefficients (μ) are very similar, so there is very little difference in the intensity of X-rays passing through them.

  4. 4

    Step 4: Explain the role of the contrast medium. A barium meal, a contrast medium, is ingested. Barium has a high atomic number and a much larger attenuation coefficient than soft tissue.

  5. 5

    Step 5: The barium fills the intestine, causing significant attenuation of X-rays passing through it. This creates a high contrast between the intestine and the surrounding tissues, making its outline clearly visible on the image.

Recap

  • Contrast media like barium increase attenuation and improve image clarity for soft tissues.
  • Image intensifiers create brighter images from lower X-ray doses.
  • CT scans use a rotating X-ray source to create multiple 2D slices of the body.
  • A computer reconstructs the CT slices into a detailed 3D image.
  • CT scans provide excellent soft tissue detail and avoid the problem of overlapping structures.
  • The main disadvantage of CT scans compared to a single X-ray is the significantly higher radiation dose to the patient.

Quick check

  1. State one advantage of a CT scan over a conventional X-ray.1 mark
  2. Name an element commonly used as a contrast medium.1 mark

4. Ultrasound Production and Detection

Ultrasound refers to sound waves with frequencies above the range of human hearing, typically greater than 20 kHz. In medical imaging, frequencies in the range of 2-10 MHz are used. These high-frequency waves are produced and detected by a device called a transducer, which relies on the piezoelectric effect. A piezoelectric crystal (like quartz or certain synthetic ceramics) has a unique property: when a potential difference is applied across it, it mechanically deforms (changes shape). If an alternating p.d. with a high frequency is applied, the crystal vibrates rapidly, producing ultrasound waves. Conversely, the effect is reversible. If an incoming ultrasound wave causes the crystal to vibrate, it generates an alternating potential difference across its faces. The transducer therefore acts as both a transmitter and a receiver of ultrasound, sending out a short pulse and then 'listening' for the returning echoes.

Key term

Piezoelectric Effect: The ability of certain materials (piezoelectric crystals) to generate an electric potential in response to applied mechanical stress, and conversely, to change shape when an electric field is applied.

Examiner insight

A good answer will clearly separate the process of generation (voltage in, sound out) from detection (sound in, voltage out) and explicitly name the piezoelectric effect for both.

Worked example 15 marks

Describe how a piezoelectric transducer is used to both produce and detect ultrasound waves.

  1. 1

    Step 1: To produce ultrasound, an alternating potential difference is applied across the faces of a piezoelectric crystal in the transducer.

  2. 2

    Step 2: The frequency of the alternating p.d. is matched to the crystal's natural frequency of vibration to achieve resonance.

  3. 3

    Step 3: Due to the piezoelectric effect, the alternating p.d. causes the crystal to contract and expand, vibrating at this high frequency.

  4. 4

    Step 4: These vibrations produce and transmit high-frequency longitudinal sound waves (ultrasound) into the surrounding medium.

  5. 5

    Step 5: To detect ultrasound, returning echo waves strike the crystal, causing it to vibrate.

  6. 6

    Step 6: Due to the reverse piezoelectric effect, this mechanical vibration induces an alternating potential difference across the crystal.

  7. 7

    Step 7: This electrical signal is then processed to form an image.

Recap

  • Ultrasound is a longitudinal wave with a frequency above 20 kHz.
  • Medical ultrasound uses frequencies in the MHz range.
  • Transducers use the piezoelectric effect to generate and detect ultrasound.
  • Applying an alternating voltage makes the crystal vibrate, producing ultrasound.
  • Returning ultrasound echoes make the crystal vibrate, generating a voltage.
  • The same transducer crystal can act as both a transmitter and a receiver.

Quick check

  1. What is the minimum frequency for a sound wave to be classified as ultrasound?1 mark
  2. What is the name of the effect that allows a transducer to generate ultrasound?1 mark

5. Acoustic Impedance and Ultrasound Reflection

When an ultrasound wave reaches a boundary between two different materials, some of it is reflected and some is transmitted. The amount of reflection depends on the difference in the acoustic impedance (Z) of the two materials. Acoustic impedance is a property of a medium that describes how much it resists the passage of sound waves. It is calculated as the product of the density (ρ) of the medium and the speed of sound(c) in that medium: Z = ρc. The greater the difference (or 'mismatch') in acoustic impedance between two materials, the greater the fraction of the wave's intensity that is reflected. The ratio of reflected intensity (Ir) to incident intensity (I₀) is given by the formula Ir/I₀ = ((Z₂ - Z₁)/(Z₂ + Z₁))². Because the impedance mismatch between air and skin is very large, almost all ultrasound would be reflected at the skin's surface. To overcome this, a coupling gel is used. The gel has an acoustic impedance similar to that of skin, which displaces the air and ensures that most of the ultrasound intensity is transmitted into the body.

Z = ρc

α = I_r / I₀ = ((Z₂ - Z₁)/(Z₂ + Z₁))²

Key term

Acoustic Impedance (Z): A physical property of a medium, defined as the product of its density and the speed of sound within it, which determines the reflection and transmission of sound waves at a boundary.

Examiner insight

Examiners expect you to be able to explain the need for a coupling gel in terms of acoustic impedance matching. Simply saying 'it removes air' is not a complete answer; you must explain *why* removing air is necessary by referring to the large impedance mismatch between air and skin.

Common pitfall

Forgetting to square the term in the reflection coefficient formula. The formula calculates the ratio of intensities, which is proportional to the square of the amplitude ratio.

Worked example 14 marks

The speed of ultrasound in fat is 1450 m/s and its density is 920 kg/m³. For muscle, the speed is 1590 m/s and the density is 1070 kg/m³. Calculate the percentage of ultrasound intensity reflected at a fat-muscle boundary.

  1. 1

    Step 1: Calculate the acoustic impedance of fat (Z_fat).

  2. 2

    Z_fat = ρ_fat * c_fat = 920 kg/m³ * 1450 m/s = 1.334 x 10⁶ kg m⁻² s⁻¹.

  3. 3

    Step 2: Calculate the acoustic impedance of muscle (Z_muscle).

  4. 4

    Z_muscle = ρ_muscle * c_muscle = 1070 kg/m³ * 1590 m/s = 1.701 x 10⁶ kg m⁻² s⁻¹.

  5. 5

    Step 3: Use the reflection coefficient formula.

  6. 6

    α = I_r / I₀ = ((Z_muscle - Z_fat)/(Z_muscle + Z_fat))²

  7. 7

    α = ((1.701 x 10⁶ - 1.334 x 10⁶) / (1.701 x 10⁶ + 1.334 x 10⁶))²

  8. 8

    α = (0.367 x 10⁶ / 3.035 x 10⁶)² = (0.1209)²

  9. 9

    α = 0.0146

  10. 10

    Step 4: Convert the fraction to a percentage.

  11. 11

    Percentage reflected = 0.0146 * 100% = 1.46%.

Recap

  • Acoustic impedance is given by Z = ρc.
  • Reflection of ultrasound occurs at boundaries between materials with different acoustic impedances.
  • A large impedance mismatch causes a strong reflection.
  • The fraction of reflected intensity is given by Ir/I₀ = ((Z₂ - Z₁)/(Z₂ + Z₁))².
  • Coupling gel is used to match the impedance of the transducer to the skin, maximising transmission.
  • The large impedance mismatch between bone and tissue causes a strong reflection, making bone appear bright but creating a shadow behind it.

Quick check

  1. What is the purpose of using a coupling gel in an ultrasound scan?2 marks
  2. If two materials have identical acoustic impedances, what percentage of ultrasound intensity is reflected at their boundary?1 mark

6. Positron Emission Tomography (PET)

Positron Emission Tomography (PET) is a nuclear medicine imaging technique that shows the metabolic or biochemical function of tissues and organs. A patient is injected with a small amount of a radiotracer, which is a biologically active molecule (like glucose) tagged with a positron-emitting radioisotope. A common tracer is Fludeoxyglucose (FDG). The tracer travels through the body and is absorbed by tissues. Areas with high metabolic activity, such as cancer cells or active parts of the brain, absorb more of the tracer. The radioisotope undergoes beta-plus (β⁺) decay, emitting a positron (an anti-electron). This positron travels a very short distance before it encounters an electron in the tissue. The particle-antiparticle pair annihilate each other, converting their entire mass into energy in the form of two high-energy gamma photons. To conserve momentum, these two photons are always emitted in opposite directions (180° apart). The PET scanner consists of a ring of gamma detectors that detect these pairs of photons arriving simultaneously. By analysing the lines along which these photon pairs are detected, a computer can reconstruct a 3D image showing the location and concentration of the radiotracer in the body.

E = mc²

Key term

Annihilation: The process that occurs when a particle collides with its corresponding antiparticle, resulting in their mutual destruction and the conversion of their entire rest mass into energy, typically in the form of two or more photons.

Examiner insight

Clear understanding of the conservation laws is crucial. Examiners reward answers that state that two photons are produced to conserve momentum (hence they travel in opposite directions) and that their energy comes from the conversion of the mass of the electron-positron pair (conservation of mass-energy).

Common pitfall

Stating that the PET scanner detects positrons. The scanner detects the gamma photons that result from the positron's annihilation; the positron itself travels only a few millimetres before this happens.

Worked example 14 marks

In a PET scan, a positron annihilates with an electron. Assuming both are initially at rest, calculate the energy of one of the gamma photons produced. (mass of electron/positron = 9.11 x 10⁻³¹ kg, c = 3.00 x 10⁸ m/s)

  1. 1

    Step 1: State the principle of mass-energy equivalence, E = mc².

  2. 2

    Step 2: The total mass being converted to energy is the mass of the electron plus the mass of the positron. m_total = m_e + m_p = 2 * (9.11 x 10⁻³¹ kg).

  3. 3

    m_total = 1.822 x 10⁻³⁰ kg.

  4. 4

    Step 3: Calculate the total energy released in the annihilation.

  5. 5

    E_total = m_total * c² = (1.822 x 10⁻³⁰ kg) * (3.00 x 10⁸ m/s)²

  6. 6

    E_total = 1.64 x 10⁻¹³ J.

  7. 7

    Step 4: This total energy is shared equally between the two gamma photons produced.

  8. 8

    Energy per photon = E_total / 2 = (1.64 x 10⁻¹³ J) / 2 = 8.2 x 10⁻¹⁴ J.

  9. 9

    Step 5 (Optional): Convert to MeV. 1 MeV = 1.60 x 10⁻¹³ J. Energy = (8.2 x 10⁻¹⁴ J) / (1.60 x 10⁻¹³ J/MeV) = 0.51 MeV.

Recap

  • PET scans map metabolic function using a radiotracer.
  • The tracer contains a positron-emitting isotope.
  • The emitted positron annihilates with an electron in the tissue.
  • Annihilation produces two gamma photons travelling in opposite directions.
  • A ring of detectors identifies these photon pairs to pinpoint the annihilation site.
  • PET images show function, whereas X-ray and CT images primarily show structure.

Quick check

  1. What type of radioactive decay is essential for PET scanning?1 mark
  2. Why are two gamma photons produced during annihilation in a PET scan, and why do they travel in opposite directions?2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Describe the basic principles of how a CT scanner produces a three-dimensional image of a patient. Outline one advantage and one disadvantage of a CT scan compared with a conventional X-ray image.5 marks
  2. A beam of X-rays of initial intensity I₀ is incident on a sheet of lead. The attenuation coefficient for lead is 4.6 cm⁻¹. Calculate the thickness of lead required to reduce the intensity of the transmitted beam to 1.0% of its initial value.4 marks
  3. Explain the role of acoustic impedance in ultrasound imaging and why a coupling gel is necessary for a scan.4 marks
  4. A piezoelectric crystal in a transducer is used to generate ultrasound for a medical scan. Explain the physical principle that allows the crystal to both generate and detect ultrasound waves.4 marks
  5. A radiotracer used in a PET scan undergoes beta-plus decay. Describe the sequence of events from this decay to the detection of signals by the PET scanner.6 marks
  6. The acoustic impedance of bone is 7.8 x 10⁶ kg m⁻² s⁻¹ and for soft tissue it is 1.6 x 10⁶ kg m⁻² s⁻¹. Calculate the fraction of ultrasound intensity that is reflected at a boundary between soft tissue and bone.3 marks
  7. An X-ray tube operates with an accelerating potential of 120 kV. Only 0.5% of the electron's energy is converted into X-rays. If the electron beam current is 5.0 mA, calculate (a) the number of electrons hitting the target per second, and (b) the power of the emitted X-ray beam.5 marks
  8. Distinguish between an A-scan and a B-scan in the context of medical ultrasound.3 marks
  9. In a PET scan, the annihilation of an electron and a positron produces two gamma photons. State the energy of each photon in MeV and justify why two photons are produced travelling in opposite directions.3 marks
  10. A doctor needs to examine a patient's kidneys. Compare the use of a CT scan with an ultrasound scan for this purpose, considering the principles, image quality, and patient safety for each technique.6 marks

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