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9702 · 24.1

Production and use of ultrasound flashcards

Revision flashcards for Cambridge 9702 Production and use of ultrasound (syllabus 24.1). Flip, recall, then mark a real past-paper question.

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    What defines ultrasound?

    Sound waves with frequencies above 20 kHz, beyond the range of human hearing, typically 1–20 MHz for medical uses.

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    Explain the piezoelectric effect.

    Certain crystals (e.g., quartz) deform mechanically when an alternating p.d. is applied, and conversely, generate a p.d. when mechanically deformed.

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    How does a transducer generate and detect ultrasound?

    It generates ultrasound by applying an alternating p.d. to a piezoelectric crystal, causing it to vibrate. It detects echoes by sensing the p.d. generated when returning waves deform the crystal.

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    What is acoustic impedance (Z) and its formula?

    A measure of a medium's resistance to sound wave propagation, calculated as $Z = \rho c$ (density × speed of sound).

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    When does significant reflection occur at a boundary?

    When there is a large difference in the acoustic impedances ($Z_1$ and $Z_2$) of the two media.

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    What is the purpose of a coupling gel in ultrasound?

    To eliminate air between the transducer and skin. Air has a vastly different acoustic impedance from tissue, which would reflect almost all the ultrasound, preventing it from entering the body.

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    What is ultrasound attenuation?

    The decrease in ultrasound intensity and amplitude as it travels through a medium, caused by absorption (conversion to heat) and scattering.

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    How is ultrasound intensity affected by distance due to attenuation?

    It decreases exponentially with distance ($x$): $I = I_0 e^{-\mu x}$, where $\mu$ is the linear attenuation coefficient.

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    Define half-value thickness ($x_{1/2}$).

    The distance an ultrasound wave travels through a medium for its intensity to be reduced to half its initial value.

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    What does 'damping' achieve in an ultrasound transducer?

    It rapidly stops the crystal's vibrations, producing short ultrasound pulses essential for high axial resolution (distinguishing closely spaced objects).

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    What is the relationship between half-value thickness ($x_{1/2}$) and the linear attenuation coefficient ($\mu$)?

    $x_{1/2} = \frac{\ln(2)}{\mu}$. A larger attenuation coefficient means a smaller half-value thickness, as the intensity decreases more rapidly.

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    Why are short ultrasound pulses necessary for good image resolution?

    Short pulses improve axial resolution, which is the ability to distinguish between two structures that are close together along the beam's axis. Long pulses would cause the echoes from these structures to overlap, blurring the image.

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    What are the two main causes of ultrasound attenuation in tissue?

    1. **Absorption:** The energy of the ultrasound wave is converted into thermal energy (heat) in the tissue. 2. **Scattering:** The wave is reflected and refracted in many directions by small structures within the tissue.

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    Calculate the acoustic impedance of water if its density is 1000 kg m⁻³ and the speed of sound in it is 1500 m s⁻¹.

    $Z = \rho c = 1000 \text{ kg m}^{-3} \times 1500 \text{ m s}^{-1} = 1.5 \times 10^6 \text{ kg m}^{-2} \text{s}^{-1}$ (or 1.5 MRayl).