When switching from a 3.5 MHz transducer to a 7.0 MHz transducer imaging a superficial structure, compared to the 3.5 MHz transducer what will the 7.0 MHz attenuation rate and wavelength be?

When switching from a 3.5 MHz transducer to a 7.0 MHz transducer imaging a superficial structure, compared to the 3.5 MHz transducer what will the 7.0 MHz attenuation rate and wavelength be?

• A. Double the attenuation rate, one half the wavelength.
• B. Double the attenuation rate, double the wavelength.
• C. One half the attenuation rate, double the wavelength.
• D. One fourth the attenuation rate, one half the wavelength.

Answer: (A)

Doubling the frequency doubles the attenuation rate and halves the wavelength. Attenuation in tissue increases with frequency, so when you go from 3.5 to 7.0 MHz, the energy loss per centimeter doubles, reducing how deeply the sound can penetrate. The wavelength is inversely related to frequency (wavelength = speed of sound / frequency), so increasing frequency halves the wavelength. For soft tissue with about 1540 m/s speed, 3.5 MHz gives roughly 0.44 mm, while 7.0 MHz gives about 0.22 mm. This higher-frequency probe thus provides better axial resolution but shallower penetration.

Doubling the frequency doubles the attenuation rate and halves the wavelength. Attenuation in tissue increases with frequency, so when you go from 3.5 to 7.0 MHz, the energy loss per centimeter doubles, reducing how deeply the sound can penetrate. The wavelength is inversely related to frequency (wavelength = speed of sound / frequency), so increasing frequency halves the wavelength. For soft tissue with about 1540 m/s speed, 3.5 MHz gives roughly 0.44 mm, while 7.0 MHz gives about 0.22 mm. This higher-frequency probe thus provides better axial resolution but shallower penetration.