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?

Sharpen your skills for the Davies Publishing SPI Test with targeted flashcards and multiple-choice questions, complete with hints and clarifications. Prepare thoroughly for success!

Multiple Choice

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?

Explanation:
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.

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy