5 (i) Calculate the wavelength of ultrasound travelling in soft tissue if the frequency of the sound is 5 MHz and its speed in the soft tissue is \( 1550 \mathrm{~m} \mathrm{~s}^{-1} \). (ii) How many times as long will the wavelength be if the frequency were reduced to 50 kHz ? (iii) Hence describe the advantages of using high-frequency ultrasound for ultrasound scanning in medical diagnosis instead of low-frequency ultrasound.

To find the wavelength of ultrasound in soft tissue, we can use the formula \( \lambda = \frac{v}{f} \), where \( \lambda \) is the wavelength, \( v \) is the speed of sound in soft tissue, and \( f \) is the frequency. For 5 MHz (or \( 5 \times 10^6 \) Hz) and a speed of \( 1550 \, \text{m/s} \): \[ \lambda = \frac{1550 \, \text{m/s}}{5 \times 10^6 \, \text{Hz}} = 0.00031 \, \text{m} \text{ or } 0.31 \, \text{mm}. \] If the frequency is reduced to 50 kHz (or \( 5 \times 10^4 \) Hz), the wavelength will be: \[ \lambda_{50 \, \text{kHz}} = \frac{1550 \, \text{m/s}}{50 \times 10^3 \, \text{Hz}} = 0.031 \, \text{m} \text{ or } 31 \, \text{mm}. \] The wavelength at 50 kHz is \( \frac{0.031 \, \text{m}}{0.00031 \, \text{m}} = 100 \) times longer than at 5 MHz. Using high-frequency ultrasound, such as 5 MHz, has significant advantages for medical diagnosis. High frequencies result in better resolution of the images because they provide more cycles per unit length, leading to clearer and more detailed visuals of soft tissues and organs. This detail is paramount for accurately identifying structures and abnormalities, facilitating better patient outcomes. On the other hand, lower frequencies, like 50 kHz, penetrate deeper into the body, but this comes at the trade-off of reduced image resolution. Consequently, while low-frequency ultrasound can visualize organs at greater depths, it may miss finer details crucial for diagnosis, making high-frequency ultrasound the preferred choice in many diagnostic situations where clarity and detail are essential.