question_answer

A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of the sound in the tissue is\[\text{1}.\text{7}\,\text{km}{{\text{s}}^{-\text{1}}}.\]The wavelength of sound in the tissue is closed to (a) \[4\times {{10}^{-4}}m\]                        (b) \[8\times {{10}^{-4}}m\] (c) \[4\times {{10}^{-3}}m\]                        (d) \[8\times {{10}^{-3}}m\]

Answer:

Given \[n=4.2MHz=4.2\times {{10}^{6}}Hz\] \[v=1.7\times {{10}^{3}}m/s\] \[\therefore \] \[\lambda =v/n\] \[=4.2\times {{10}^{6}}/1.7\times {{10}^{3}}\] \[=2.47\times {{10}^{3}}m.\]