Question

(a) What is the wavelength of 100 MHz radio waves used in an MRI unit? (b) If the frequencies are swept over a $\mathbf{\pm }\mathbf{1}\mathbf{.}\mathbf{00}$range centered on 100MHz , what is the range of wavelengths broadcast?

Short answer

(a) The wavelength of 100 MHz radio waves used in an MRI unit is 3m.

(b) The range of the wavelength broadcast is 2.97 m to 3.03 m .

Step-by-step solution

Definition of Concept

Wavelength: The distance between identical points (adjacent crests) in adjacent cycles of a waveform signal propagated in space or along a wire is defined as the wavelength. This length is typically specified in wireless systems in metres (m), centimetres (cm), or millimetres (mm).

Given data

The radio waves have a frequency of $f=100\mathrm{MHz}\left(\frac{{10}^6\mathrm{Hz}}{1\mathrm{MHz}}\right)=1.00\times {10}^8\mathrm{Hz}$.

The speed of the light is$c=3.00\times {10}^8\mathrm{m}/\mathrm{s}$

Find the wavelength of 100MHz radio waves used in an MRI unit

(a)

The relationship between the wavelength and frequency can be expressed as,

$$\begin{gathered} f=\frac{c}{\lambda }..................\left(1\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{will}\mathrm{get}, \\ \mathrm{\lambda }=\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.00\times {10}^8\mathrm{Hz}} \\ =3.00\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{wavelength}\mathrm{is}3.00\mathrm{m} \end{gathered}$$………………(1)

Find the range of the wavelength broadcast

(b)

Considering the range, the lower bound of the frequency is,

$$\begin{gathered} f_1=\left(1.00\times {10}^8\right)-\left(\frac{1}{100}\right)\left(1.00\times {10}^8\right) \\ =9.9\times {10}^7\mathrm{Hz} \\ \mathrm{Therefore}\mathrm{using}\mathrm{the}\mathrm{wavelength}’\mathrm{s}\mathrm{higher}\mathrm{limit}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{is}, \\ {\mathrm{\lambda }}_1=\frac{\mathrm{c}}{{\mathrm{f}}_1} \\ =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{9.9\times {10}^7\mathrm{Hz}} \\ =3.03\mathrm{m} \\ \mathrm{The}\mathrm{frequency}’\mathrm{s}\mathrm{higher}\mathrm{limit}\mathrm{is}, \\ {\mathrm{f}}_2=\left(1.00\times {10}^8\right)+\left(\frac{1}{100}\right)\left(1.00\times {10}^8\right) \\ =1.01\times {10}^8\mathrm{Hz} \\ \mathrm{Therefore}\mathrm{using}\mathrm{the}\mathrm{wavelength}’\mathrm{s}\mathrm{higher}\mathrm{limit}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{is}, \\ {\mathrm{\lambda }}_2=\frac{\mathrm{c}}{{\mathrm{f}}_2} \\ =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.01\times {10}^8\mathrm{Hz}} \\ =2.97\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{range}\mathrm{is}2.97\mathrm{to}3.03\mathrm{m}. \end{gathered}$$