Question

(a) What is the wavelength of$\mathbf{100}\mathbf{}\mathbf{MHz}$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$\mathbf{100}\mathbf{}\mathbf{MHz}$, what is the range of wavelengths broadcast?

Short answer

(a) The wavelength of$100\mathrm{MHz}$radio waves used in an MRI unit is$3\mathrm{m}$

(b) The range of the wavelength broadcast is $2.97\mathrm{m}$to $3.03\mathrm{m}$.

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$=100\mathrm{MHz}\left(\frac{{10}^6\mathrm{Hz}}{1\mathrm{MHz}}\right)=1.00\times {10}^8\mathrm{Hz}$.

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

Find the wavelength of100MHz radio waves used in an MRI unit

(a)

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

$\mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }}$………………(1)

Substituting the given values in the equation (1), we will get,

$$\begin{gathered} \lambda =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.00\times {10}^8\mathrm{Hz}} \\ =3.00\mathrm{m} \end{gathered}$$

Therefore, the required wavelength is$3.00\mathrm{m}$

Find the range of the wavelength broadcast

(b)

Considering the $\pm 1.00$range, the lower bound of the frequency is,

$$\begin{gathered} {\mathrm{f}}_1=\left(1.00\times {10}^8\right)-\left(\frac{1}{100}\right)\left(1.00\times {10}^8\right) \\ =9.9\times {10}^7\mathrm{Hz} \end{gathered}$$

Therefore using the wavelength’s higher limit using equation (1) is,

$$\begin{gathered} {\lambda }_1=\frac{\mathrm{c}}{f_1} \\ =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{9.9\times {10}^7\mathrm{Hz}} \\ =3.03\mathrm{m} \end{gathered}$$

The frequency’s higher limit is,

$$\begin{gathered} {\mathrm{f}}_2=\left(1.00\times {10}^8\right)+\left(\frac{1}{100}\right)\left(1.00\times {10}^8\right) \\ =1.01\times {10}^8\mathrm{Hz} \end{gathered}$$

Therefore using the wavelength’s higher limit using equation (1) is,

$$\begin{gathered} {\lambda }_2=\frac{\mathrm{c}}{{\mathrm{f}}_2} \\ =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.01\times {10}^8\mathrm{Hz}} \\ =2.97\mathrm{m} \end{gathered}$$

Therefore, the required range is $2.97\mathrm{m}$to $3.03\mathrm{m}$.