(a) Calculate the range of wavelengths for AM radio given its frequency range is \(540\) to \(1600{\rm{ }}kHz\). (b) Do the same for the FM frequency range of \(88.0\) to \(108{\rm{ }}MHz\).

Wavelength: The separation between two waves' subsequent locations that exhibit the same oscillation phase, measured in the wave's propagation direction.

Frequency: The quantity of waves that pass a set location in a predetermined period of time is known as frequency.

• The frequency range for AM radio waves:\(540 - 1600{\rm{ }}kHz\)
• The frequency range for FM radio waves: \(88.0 - 108{\rm{ }}MHz\)

Consider the given values,

We get \(\lambda = \frac{c}{f}\)

Where \(\lambda \) stands for wavelength, \(f\) frequency, and \(c = 3 \times {10^8}{\rm{ }}m{s^{ - 1}}\) is the speed of propagation (speed of light).

Plugging in \({f_1} = 540{\rm{ }}kHz\) and \({f_2} = 1600{\rm{ }}kHz\) we get

\(\begin{aligned} {\lambda _1} &= \frac{c}{{{f_1}}}\\{\lambda _1} &= \frac{{\left( {3 \times {{10}^8}{\rm{ }}m{s^{ - 1}}} \right)}}{{\left( {540{\rm{ }}kHz} \right)}}\\ &= 5.56 \times {10^2}{\rm{ }}m\end{aligned}\)

\(\begin{aligned} {\lambda _2} &= \frac{c}{{{f_2}}}\\ &= \frac{{\left( {3 \times {{10}^8}{\rm{ }}m{s^{ - 1}}} \right)}}{{\left( {1600{\rm{ }}kHz} \right)}}\\ &= 1.88 \times {10^2}{\rm{ }}m\end{aligned}\)

Therefore, the range of wavelength for AM radio is \(\lambda = (1.88 - 5.56) \times {10^2}m\).

Plugging in \({f_3} = 88.0{\rm{ }}MHz\)

And \({f_4} = 108{\rm{ }}MHz\)we get

\(\begin{aligned} {\lambda _3} &= \frac{c}{{{f_3}}}\\ &= \frac{{\left( {3 \times {{10}^8}{\rm{ }}m{s^{ - 1}}} \right)}}{{\left( {88.0{\rm{ }}MHz} \right)}}\\\lambda &= 3.41 \times {10^0}{\rm{ }}m\end{aligned}\)

\(\begin{aligned} {\lambda _4} &= \frac{c}{{{f_4}}}\\ &= \frac{{\left( {3 \times {{10}^8}{\rm{ }}m{s^{ - 1}}} \right)}}{{\left( {108{\rm{ }}MHz} \right)}}\\ &= 2.78 \times {10^0}{\rm{ }}m\end{aligned}\)

Hence, the FM frequency range is \(\lambda = (2.78 - 3.41)m\).