Question

Radio city broadcasts on a frequency of \[5090{\text{kHz}}\]. What is the wavelength of electromagnetic radiation emitted by the transmitter?

Answer 1

Hint: The wavelength of a wave is inversely proportional to the frequency and directly proportional to the velocity of the wave. In this case, we are given an electromagnetic wave that the radio city broadcasts and we need to find the wavelength of the electromagnetic radiation.

Complete step by step solution:
Frequency of an electromagnetic wave indicates how often the particles move back and forth in the medium. Frequency is measured as the total number of complete vibrations in unit time. And wavelength is defined as the distance between two consecutive crests and troughs in a wave. A high-frequency wave has a shorter wavelength than a wave with a lesser frequency. This means that the wavelength and the frequency are inversely related to each other. Therefore we have
$\nu = \dfrac{c}{\lambda }$
where $\nu $ is the frequency of the wave, $\lambda $ is the wavelength and $c$ is the speed of light.
We will consider the speed of the electromagnetic wave as the speed of light in a vacuum.
Substituting the values of frequency and speed we get
$\lambda = \dfrac{{3 \times {{10}^8}}}{{5090 \times {{10}^3}}} = 0.589 \times {10^2} = 58.9{\text{m}}$ Which falls in the range of the radio spectrum.
Radio waves are electromagnetic radiation that has a wavelength longer than the infra-red waves in the electromagnetic spectrum and has very frequencies. These waves are generated by the transmitters and received by radio receivers in radio stations using an antenna.

Note: Electromagnetic waves consist of small packets of energy known as a photon. The energy carried by a single photon is directly proportional to the photon’s electromagnetic frequency and thus equivalently proportional to the wavelength. The higher the frequency of the photon, the higher is its energy.