
The Vivid Bharati station of All India Radio, Delhi broadcasts on a frequency of 1,368kHz(kilohertz). Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong to?
Answer
560.7k+ views
Hint: Frequency and wavelength are inversely proportional to each other. The electromagnetic spectrum having greater frequency occurs on low wavelengths. Also, Energy and frequency are directly proportional to each other.
Complete answer:
Frequency of a wave is the number of times by which the wave passes through a particular point. The frequency of the transmitter in the Vivid Bharati station of All India Radio is 1,368kHz.
Frequency is measured in Hertz which is equal to per second (${{s}^{-1}}$)
As we know, Frequency and wavelength are related as,
\[v=\dfrac{c}{\lambda }\] … (i)
where, v = is the frequency (Hz)(${{s}^{-1}}$)
c= speed of light (m/s)
λ= wavelength (m)
we know , speed of light is $3\times {{10}^{8}}m/s$.
And the frequency is given in kiloHertz so we will convert it in Hertz by multiplying the value with${{10}^{3}}$, because the value of speed of light is in meters per second.
Substitution the values in the equation (i) we get,
\[\begin{align}
& \Rightarrow \lambda =\dfrac{c}{v} \\
& \Rightarrow \lambda =\dfrac{3\times {{10}^{8}}}{1368\times {{10}^{3}}} \\
& \Rightarrow \lambda =21.9m \\
\end{align}\]
The wavelength of the transmitter is 21.9m. Since the obtained value of the transmitter is in meters, therefore, it will belong to the Radio Wave spectrum.
Note:
Radio Wave wavelength is from 1 meter to ${{10}^{3}}$. Below 1 meter to ${{10}^{-3}}$m is the microwave spectrum. Wavenumber is the inverse of wavelength. Wavelength is measured in meter or centimeter and wavenumber is measured in ${{m}^{-1}}$or $c{{m}^{-1}}$.
Complete answer:
Frequency of a wave is the number of times by which the wave passes through a particular point. The frequency of the transmitter in the Vivid Bharati station of All India Radio is 1,368kHz.
Frequency is measured in Hertz which is equal to per second (${{s}^{-1}}$)
As we know, Frequency and wavelength are related as,
\[v=\dfrac{c}{\lambda }\] … (i)
where, v = is the frequency (Hz)(${{s}^{-1}}$)
c= speed of light (m/s)
λ= wavelength (m)
we know , speed of light is $3\times {{10}^{8}}m/s$.
And the frequency is given in kiloHertz so we will convert it in Hertz by multiplying the value with${{10}^{3}}$, because the value of speed of light is in meters per second.
Substitution the values in the equation (i) we get,
\[\begin{align}
& \Rightarrow \lambda =\dfrac{c}{v} \\
& \Rightarrow \lambda =\dfrac{3\times {{10}^{8}}}{1368\times {{10}^{3}}} \\
& \Rightarrow \lambda =21.9m \\
\end{align}\]
The wavelength of the transmitter is 21.9m. Since the obtained value of the transmitter is in meters, therefore, it will belong to the Radio Wave spectrum.
Note:
Radio Wave wavelength is from 1 meter to ${{10}^{3}}$. Below 1 meter to ${{10}^{-3}}$m is the microwave spectrum. Wavenumber is the inverse of wavelength. Wavelength is measured in meter or centimeter and wavenumber is measured in ${{m}^{-1}}$or $c{{m}^{-1}}$.
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