The Vividh Bharati Station of All India Radio Delhi broadcasts on a frequency of 1,368 kHz. Calculate the wavelength ( λ) of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong to?
A. 319.4m and X- rays
B. 319.4m and radio wave
C. 219.3m and microwave
D. 219.3m and radio wave
Answer
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Hint: Electromagnetic waves or EM waves are a result of vibrations between an electric and magnetic field. It has been found that their wavelength is inversely proportional to the frequency of these waves, i.e.
$\lambda \propto \dfrac{1}{\nu }$
The proportionality constant here is the velocity of light, c.
The typical electromagnetic spectrum of wavelengths of EM waves ranges from 102Hz to 1024 Hz. The various components of this spectrum are as follows:
Radio waves \[\left( {{10}^{2}}-{{10}^{6}}Hz \right)\]
Microwaves \[\left( {{10}^{7}}-{{10}^{12}}Hz \right)\]
Infrared waves \[\left( {{10}^{12}}\text{ }4\text{ }\times \text{ }{{10}^{14}}Hz \right)\]
Visible light \[\left( 4\text{ }\times{{10}^{14}}-\text{ }7\text{ }\times {{10}^{14}}Hz \right)\]
Ultra Violet rays \[\left( 7\text{ }\times \text{ }{{10}^{14}}-{{10}^{16}}Hz \right)\]
X rays \[\left( {{10}^{17}}-{{10}^{21}}Hz \right)\]
Gamma rays \[\left( {{10}^{21}}Hz\text{ }+ \right)\]
Complete answer:
The frequency (υ) of the electromagnetic radiation emitted by the radio station:\[1368\text{ }kHz=1368\text{ }\times{{10}^{3}}Hz\]
Since, \[\lambda =\dfrac{c}{\nu }\]
where: c stands for speed of light in vacuum/air \[\left( 3\text{ }\times \text{ }{{10}^{8}}m/s \right)\]
λ Stands for wavelength of the radiation
On substituting the values in the given equation:
$\lambda =\dfrac{3\times {{10}^{8}}}{1368\times {{10}^{3}}}$
$\lambda =219.3$ m.
The wavelength of the given radiation is 219.3 m.
Since the frequency of the radiation \[\left( 1368\text{ }\times{{10}^{3}}Hz \right)\]is of order \[{{10}^{3}}\], it is a radio wave.
So, the correct answer is “Option D”.
Note:
Conversion of the units to the standard form should be carefully implemented as lack of attentiveness and accuracy here, may cause calculation errors thereby leading to wrong results.
$\lambda \propto \dfrac{1}{\nu }$
The proportionality constant here is the velocity of light, c.
The typical electromagnetic spectrum of wavelengths of EM waves ranges from 102Hz to 1024 Hz. The various components of this spectrum are as follows:
Radio waves \[\left( {{10}^{2}}-{{10}^{6}}Hz \right)\]
Microwaves \[\left( {{10}^{7}}-{{10}^{12}}Hz \right)\]
Infrared waves \[\left( {{10}^{12}}\text{ }4\text{ }\times \text{ }{{10}^{14}}Hz \right)\]
Visible light \[\left( 4\text{ }\times{{10}^{14}}-\text{ }7\text{ }\times {{10}^{14}}Hz \right)\]
Ultra Violet rays \[\left( 7\text{ }\times \text{ }{{10}^{14}}-{{10}^{16}}Hz \right)\]
X rays \[\left( {{10}^{17}}-{{10}^{21}}Hz \right)\]
Gamma rays \[\left( {{10}^{21}}Hz\text{ }+ \right)\]
Complete answer:
The frequency (υ) of the electromagnetic radiation emitted by the radio station:\[1368\text{ }kHz=1368\text{ }\times{{10}^{3}}Hz\]
Since, \[\lambda =\dfrac{c}{\nu }\]
where: c stands for speed of light in vacuum/air \[\left( 3\text{ }\times \text{ }{{10}^{8}}m/s \right)\]
λ Stands for wavelength of the radiation
On substituting the values in the given equation:
$\lambda =\dfrac{3\times {{10}^{8}}}{1368\times {{10}^{3}}}$
$\lambda =219.3$ m.
The wavelength of the given radiation is 219.3 m.
Since the frequency of the radiation \[\left( 1368\text{ }\times{{10}^{3}}Hz \right)\]is of order \[{{10}^{3}}\], it is a radio wave.
So, the correct answer is “Option D”.
Note:
Conversion of the units to the standard form should be carefully implemented as lack of attentiveness and accuracy here, may cause calculation errors thereby leading to wrong results.
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