question_answer 3)

The Vividh Bharati Station of All India Radio, Delhi broadcasts on a frequency 1.368 kHz. (kilohertz). Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong to?

Answer:

Information shadow: Programmes of a particular radiostation are broadcasted on a particular frequency. In the problem, the frequency on which programmes of All India Radio, Delhi is broadcasted, is given. Frequency, \[v=1.368kHz\] \[=1368\,Hz\] (i) (1 kHz = 1000 Hz: conversion factor used) All electromagnetic radiations propagate with constant speed through a medium, i.e., velocity of light. Velocity, \[c=3\times {{10}^{8}}m/\sec \] (ii) Problem solving strategy: In the given problem, wavelength of the radiation is required. It can be calculated using the following relation: \[\lambda =\frac{c}{v}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(iii)\] Working it out: Substituting the values of c and v in equation (iii) we get \[\lambda =\frac{3\times {{10}^{8}}}{1368}=2.192\times {{10}^{5}}m.\] The radiation of wavelength \[2.192\times {{10}^{5}}m\] lies in the radiowave region of the electromagnetic spectrum.