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Power of the medium wave transmitter, P= 10 kW = 104 W = 104J/s
Hence, energy emitted by the transmitter per second, E= 104
Wavelength of the radio wave, = 500 m
The energy of the wave is given as:
h* c/Lambda
Where,
h= Planck’s constant = 6.6* 10^-34 J sec
c= Speed of light = 3 × 108m/s
E1 = 6.6*10 ^-34 * 3* 10^8/500 = 3.96* 10^-28
Let n be the number of photons emitted by the transmitter.
∴nE1= E
n = E/E1
= 10^4/3.96 * 10^-28 = 2.525* 10^31
The energy (E1) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large.
The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.
Regards
Arun (askIITians forum expert)
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