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Question
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in the barely detectable light.
The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
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Solution
Power of the medium wave transmitter, P = 10 kW = 104 W = 104 J/s
Hence, energy emitted by the transmitter per second, E = 104
Wavelength of the radio wave, λ = 500 m
The energy of the wave is given as:
`"E"_1 = "hc"/lambda`
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
∴ `"E"_1 = (6.6 xx 10^(-34) xx 3 xx 10^8)/500`
= 3.96 × 10−28 J
Let n be the number of photons emitted by the transmitter.
∴ nE1 = E
= `10^4/(3.96 xx 10^(-28))`
= 2.525 × 1031
≈ 3 × 1031
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.
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