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 barely detectable light.
The number of photons emitted per second by a medium wave transmitter of 10 kW power emitting radio waves of length 500 m.

Solution

Given,
Wavelength of radio waves, $λ = 500 m$
Energy of photon, $E = hv$

Now, using the formula for energy of a photon,

$E = \frac{hc}{λ} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{500} J$

$E = 3.98 \times 10^{- 28} J$

Number of photons emitted per second,

$= \frac{Power of transmitter}{Energy of one photon} ∵ P = ne$

$= \frac{10^{4} {Js}^{- 1}}{3.98 \times 10^{- 28} J}$

$= 3 \times 10^{31} s^{- 1}$

We can see, that the energy of a radiophoton is exceedingly small, and the number of photons emitted per second from a radio beam is enormously large. Thus, there is negligible error involved in ignoring the existence of a minimum quantum of energy of photon and treating the total energy of a radiowave as continuous.

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