Estimate the following the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about a photons. The second number tells you why our eye can never "count photon" even in barely detectable light.
(i) The number of photons emitted per second by a MW transmitter of 10kW power emitting radiowaves of length 500m.
(ii) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive `(~10^(-10)Wm^(-2))`. Take the area of the pupil to be about `0.4cm^(2)`, and the average frequency of white light to be about `6xx10^(4)Hz. (h=6.6xx10^(-34)J)`

Explanation

Given , P = 10 kW ` = 10 xx 10^(3) W , lambda = 500 m , h = 6.63 xx 10^(-34) J - s, c = 3 xx 10^(8)`
The no . Of photons emitted per second , `N = P/E = P/((hc)/lambda ) = (plambda )/(hc) = (10 xx 10^(3) xx 500 )/(6.63 xx 10^(-34) xx 3xx 10^(8) )`
` :. N = 2.51 xx 10^(31) ` photons/s