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.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) 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) W m^(-2))`. Take the area of the pupil to be about 0.4 cm2 , and the average frequency of white light to be about `6 xx 10^(14) Hz`

Explanation

Here `lambda`=500 m,P=10 kW=`10^(4)` W
`h=6.63xx10^(-34)Js,C=3xx10^(8)ms^(-1)`
(a) Energy of each photon ,
E=hv
`therefore E=(hc)/(lambda)=(6.63xx10^(-34)xx3xx10^(8))/(500)`
`therefore E=0.0398xx10^(-26)`
`therefore E~~3.98xx10^(-28)J`
`implies`Photon emitted per second =`("power")/("Energy of one photon")`
`N=(P)/(E)`
`=(10^(4))/(3.98xx10^(-28))`
`0.251xx10^(32)`
`~~0.251xx10^(31)`photon/sec
`~~3.0xx10^(31)` photon/sec
`implies` We can see that energy of photon of radio wave is very less and in a beam of radio waves no.of photon emitted every second are very large.
Thus if minimum quantum energy of photon is neglected and total energy of radio waves is considered continuous then negligible error is produced.
(b)`v=6xx10^(14)Hz,h=6.63xx10^(-34)Js`
Energy of photon,
`E=hv=6.63xx10^(-34)xx6xx10^(14)`
`therefore E=39.78xx10^(-20)`
`therefore E=~~4xx10^(-19)`
Photon flux corresponding to minimum intensity,
`=(P)/(E)`
`=(10^(-10))/(4xx10^(-19))`
`=2.5xx10^(8)m^(2)s^(-1)`
`implies`Area of pupil of eye `therefore` No. of photons entering into eye every sec=`"flux"xx"area"`
`=2.5xx10^(8)xx0.4xx10^(-4)`
`=1xx10^(4)` photon/sec
`implies`This number is very less compared to `2.51xx10^(31)` but not large enough which can produce sensation in our eye.