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

Power of the medium wave transmittter, `P = 10 KW = 10 ^(4)W = 10^(4) J//s`
Hence energy emitted by the transmitter per second, `E= 10^(4)`
Wavelength of the radio wave `lambda =500m`
The energy of the wave is given as : `E_(1)=(hc)/(lambda)`
Where,
h=Planck's constant `= 6.6 xx10^(-34)Js`
C= Speed of light `=3xx10^8m//s`
`therefore E_(1)=(6.6xx10^(-34)xx3xx10^(8))/(500)=3.96xx10^(-28)J`
Let n be number of photons emitted by transmitter `therefore nE_(1)=E`
`n=(E)/(E_(1))`
`=(10^(4))/(3.96xx10^(-28))=2.525xx10^(34)approx3xx10^(34)`
The enregy `(E_(1))` of a radio photon is very less, but the number of photos (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 continous.
(b) Intensity of light perceived by the human eye,`I=10^(-10)Wm^(-2)`
Area of a pupil , `A=0.4cm^(2)=0.4xx10^(-4)m^(2)`
Frequency of white light, `v= 6xx10^(14)Hz`
The energy emitted by a photon is given as :
`E=hv`
Where,
h=Planck's constant =`6.6xx10^(-34)Js`
`therefore E=6.6xx10^(-34)xx6xx10^(14) =3.96xx10^(-19)J`
Let n be total number of photons falling per second, per unit area of the pupil.
The total energy per unit for n falling photons is given as :
`E=nxx3.96xx10^(-19)JS^(-1)M^(-1)`
The energy per unit area per second is the intensity of light.
`therefore E=1`
`nxx3.96xx10^(-19)=10^(-10)`
`n=(10^(-10))/(3.96xx10^(-19))`
`2.52xx10^(8)m^(2)s^(-1)`
The total number of photons entering the pupil per second is given as:
`n_(A)=nxxA`
`=2.52xx10^(8)xx0.4xx10^(-4)`
`=1.008xx10^(4)s^(-1)`
This number is not as large as the one foun d in problem (a), but it is large enough for the human eye to never see the individual photons.