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.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves 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 × 1014 Hz.
Given:
Power of transmitter, E’ = 10 KW = 10000 Js-1
Wavelength of radio waves emitted = 500 m
(a) We know that Energy in a wave is given by,
…(1)
Where,
E = Energy of photon
h = Planck’s constant = 6.6 × 10-34 Js
c = 3 × 108 ms-1
λ = wavelength
E = 3.9 × 10-28 J
Total power = number of photons emitted × Energy of photon
E’ = n × E
n = 
n = 2.55 × 1031 s-1
n = 3 × 10-31 s-1
Here we see that the number of radio waves emitted per second is very high.
(b) Given:
Intensity of light perceived by human eyes, I = 10-10Wm-2
Area of pupil, A = 0.4 × 10-4 m2
Frequency of white light, v = 6 × 1014Hz
The energy of each photon is given by,
E = hv
Where,
E = energy of photon
h = Planck’s constant = 6.6 × 10-34Js
v = 6 × 1014Hz
E = 6.6 × 10-34Js × 6 × 1014 s-1
E = 3.96 × 10-19J
Energy of each photon is 3.96 × 10-19J
Let total number of photons being emitted per second, falling on unit area = n
We define intensity as the amount of energy falling on unit area in unit time so we can write,
I = n × 3.96 × 10-19J
10-10 Jm-2s-1 = n × 3.96 × 10-19J
n = 2.52 × 108 m-2s-1
Number of photons entering pupil = area of pupil × n
Number of photons entering pupils = 0.4 × 10-4m2 × 2.52 × 108 m-2s-1
→ np = 1.0082 × 104s-1
Almost 10000 photons enter our pupil per second.