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Question 11.25: 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 Mediumwave 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 x 1014 Hz.

(a)
Hint: \(P=nE\)
Step 1: Find the energy emitted by the transmitter per second.
Power of the medium wave transmitter, P = 10 kW = 104  W = 104 J/s
Hence, the energy emitted by the transmitter per second, E = 104
Step 2: Find the energy of wave.
The energy of the wave is given as:
E1 = hcλ
∴E1=6.6×10−34×3×108500=3.96×10−28 J
Step 3: Find the number of photons emitted per second.
n=EE1=1043.96×10−28=2.525×1031≈3×1031 
The energy (E1) of a radio photon is very less, but the number of photons (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 continuous.
(b)
Hint: The energy per unit area per second is the intensity of light.
Step 1: Find the energy emitted by a photon.
Frequency of white light, \(\nu\): 6 x 1014 Hz
The energy emitted by a photon is given as:
E = h\(\nu\) Where,
h = Planck's constant 6.6 x 10-34 Js
∴E=6.6×10−34×6×1014=3.96×10−19 J

Step 2: Find the total energy per unit for n falling photons.
Let n be the 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=n×3.96×10−19 J s−1 m−2

Step 3: Find the total number of photons entering the pupil per second.
Intensity of light perceived by the human eye, I = 10-10 W m-2
Area of a pupil, A = 0.4 cm2= 0.4 x 10-4m2
As, the energy per unit area per second is the intensity of light.
∴E=In×3.96×10−19=10−10n=10−103.96×10−19=2.52×108 m2 s−1
The total number of photons entering the pupil per second is given as:
nA=2.52×108×0.4×10−4=1.008×104 s−1
This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

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