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 × 1014 Hz.

Question

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 × 1014 Hz.

Harshit Bhardwaj · Accepted Answer

(a) Power of the medium wave transmitter, P = 10 kW = 10⁴W = 10⁴ J/s

Hence, energy emitted by the transmitter per second, E = 10⁴

Wavelength of the radio wave, λ = 500 m

The energy of the wave is given as:

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

Let n be the number of photons emitted by the transmitter.

∴nE₁ = E

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) Intensity of light perceived by the human eye, I = 10⁻¹⁰ W m⁻²

Area of a pupil, A = 0.4 cm²= 0.4 × 10⁻⁴ m²

Frequency of white light, ν= 6 × 10¹⁴ Hz

The energy emitted by a photon is given as:

E = hν

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

∴E = 6.6 × 10⁻³⁴ × 6 × 10¹⁴

= 3.96 × 10⁻¹⁹ J

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

The energy per unit area per second is the intensity of light.

∴E = I

n × 3.96 × 10⁻¹⁹ = 10⁻¹⁰

= 2.52 × 10⁸ m² s⁻¹

The total number of photons entering the pupil per second is given as:

n_A= n × A

= 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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