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 x 10¹⁴Hz.

Answer 1

(a) Power of energy transmitted per second,
E = 10 kW = 10⁴ W = 10⁴ Js^-1
λ = 500 m
Energy of one photon,

(b) Minimum intensity of white light = 10¹⁰ Wm²
Area of the pupil = 0.4cm² = 0.4 x 10⁴ m²
∴ Light energy falling on pupil per second
= 10^-10 x 0.4 x 10^-4 = 0.4 x 10^-14W
Frequency of white light, u = 6x 10¹⁴ Hz
∴ Energy of the photon of the white light,
hV = 6.62 x 10^-34 x 6 x 10¹⁴J
∴ Number of photons entering the pupil of our eye per second