Estimating the following two numbers should be interesting.

Solution:

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

$E_{1}=\frac{h c}{\lambda}$

Where,

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

c = Speed of light = 3 × 108 m/s

$\therefore E_{1}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{500}=3.96 \times 10^{-28} \mathrm{~J}$

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

 

∴nE1 = E

$n=\frac{E}{E_{1}}$

$=\frac{10^{4}}{3.96 \times 10^{-28}}=2.525 \times 10^{31}$

$\approx 3 \times 10^{31}$

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

 

$n \times 3.96 \times 10^{-19}=10^{-10}$

$n=\frac{10^{-10}}{3.96 \times 10^{-19}}$

= 2.52 × 108 m2 s−1

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

nA = 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.