Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
Let n be the number ofphotons emitted by the transmitter.
∴nE1 = E
The energy (E1) of a radiophoton is very less, but the number of photons (n) emitted per second ina radio wave is very large.
The existence of a minimumquantum of energy can be ignored and the total energy of a radio wave can betreated as being continuous.
(b) Intensity of lightperceived 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 aphoton 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 numberof photons falling per second, per unit area of the pupil.
The total energy per unit for n fallingphotons is given as:
E = n × 3.96 × 10−19 Js−1 m−2
The energy per unit areaper second is the intensity of light.
∴E = I
n ×3.96 × 10−19 = 10−10
= 2.52 × 108 m2 s−1
The total number ofphotons entering the pupil per second is given as:
nA = n × A
= 2.52 × 108 × 0.4 × 10−4
= 1.008 × 104 s−1
Thisnumber is not as large as the one found in problem (a), but it is large enough for the human eye to never see theindividual photons.