Answer :

Given:


Power of transmitter, E’ = 10 KW = 10000 Js-1


Wavelength of radio waves emitted = 500 m


(a) We know that Energy in a wave is given by,


…(1)


Where,


E = Energy of photon


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


c = 3 × 108 ms-1


λ = wavelength



E = 3.9 × 10-28 J


Total power = number of photons emitted × Energy of photon


E’ = n × E


n =


n = 2.55 × 1031 s-1


n = 3 × 10-31 s-1


Here we see that the number of radio waves emitted per second is very high.


(b) Given:


Intensity of light perceived by human eyes, I = 10-10Wm-2


Area of pupil, A = 0.4 × 10-4 m2


Frequency of white light, v = 6 × 1014Hz


The energy of each photon is given by,


E = hv


Where,


E = energy of photon


h = Planck’s constant = 6.6 × 10-34Js


v = 6 × 1014Hz


E = 6.6 × 10-34Js × 6 × 1014 s-1


E = 3.96 × 10-19J


Energy of each photon is 3.96 × 10-19J


Let total number of photons being emitted per second, falling on unit area = n


We define intensity as the amount of energy falling on unit area in unit time so we can write,


I = n × 3.96 × 10-19J


10-10 Jm-2s-1 = n × 3.96 × 10-19J


n = 2.52 × 108 m-2s-1


Number of photons entering pupil = area of pupil × n


Number of photons entering pupils = 0.4 × 10-4m2 × 2.52 × 108 m-2s-1


np = 1.0082 × 104s-1


Almost 10000 photons enter our pupil per second.


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