Q. 253.6( 7 Votes )

# Estimating the fo

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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