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 × 10^{8} 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 × 10^{31} 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^{-10}Wm^{-2}

Area of pupil, A = 0.4 × 10^{-4} m^{2}

Frequency of white light, v = 6 × 10^{14}Hz

The energy of each photon is given by,

E = hv

Where,

E = energy of photon

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

v = 6 × 10^{14}Hz

E = 6.6 × 10^{-34}Js × 6 × 10^{14} s^{-1}

E = 3.96 × 10^{-19}J

Energy of each photon is 3.96 × 10^{-19}J

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^{-19}J

10^{-10} Jm^{-2}s^{-1} = n × 3.96 × 10^{-19}J

n = 2.52 × 10^{8} m^{-2}s^{-1}

Number of photons entering pupil = area of pupil × n

Number of photons entering pupils = 0.4 × 10^{-4}m^{2} × 2.52 × 10^{8} m^{-2}s^{-1}

→ n_{p} = 1.0082 × 10^{4}s^{-1}

Almost 10000 photons enter our pupil per second.

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