Advertisement Remove all ads

Estimating the Following Two Numbers Should Be Interesting. the First Number Will Tell You Why Radio Engineers Do Not Need to Worry Much About Photons! the Second Number Tells You Why Our Eye Can Never ‘Count Photons’, Even in the Barely Detectable Light. the Number of Photons Emitted per Second by a Medium Wave Transmitter of 10 Kw Power, Emitting Radiowaves of Wavelength 500 M - Physics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in the barely detectable light.

The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.

Advertisement Remove all ads

Solution

Power of the medium wave transmitter, P = 10 kW = 10W = 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 = "hc"/lambda`

Where,

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

c = Speed of light = 3 × 108 m/s

`:. E_1 = (6.6 xx 10^(-34) xx 3 xx 10^8)/500 = 3.96 xx 10^(-28) J`

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

nE1 = E

= `10^4/(3.96 xx 10^(-28)) = 2.525 xx 10^31`

`~~ 3 xx 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.

Concept: Particle Nature of Light: the Photon
  Is there an error in this question or solution?

APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 25.1 | Page 409
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×