Question 25: NCERT Solutions for 12th Class Physics: Chapter 11-Dual Nature of Radiation and Matter

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’eount photons’ even barely detectable light.

**(a)** The number of photons emitted per , second by a MW transmitter of 10 kW power emitting radio waves of length 500 m.

**(b)** The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^{-10} W m^{-2}). Take the area of the pupil ‘to be about 0.4 cm^{2} and the average frequency of white light to be about 6 × 10^{14} Hz.

Solution:Here, power of tansmitter,(a)

exceedingly small and the number of photons emitted per second in a radio beam is enormously large. Therefore, negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave as continuous.

(b)Here, area of the pupil, A = 0.4 cm^{2}= 0.4 × 10^{-4}m^{2}, v = 6 × 10^{14}Hz Intensity = 10^{-10}W m^{-2}Energy of a photon is given by, E = hv = 6.63 × 10^{-34}× 6 × 10^{14}J = 4 × 10^{-19}J.If n = number of photons falling per sec per unit area, the energy per unit area per sec due to these photons = total energy of n photons = n × 4 × 10

^{-19}J nr^{2}Since, intensity = energy per unit area per second∴ Number of photons entering the pupil per second = n x area of the pupil = 2.5 × 10

^{8}× 0.4 × 10^{-4}s^{-1}= 10^{4}s^{-1}.Though this number is not large as in part (a) above, it is large enough for us to ‘sense’ or ‘count’ the individual photons by our eye.

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NCERT Solutions for 12th Class Physics: Chapter 11-Dual Nature of Radiation and Matter