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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 barely detectable light.

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Anonym0us
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(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 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 (\sim110^{-10}Wm^{-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.

Qalaxia QA Bot
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I found an answer from worldbuilding.stackexchange.com

Stealth in Space: How realistic is it? - Worldbuilding Stack Exchange


Aug 25, 2015 ... Heat radiation can be detected by a distance of 13.4⋅√AT[K]2. ... Hope that you are literally lost in space and the sensors do not pick your ship up. ... range that means that even a human entering the laboratory gives off too much heat. ... That could work but stealth in space has always time as enemy.


For more information, see Stealth in Space: How realistic is it? - Worldbuilding Stack Exchange

Qalaxia Knowlege Bot
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I found an answer from www.scientificamerican.com

The Bicycle Problem That Nearly Broke Mathematics - Scientific ...


Jul 20, 2016 ... Jim Papadopoulos has spent a lifetime pondering the maths of bikes in motion. ... His time at the US Geological Survey was part of an internship, not a full-time job. ... He could never ride a bike without pondering the mathematical ... Whipple modelled the bicycle as four rigid objects — two wheels, a frame ...


For more information, see The Bicycle Problem That Nearly Broke Mathematics - Scientific ...

Qalaxia Master Bot
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I found an answer from ncert.nic.in

Chapter 11.pmd


This unit of energy is commonly used in atomic and nuclear physics. The work ... ( iii) Photo-electric emission: When light of suitable frequency illuminates a metal ... photons per second crossing a given area, with each photon having the same ... The dual (wave-particle) nature of light (electromagnetic radiation, in general) ...


For more information, see Chapter 11.pmd

Pravalika
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a)

Power P = 10 kW = 10 * 10^{3}

Wavelength \lambda = 500 m


Step 1: Finding the energy of photon

                Energy of proton E=h\upsilon\ =\ \frac{hc}{\lambda}

                   Where, h - Planck's constant( 6.63*10^{-34} Js}), speed of light c = 3*10^{8} m/s   and \upsilon - frequency.


                   E = \frac{6.63* 10^{-34} * 3*10^{8}}{500}

                   E = 3.98 * 10^{-28} J


Step 2: Determining the number of photons per second, on the average

            Relation between power(energy flux) and energy of photon

                               P = nE

                 Number of photons per second  n = \frac{P}{E}   

                             n = \frac{10 * 10^{3}}{3.98 * 10^{-28}}

                             n = 3 * 10^{31} s^{-1}


               Hence, number of photons per second  n = 3 * 10^{31} s^{-1}


b)

Frequency of white light \upsilon = 6* 10^{14} Hz

The intensity of light detected by the human eye. I = 10^{-10} Wm^{2}

The area of a pupil A = 0.4 cm^2 = 0.4* 10^{-4} m^2


Step 1: Calculating the energy released by a photon.

              Energy of proton E=h\upsilon

               E = 6.63* 10^{-34} * 6* 10^{14}

               E = 3.96 * 10^{-19} J


Step 2: Finding the total photon droppings per second, per pupil unit area

             Intensity of light I = n E

             Total photon droppings per second n = \frac{I}{E}

               n = \frac{10^{-10} }{3.96 * 10^{-19}}

               n = 2.5 * 10^8 m^2/s


           Photons per second entering into the pupil

             n_A = \text{ area } * n

             n_A = 0.4* 10^{-4} * 2.5 * 10^8

             n_A = 1.008 * 10^{4} s^{-1}


            Hence, total photon droppings per second, per pupil unit area n_A=1.008*10^4\ s^{-1}