A $10\, kW$ transmitter emits radio waves of wavelength $500\, m$. The number of photons emitted per second by the transmitter is of the order of

The energy of a photon of light with wavelength $5000\,\mathop A\limits^o $ is approximately $2.5\, eV$. This way the energy of an $X-$ ray photon with wavelength $1\,\mathop A\limits^o $ would be

A beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10 \%$ of the incident photons effectively produce photoelectron, then find current due to these electrons $...........\mu A$

[given, $hc =1240\,eV - nm , e =1.6 \times 10^{-{ }^{19}\,C }$ )

A $2\,mW$ laser operates at a wavelength of $500\,nm.$ The number of photons that will be emitted per second is [Given Planck’s constant $h = 6.6 \times 10^{-34}\,Js,$ speed of light $c = 3.0\times 10^8\,m/s$ ]

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.

$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power, emitting radiowaves 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 percetve $(-10^{-10}\; W m ^{-2}$). Take the area of the pupil to be about $0.4 \;cm ^{2}$, and the average frequency of white light to be about $6 \times 10^{14}\; Hz$