An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3\, {eV}$. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000\, \mathring {{A}}$. What is the maximum kinetic energy of the emitted photoelectron ? (In ${eV}$)

According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

In a photoelectric effect experiment a light of frequency $1.5$ times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:

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 $6 \times 10^{14}\; Hz$