Question

(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) Write the basic features of photon picture of electromagnetic radiation on which Einstein's photoelectric equation is based.

Answer 1

$a)$ According to Photoelectric effect experiments, there exist a threshold frequency below which, no electrons are emitted.

According to the wave picture of light the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. the greater the intensity of radiation. the greater are the amplitude of electric and magnetic fields consequently. the greater the intensity. the greater should be the energy absorbed by each electron in this picture the maximum kinetic energy of the photo electrons on the surface is then expected to increase with increase in intensity. Also no matter what the frequency of radiation is a sufficiently intense beam of radiation(over sufficient time) should be able to impart enough energy to the electrons, so that they exceeds the minimum energy needed to escape from the metal surface.A threshold frequency therefore should not exist.

Further we should note that in the wave picture, the absorption of energy by electron takes place continuously over the entire wave front of the radiation. since a large number of electrons absorb energy. the energy absorbed per electron per unit time turns out to be small. explicit calculation estimate that it can take hours or more for single electron to pick up sufficient energy to overcome the work function and come out of the metal. this conclusion is again in streaking contrast to observation that the photoelectric emission is instantaneous. in short, the wave picture is unable to explain the most basic features of photoelectric emission.

$b)$The basic features of photon picture of light are:-

$1)$ In interaction of radiation with matter radiation behaves as if it is made up of particles called photons.

$2)$Each Photon has energy $E$$\left(h\upsilon \right)$and Momentum $p$($h\upsilon /c)$and speed $c$ the speed of light.

$3)$All photons of light of a particular frequency $\upsilon$, or wavelength$\lambda$, have the same energy$E=(h\upsilon =hc/\lambda )$ and momentum$p=(h\upsilon /c=h/\lambda )$whatever the intensity of radiation maybe by increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each Photon having the same energy thus, Photon energy is independent of intensity of radiation.

$4)$Photons are electrically natural and are not defected by electric and magnetic field.

$5)$In a photon particle collision (such as Photon-electron collision). the total energy and total momentum are conserved. however, the number of photons may not be conserved in a collision. the photon may be observed on a new Photon may be created.