Consider figure for photoemission.
How would you reconcile with momentum-conservation ? Note light (photons) have momentum in a different direction than the emitted electrons.
When incident photon is absorbed completely, its momentum becomes zero. This momentum is gained by atoms of metal and so these atoms get excited. As a result, electrons in them, transit from lower to higher energy levels and along with this, some free electrons get emitted in the form of photoelectrons. Here, total momentum is definitely conserved, irrespective of type of collision.
A radiation of energy $'E'$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $( C =$ Velocity of light $)$
What is relation between photon and intensity of radiation ?
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 ${eV}$)
Monochromatic light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$. How many photons per second on an average, are emitted by the source?
(Given $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ )