Question

The necessary condition for photoelectric emission is
A) $h\nu < h{\nu _o}$
B) $h\nu > h{\nu _o}$
C) ${E_r} < h{\nu _o}$
D) ${E_r} > h{\nu _o}$

Answer 1

Hint
These types of questions can be solved by using the photoelectric emission equation. In a photoelectric emission, the minimum condition required for emission of electrons from the outermost shell of an atom is the frequency of incident rays should be very high to provide energy to the electron so that it can leave from their outermost shell.
The equation for photoelectric emission is given by-
$h\nu = h{\nu _o} + {E_r}$

Complete Step-by-step solution
As ${\nu _o}$is the threshold frequency which is the minimum frequency required for emission of electrons from the outermost shell. The light with frequencies lower than threshold would not eject photoelectrons from the shell of an atom.
As for the emission kinetic energy (${E_r}$) of an electron should be greater than or equal to zero. Hence, the photoelectric emission will take place only when the energy of a photon is greater than the energy possessed by the electron.
Therefore, the necessary condition for photoelectric emission is $h\nu > h{\nu _o}$
Option (B) is correct.

Additional information
Other conditions required for photoelectric emission:
The electrons emitted from the metal surface is directly proportional to the light flux of the incident light. If the light flux increases then the amount of electrons emitted also will increase.
The emission will happen only if the energy of a photon is more than or equal to the energy possessed by the electron of the metal.
The maximum energy state of the emitted electrons depends on the frequency of the incident light but don't depend upon the amount of light incident.

Note
It should be noted that higher the frequency of light, the more will be the energy of light. Hence, the more energy will be imparted to photoelectrons so that it can be converted into kinetic energy.