Light emitted during the de-excitation of an electron from $n=3$ to $n=2$ is incident on a metal, and photoelectrons are just emitted from that metal. In which of the following de-excitations is the photoelectric effect $NOT$ possible?

Which one of the following four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident light $(I)$ is correct?

When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by the incidence of:

Photons with energy $5\, eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photoelectrons is $2\, eV.$ When photons of energy $6\, eV$ are incident on $C,$ no photoelectrons will reach the anode $A,$ if the stopping potential of $A$ relative to $C$ is ............ $V$.

The electron in the hydrogen atom jumps from excited state $(n = 3)$ to its ground state $(n = 1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \, eV$,the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n = \frac{-13.6}{n^2} \, eV$) (in $, V$)

The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?