Photons of energy $6\, eV$ are incident on a metal surface whose work function is $4\, eV$. The minimum kinetic energy of the emitted photo-electrons will be :-

The threshold frequency for a certain metal is $3.3 \times 10^{14} \;Hz$. If light of frequency $8.2 \times 10^{14} \;Hz$ is incident on the metal,predict the cutoff voltage (in $eV$) for the photoelectric emission.

The threshold frequency of a photoelectric metal is $\nu_0$. If light of frequency $4\nu_0$ is incident on this metal,then the maximum kinetic energy of emitted electrons will be $.......h\nu_0$.

When a photosensitive metal surface is illuminated with radiation of wavelength $\lambda_1$,the stopping potential is $V_1$. If the same surface is illuminated with radiation of wavelength $3\lambda_1$,the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is:

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons emitted are $v_{1}$ and $v_{2}$ respectively,then: