When photons of energy $5 eV$ are incident on a metal surface with a work function of $4.36 eV$,photoelectrons are emitted. What is the maximum momentum of the emitted photoelectrons?

When a metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

$A$ point source of light is used in a photoelectric effect. If the source is moved farther from the emitting metal, the stopping potential:

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency) is incident on a metal plate,the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$,the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

The maximum velocity of emitted photoelectrons is $4.8 \ m/s$. If the $e/m$ ratio of an electron is $1.76 \times 10^{11} \ C/kg$,then the stopping potential is ......

When radiation of wavelength $\lambda$ is incident on a photocell,the maximum velocity of the photoelectrons is $v$. What will be the maximum velocity of the photoelectrons when radiation of wavelength $3\lambda/4$ is incident on the photocell?