If light of wavelength $6600 \ \mathring A$ is incident on a metal surface with a work function of $2 \ eV$,what will be the maximum kinetic energy of the emitted photoelectrons?

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

When a photon of energy $3.8 \,eV$ falls on a metallic surface of work function $2.8 \,eV$,then the kinetic energy of the emitted electrons is .......... $eV$.

$UV$ light of $4.13 eV$ is incident on a photosensitive metal surface having work function $3.13 eV$. The maximum kinetic energy of ejected photoelectrons will be : (in $eV$)

According to Einstein's photoelectric equation,the graph between the kinetic energy of emitted photoelectrons and the frequency of incident radiation is:

For photoelectric emission from a certain metal,the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate,the maximum possible velocity of the emitted electron will be ($m$ is the electron mass).