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

The work function of a photosensitive metal surface is $hv_0$. If a photon of energy $2hv_0$ is incident on this surface,electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. What will be the maximum velocity of the photoelectrons when the energy of the incident photon is increased to $5hv_0$?

$A$ photon of energy $3.4 \ eV$ is incident on a metal surface with a work function of $2 \ eV$. The maximum kinetic energy of the emitted photoelectrons is .......... $eV$.

The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$,the maximum velocity of the ejected photoelectrons will be:

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

The stopping potential $V_0$ (in $volt$) as a function of frequency $(\nu)$ for a sodium emitter is shown in the figure. The work function of sodium,from the data plotted in the figure,will be: ................. $eV$
(Given: Planck's constant $(h) = 6.63 \times 10^{-34} \, Js$,electron charge $e = 1.6 \times 10^{-19} \, C$)