The maximum kinetic energy of the photo-electrons depends only on

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

$A$ light of wavelength $310 \,nm$ is used in a photoelectric experiment. The metal electrode of work function $2.5 \,eV$ is used in the experiment. The stopping potential for the photoelectrons will be (assume $hc = 1240 \,eV-nm$): (in $V$)