If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be:

The threshold wavelength for photoelectric effect from a metal is $200 \ nm$. The maximum kinetic energy of the emitted photoelectrons when radiation of wavelength $100 \ nm$ is incident on the metal is .......... $eV$.

In the experiment of the photoelectric effect,the stopping potential for a given metal is $V_{0}$ volts when radiation of wavelength $\lambda_{0}$ is used. If radiation of wavelength $2\lambda_{0}$ is used for the same metal,then the stopping potential (in volts) will be: [$e=$ charge on electron,$c=$ speed of light,$h=$ Planck's constant.]

In a photoelectric experiment,light of energy $2.48 eV$ irradiates a photosensitive material. The stopping potential was measured to be $0.5 V$. The work function of the photosensitive material is: (in $eV$)

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $6 eV$ fall on it is $4 eV$. The stopping potential in volts is