Light of wavelength $4000 \ \mathring A$ is incident on a photosensitive surface. If a potential of $-2 \ V$ is required to stop the emitted electrons,the work function of the material is: $(h = 6.6 \times 10^{-34} \ J \cdot s, e = 1.6 \times 10^{-19} \ C, c = 3 \times 10^8 \ m/s)$ (in $eV$)

$A$ photoelectric cell is connected to a source of variable potential difference and the resulting photoelectric current $(\mu A)$ is plotted against the applied potential difference $(V)$. The graph with the broken line represents one situation for a given frequency and intensity of the incident radiation. If the frequency is increased and intensity is reduced,which of the following graphs of unbroken line represents the new situation?

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 .......... $eV$.

Light of two different frequencies with photon energies of $1 \ eV$ and $2.5 \ eV$ are incident one after another on a metal surface having a work function of $0.5 \ eV$. The ratio of the maximum kinetic energy of the emitted photoelectrons will be:

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-fourth that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).

$A$ photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_c$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de-Broglie wavelength is given by ($m$: mass of the electron,$h$: Planck's constant,and $c$: speed of light).