The work functions of metals are given as $Na = 1.92 \, eV$,$K = 2.15 \, eV$,$Mo = 4.17 \, eV$,and $Ni = 5.0 \, eV$. Which of these metals will not show photoelectric emission for radiation of wavelength $3300 \, \mathring{A}$ from a $He-Cd$ laser?

Cut-off potentials for a metal in the photoelectric effect for light of wavelengths $\lambda_1$,$\lambda_2$,and $\lambda_3$ are found to be $V_1$,$V_2$,and $V_3$ volts. If $V_1$,$V_2$,and $V_3$ are in Arithmetic Progression,then $\lambda_1$,$\lambda_2$,and $\lambda_3$ will be in:

$A$ photon of energy $6 \ eV$ is incident on a metal surface. The work function of the metal is $2 \ eV$. The minimum reverse potential required to stop the emission of electrons is ........ $V$.

When a photosensitive surface is irradiated by light of wavelengths $\lambda_1$ and $\lambda_2$,the kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is:

The photoelectric cut-off voltage in a certain experiment is $1.5 \text{ V}$. The maximum kinetic energy of photoelectrons emitted will be . . . . . . .

When light of wavelength $\lambda$ incidents on a photosensitive material,photoelectrons are emitted. If the wavelength of the incident light is reduced by $50 \%$,the maximum kinetic energy of the emitted photoelectrons becomes $3$ times the initial maximum kinetic energy. The work function of the material is ($h$ - Planck's constant,$c$ - Speed of light in vacuum).