When radiation of wavelength $400 \ nm$ is incident on a metal surface,the kinetic energy of the emitted photoelectrons is $1.68 \ eV$. The work function of the metal is .......... $eV$.

Two identical capacitors are arranged as shown. The work function of plate $1$ is $\phi \ eV$. The $emf$ of the battery is $\frac{\phi}{e}$. If the energy of the incident photon is $hv$,then what is the maximum kinetic energy of the $e^-$ reaching plate $2$?

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum velocities of photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is ($h=$ Planck's constant,$c=$ velocity of light,$\lambda_1 > \lambda_2$).

$A$ photoelectric cell is connected to a source of variable potential difference. The photoelectric current is plotted against the applied potential difference. The graph with the broken line represents the current for a given frequency and intensity of the incident radiation. If the frequency is increased and the intensity is reduced, which curve now represents the situation?

Two streams of photons,possessing energies equal to $5$ and $10$ times the work function of a metal,are incident on the metal surface successively. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases,respectively?

$K_1$ and $K_2$ are the maximum kinetic energies of photoelectrons emitted from a surface of a given material for light of wavelengths $\lambda_1$ and $\lambda_2$,respectively. If $\lambda_1 = 2\lambda_2$,then the work function of the material is given by: