The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \; \mathring{A}$ is $0.42 \; V$. If the threshold frequency is $x \times 10^{13} \; s^{-1}$,where $x$ is ... (nearest integer).
(Given: speed of light $= 3 \times 10^{8} \; m/s$,Planck's constant $= 6.63 \times 10^{-34} \; J \cdot s$)

Assuming photoemission to take place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is

For the photoelectric effect,which of the following statements are true?
$I$ The kinetic energies of the photoelectrons do not depend on the frequency of light.
$II$ The photoelectric effect will always occur for highly intense light.
$III$ The maximum kinetic energy of a photoelectron does not depend upon the intensity of the light.
$IV$ The escaping electron's kinetic energy is larger for a larger frequency.

The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

The work function of Cesium is $2.27 eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 nm$ wavelength is