When light is incident on a photosensitive surface,electrons are emitted from the surface. The kinetic energy of these electrons does not depend on:

In a photoelectric experiment,if the wavelength of incident radiation is reduced from $6000 \ \mathring{A}$ to $4000 \ \mathring{A}$ while keeping the intensity of radiation constant,then:

The electric field associated with a light wave is given by $E = E_0 \sin [1.57 \times 10^7 (x - ct)]$,where $x$ is in meters and $t$ is in seconds. If this light is used to produce photoelectric emission from a metal surface with a work function of $1.9 \ eV$,what will be the stopping potential in $V$?

The variation of stopping potential $(V_0)$ as a function of the frequency $\nu \ (\times 10^{14} \ Hz)$ of the incident light for a metal is shown in the figure. The work function of the surface is $........... \ eV$.

The maximum velocity of photoelectrons emitted from a metal surface is $1.2 \times 10^6 \ m/s$. If the specific charge of the electron is $1.8 \times 10^{11} \ C/kg$,then the value of the stopping potential in volts is:

Two photons with energies three times and nine times the work function of a metal surface are incident on it. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases?