$A$ 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 ........ $V$.

The work function of a photosensitive metal surface is $1.1 \ eV$. Two light beams of energies $1.5 \ eV$ and $2 \ eV$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

The photoelectric threshold wavelength for a metal surface is $6600 \, \mathring{A}$. The work function for this is:

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

The variation of photo-current with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph. Then: