The stopping potential required to reduce the photoelectric current to zero is . . . . . .

Light of frequency $4 \times 10^{14} \,Hz$ is incident on a metal surface of work function $2.14 \,eV$, resulting in photoemission of electrons. The maximum kinetic energy of the emitted electrons is $\left[h=6.63 \times 10^{-34} \,J-s\right]$ (in $\,eV$)

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be:

Which one of the four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident radiation $(I)$ is correct?

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)